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12-18-2000, 08:48 PM
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#51 | ||
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Coordinator
Join Date: Oct 2000
Location: Big Ten Country
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The only thing I think that's wrong with that is where you said that the order of the cards is not significant. It is -- that's how I got to my incorrect answer earlier (if you can glean anything through the rampant editing done by QS and I -- maybe it was just a "had-to-be-there-to-understand" sort of thing -- literally, because most of what we said is gone now. But the order of the cards does matter, so you need to multiply your probablity for D by 5, to allow for each position that the ace could be in.
------------------ "If we can put a man on the moon, we can grow grass indoors." |
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12-18-2000, 09:03 PM
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#52 | |
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lolzcat
Join Date: Oct 2000
Location: Annapolis, Md
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Grunion, your analysis does make sense. However, the flaw is not in your math per se, but in the translation of the lettered conditions (as set forth) into mathematics.
Your discussion of A,B,C, and E are all right on, and in complete agreement with several other accurate discussions previously on this thread. All good. As for D, here is your narrative: Quote:
While this has some surface logic, but it essentially boils down to a restatement of the conditions of item B, rather than an accurate representation of D. This is even suggested when you say "the next four cards" - the seeming differentiation between *the first card* in B and *the flipped card* in your example is illusory. It's the same construct exactly-- this is just case B. And thus, the calculation works out the same. The flaw lies in the assumption that the narrative above is equivalent, in practice, to the conditions of D in this puzzle. The difference is subtle, but meaningful. In your narrative for D, we now know one thing about the hand-- that it is a hand sonewhere within the subset of hands that contain at least one ace. That is true. However, that's not all we "know" about the hand at this point. We also know that the inherent condition about having at least one ace has already been met. I realize this sounds picky, but it's the difference between right and wrong here. Consider again all the hands that have at least one ace. Let's break them down into two groups-- based on the outcome of your suggested "misdealt" card. D1 = the turned card is an ace D2 = the turned card is not an ace D1 is the case you describe above. D2 is the rest of the hands from the entire subset D. Like you said, in D1, we now have no knowledge about the other cards in the hand-- particularly about their rank. Recall, in predicting a flush, it's the duplicate ranks that determine the chances. So, D1 (again, as you have said) has the same probability as A,B,C, or E. However, D2 is different. For some subset of the D hands, we would have one card turned over accidentally, and we would know that it was a non-ace... say a 4. Aha! Now we have some information we can use. We know that this hand has at least one ace, so we know that at least one of hte other cards [b]will not match the rank of the exposed card[b]. This knowledge-- reducing the overall chances of a duplicate rank (I trust that is intuitively obvious)-- therefore increases the chance that the hand is a flush. And as a result, we now know that: P (D=flush) is a weighted average of P(D1=flush) and P(D2=flush). And since P(D1=flush) = P(A=flush), and P(D2=flush) > P(A=flush), we now know that P(D=flush) > P(a=flush). q.e.d. This all comes back to a fairly tenuous argument (that I fear I have not articulated well) that your narrative description of D above is not in fact a description of D, but instead a description of what I call D1. By excluding the other cases within D, you eliminate the remainder of its probability-- which happens to be the portion that favors a flush. Hope that helps... |
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12-18-2000, 09:14 PM
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#53 |
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Coordinator
Join Date: Oct 2000
Location: Big Ten Country
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Um....Go Bucs!
------------------ "If we can put a man on the moon, we can grow grass indoors." |
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12-18-2000, 10:32 PM
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#54 |
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Mascot
Join Date: Oct 2000
Location: Richmond, VA
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QuikSand,
I think we are getting somewhere, however I still disagree. You have broken D in to D1 & D2 to illustrate your logic. Your explanation was quite clear, I believe I fully understand it. Our given is that there is at least one ace in our five card hand. D1: We expose the first card and it is an Ace. No further discussion needed, as we are in agreement. D2: We expose the first card and it is not an ace. Since an Ace, and therefore a guaranteed non-match to our first card, is certain there is an improved probability of a flush. This is because situations where the first two cards match would be eliminated, enhancing flush possibilities. However, lets continue exposing cards. I believe the probability for exposing a pair of Aces will be increased, so that while a second card ace will be guaranteed not to match rank with the first card, there will be a much higher chance for there to be two or more aces in the remainder of the hand, levelling the percentages out. In case D2, where the first card pulled is not an ace and at least one ace is in the deck the odds of not pulling a second ace is 47/50*46/49*45/48 = 82.7% So a pair of aces are 17.3% likely If you remove the mandatory ace condition the odds of pulling two aces are going to decrease 48/51*47/50*46/49*45/48 = 77.9% chance of pulling no aces. Which means the probability of pulling one ace is 22.1%. Out of that 22.1% the odds of not pulling a second ace is identical to the situation above, meaning that the odds of pulling a pair of aces are (0.221*0.173)= 3.82% (I know the above is an awkward way to present the calculations, but my statistics book is in my office. I am fairly sure that my logic is sound.) Anyway, I think the dramatically increased chance of pulling a second ace cancels out the gains in knowing you have an "ace in the hole" as you expose non-ace cards. Also, Assume having at least one ace in your five card hand increased the likelyhood of pulling a flush as compared to the average probability (five cards, no info). As each value is equally represented by four cards in a deck it would stand to reason that having at least one X in your five card hand increased the likelyhood of a flush, where X can be ANY valued playing card. EVERY five card hand dealt will have at least one X, where X can be any valued playing card. Therefore, every possible five card hand has a greater chance to pull a flush than the average probability, which is impossible.
__________________
Those who don't learn history are doomed to repeat it. |
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12-19-2000, 08:08 AM
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#55 | |||
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lolzcat
Join Date: Oct 2000
Location: Annapolis, Md
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I agree, we are getting somewhere.
In your argument above: Quote:
I'm not checking your math, but I will contest your logic. Your discussion of the increased likelihood of coming up with two (or more) aces is certainly on point. And, by itself, that suggests that there might be a reduced chanced of getting a flush. However, it's not complete to look at that issue in isolation. What about all the other ranks of cards? If your "shown" card is a deuce, and we add the promise of at least one ace in the hand-- doesn't that clearly mean that the chances of our seeing duplicate threes or queens in this hand decrease? Of course it does-- I'm not going to bother with the mathematics, it's intuitively obvious. So, the complete analysis of the likeihood of getting duplicate cards within each rank would tell us something like this: (Shown card = deuce, hand has at least one ace) -Chances of getting duplicate aces = higher than without second condition -Chances of getting duplicate deuces = lower than without second condition -Chances of getting duplicates of any other rank = lower than without second condition I'm not going to work through all the math here (ugh) but suffice it to say that with multiple effects in countervailing directions, it is no cinch that that we end upwith anything like the following conclusion: Quote:
- - - In your second, more intuitive, argument, you observe: Quote:
The flaw here is that this is an ex post facto condition-- categorizing the hand after we see what's inside. Once we see what's in the hand, of course we will be able (after the fact) to go back and recognize that it fits into various subsets of all the possible hands. (Hands with at least one ace, hands with at least one queen, etc.) Some of these subsets will have a higher likelihood of being a flush than others, and many will have a different flush probability than the entire set of possible hands-- but that doesn't invalidate a mathematical analysis about the flush probability of any of the given subsets. |
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12-19-2000, 08:14 AM
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#56 |
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lolzcat
Join Date: Oct 2000
Location: Annapolis, Md
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Another intuitive argument about the "hand with at least one ace" is as follows:
By describing the condition D that the hand has at least one ace, we essentially divide the universe of all possible hands (A) into two subsets. The other subset (A-D) would be the hands that do not include at least one ace. I would argue that you could characterize the set of (A-D) as being "the hands that do not make use of all the ranks in the deck." (It obviously doesn't matter whether the condition is that the hand has one ace, one queen, or whatever) As I have stated before, this subset (A-D) is exactly represented by the entire universe of hands that could be created by a 48-card deck that simply was stripped of the rank mentioned in the conditions for D. Intuitively, as you "shorten" each rank within the deck, you make it increasingly improbable that the hand will remain in suit and stay a flush. This intuition is borne out by considering the extreme case-- a deck with only 5-card suits. In that case, it's painfully clear that drawing a flush is an extreme improbability. By this logic, the subset of (A-D) has a lesser probability of being a flush than does the full set A. Since the subset D is what is "added" to the subset (A-D) to get to the full set A, it's clear that the flush probability of D must be greater than that of A. |
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12-19-2000, 09:08 AM
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#57 |
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lolzcat
Join Date: Oct 2000
Location: Annapolis, Md
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Flush Calculation (the "direct" method)
There was at least one request for this "direct" calculation, which is a little bit more tedious, but I suppose it may help to reinforce the correct answer. Condition D - Hand has at least one ace I'll use my standard method-calculating D and N where: D = total number of hands meeting the stated condition N = number of such hands that are flushes - - - Calculating D directly is a pain in the ass. First, consider the one-ace hand: 4x48x47x46x45 = 18,679,680 x 5 different positions = 93,398,400 and since there is only one card we have no duplication Second, consider the two-ace hand: 4x3x48x47x46 = 1,245,312 x 20 different positions = 24,906,240 and divide by 2! (2) to eliminate duplicates = 12,453,120 Next, consider the three-ace hand: 4x3x2x48x47 = 54,144 x 60 different positions = 3,248,640 and divide by 3! (6) to eliminate duplicates = 541,440 Finally, the four ace hand: 4x3x2x1x48 = 1,152 x 120 different positions = 138,240 and divide by 4! (24) to eliminate duplicates = 5,760 Adding the four together gives us 106,398,720 different hands that have at least one ace. As predicted, this is identical to the number reached on page two of this thread, by the more simple calculation of subtracting out the 48-card hands from the 52-card hands D = 106,398,720 - - - Calculating N directly is actually pretty easy: Start with the 4 aces, and multiply by the number of cards remaining in suit: 4 x (12 x 11 x 10 x 9) = 47,520 Then, multiply this by 5 to show each of the five positions for the ace: 47,520 x 5 = 237,600 This is the total number of flushes that contain at least one ace. (And, of course, they all contain exactly one ace-that's what makes this a simpler calculation than D above) N = 237,600 - - - To calculate the probability that a given five card hand, kown to have at least one ace, is a flush, we take N / D. N = 237,600 D = 106,398,720 N/D = 0.22331% This probability is, indeed, higher than the 0.198% or so that everyone has calculated as the correct probability for A (as well as B,C, and E). Therefore, the answer to the problem remains: D ABCE q.e.d. |
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12-19-2000, 09:32 AM
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#58 |
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Coordinator
Join Date: Oct 2000
Location: Big Ten Country
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What does q.e.d. mean again?
------------------ "If we can put a man on the moon, we can grow grass indoors." |
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12-19-2000, 09:51 AM
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#59 |
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lolzcat
Join Date: Oct 2000
Location: Annapolis, Md
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q.e.d. = (Latin) quod erat demonstrandum, meaning "that which was to be shown" or somesuch. Used mathematically to say "this is what we were after."
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12-19-2000, 03:18 PM
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#60 |
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Mascot
Join Date: Oct 2000
Location: Richmond, VA
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QuikSand,
I think I got it. # of Total flushes = 52*12*11*10*9 =617,760 # of flushes w/ an Ace = 4*12*11*10*9 =47,520 # of total hands = 52*51*50*49*48 =311,875,200 # of total hands with an ace = 4*51*50*49*48 =23,990,400 617,760/311,875,200=.00198 47,520/23,990,400=.00198 ------------------ Those who don't learn history are doomed to repeat it.
__________________
Those who don't learn history are doomed to repeat it. |
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12-19-2000, 03:24 PM
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#61 |
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lolzcat
Join Date: Oct 2000
Location: Annapolis, Md
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Once again, Grunion, your math looks fine, but it is the math for the wrong problem.
Above, you are incorrectly labeling your calculations-- your numbers are not showing the total hands with an ace, nor the total flushes with an ace. Instead, they correctly count the hands (of each type) that start with an ace. (This was either B or C in the original puzzle, I forgot which) And, as we have (all) agreed, that condition doesn't change the flush probability... as your correct (albeit mislabeled) mathematics reinforce. [This message has been edited by QuikSand (edited 12-19-2000).] |
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12-19-2000, 04:20 PM
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#62 |
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Mascot
Join Date: Oct 2000
Location: Richmond, VA
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This case describes a hand with at least one ace, with the position of the ace itself irrelevant. The placement of the known ace is irrelevant, if all other cards are unknown.
# of Total flushes = 52*12*11*10*9 =617,760 # of flushes w/ an Ace = 1/5(4*12*11*10*9)*1/5(12*4*11*10*9)*1/5(12*11*4*10*9)*1/5(12*11*10*4*9)*1/5(12*11*10*9*4) =47,520 # of total hands = 52*51*50*49*48 =311,875,200 # of total hands with an ace = 1/5(4*51*50*49*48)*1/5(51*4*50*49*48)*1/5(51*50*4*49*48)*1/5(51*50*49*4*48)*1/5(51*50*49*48*4) =23,990,400 617,760/311,875,200=.00198 47,520/23,990,400=.00198 ------------------ Those who don't learn history are doomed to repeat it.
__________________
Those who don't learn history are doomed to repeat it. |
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12-19-2000, 06:43 PM
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#63 | |
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lolzcat
Join Date: Oct 2000
Location: Annapolis, Md
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Quote:
Some very fancy numberworking here, to be sure. (I'm assuming that some of the asterisks were intended to be plus signs... I'm a little bit perplexed by the presence of all the 1/5s in the forumla, but they are in the end harmless, as they cancel out in the final ratio. Converting some * to + is the best remedy I could think of, but I'm still not sure I'm following the logic there) That said, a problem (I can't say right now if it's the only problem) with this calculation is that it counts certain hands more than once. Specifically, in the latter calculation (finding the number of hands with at least one ace), your numbers strongly suggest that you are considering five independent cases: one representing the appearance of the ace in each position of the hand. This is, of course, correct. So, you consider each of the following: a(4*51*50*49*48) - ace is in first position b(51*4*50*49*48) - ace in in second position c(51*50*4*49*48) - ace is in third position d(51*50*49*4*48) - ace is in fourth position e(51*50*49*48*4) - ace is in fifth position However, this fails to account properly for the hands in which more than one ace appears. Consider the following hand: AH-4C-5S-QH-AD Since the card in slot #1 is an ace, it's certainly being counted in group a above, but since there's an ace in slot 5 (the AD is definitely one of the 48 cards for that slot), but it's also being counted in group e above (the ace of hearts is definitely included in the 51 for that slot). By overstating the total number of hands with an ace, this then goes to understate the final ratio (since the denominator is too big). Is it coincidence that this overcounting of hands leads to exactly the same probability as in case A? No, because the construct shown above is simply an algebraic restatement of the same calculations in your earlier post, which are themselves just a calculation for case B (rather than case D, which we're trying to solve) in the original problem (case B = hand's first card in an ace). It's no coincidence, it's a direct product of that original mistranslation. |
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12-19-2000, 06:47 PM
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#64 |
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lolzcat
Join Date: Oct 2000
Location: Annapolis, Md
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Incidentally, let me apologize for the pedantic tone in my writings on this thread, in particular those directed toward Grunion. I happen to believe (quite forcefully) that I am correct in this problem. If it turns out that I am demonstrably wrong, I'll quickly and publicly acknowledge Grunion and anyone else appropriate for their assistance in showing me the error of my ways.
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12-19-2000, 07:09 PM
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#65 |
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lolzcat
Join Date: Oct 2000
Location: Annapolis, Md
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A couple more ideas for Grunion or anyone who is not yet sold on the "D is higher" answer. If you've got the patience to stick it out this far... more power to you.
Easy one first --Grunion, since you've (in part) agreed to work this through with my methopds (counting all the hands, and taking the ratio)-- how would you set that construct up for selection B? It will, I believe, come out as exactly the same setup as what you are suggesting is correct for selection D. Is this a coincidence, are they exactly the same hands? (Neither one makes sense to me) Togher one next --Another idea is this: I could have added another selection F to say "Hands with no card higher than a king." We should be able to agree that if the subset of A that has at least one ace has the same proportion of flushes as does A, then the remainder of set A should have that same proportion. (I don't expact an argument on that). Therefore, if you answer that D has the same flush probability as A, then you would have to say that F has the same probability. Wouldn't it stand to reason, then, that the flush probability in selection F it shouldn't change when we "lower the bar" by additional ranks? Under your calculations, we ought to be able to add new subsets of hands that would also have the same probability, by incrementally "throwing out" one more rank. First we throw out aces, then kings, etc. Eventually, we should be able to get all the way down to "Hands with no cards over a six." And that one should have the exact same probability as the original set A. However, that one is easy to calculate. It's just 20x4x3x2x1=480 flush hands out of 20x19x18x17x16=18,604,480 possible hands-- which calculates to 0.00258% - a MUCH smaller chance than in the full set A. This makes intuitive sense-- once the suits get smaller, it gets tougher to "stay within the suit" and keep building a flush. This is, in a sense, a reductio ad absurdum argument that also reinforced the correct answer-- that set D has a greater likelihood of generating a flush. |
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12-19-2000, 10:10 PM
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#66 |
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Mascot
Join Date: Oct 2000
Location: Richmond, VA
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QuikSand,
First off, I am considering our conversation nothing more than a friendly debate. Even though I haven't posted on them before, I enjoy your frequent OT brainteasers. Please don't take our dialog as nothing more than an exchange of ideas. Yes, you are correct, I got carried away with my *'s in my formula. Your assumpion that there should be plusses before each 1/5 is correct. I have no idea to reference part of your post as a quote, but wat follows is from an earlier post of yours. So, you consider each of the following: a(4*51*50*49*48) - ace is in first position b(51*4*50*49*48) - ace in in second position c(51*50*4*49*48) - ace is in third position d(51*50*49*4*48) - ace is in fourth position e(51*50*49*48*4) - ace is in fifth position However, this fails to account properly for the hands in which more than one ace appears. Consider the following hand: AH-4C-5S-QH-AD Since the card in slot #1 is an ace, it's certainly being counted in group a above, but since there's an ace in slot 5 (the AD is definitely one of the 48 cards for that slot), but it's also being counted in group e above (the ace of hearts is definitely included in the 51 for that slot). This ends your post. In response to the above. I believe that my equation above does properly account for hands with more than one ace. And that is because of your constraint. We know that at least one card is an ace. In no case do we ever know that more than one card is an ace. Also, we do not know where that ace is. However, we can be assured that no one position has a greater likelyhood of having an ace than any other (implied by the original constraint and stated by yourself in an earlier post). I completely agree that there is a greater than 1 in five chance of any given card being an ace, due to duplicate aces. However, the extra aces are accounted for in my equations, more specifically, the one I posted prior to my last one (4*51*50*49*48). In essence, our disagreement boils down to this. You believe that the uncertainty of not knowing exactly what ace you have or where it is somehow effects the percentage of pulling a flush. I believe that the knowledge that there is one ace provides us with no additional information which would alter the baseline probability. I also believe that whether we have additional information on the location of an ace in the hand or suit of the ace does not effect the probabilities of a flush. In response to your last post: To answer your easy question. Yes, they are identical. That is exactly my point. It is not a coincidence, they are exactly the same hands. In every case which you initially presented only one card was known (other than the baseline case A). In each of those instances, we know that the card is an ace. Here is what intuitively makes no sence to me; We agree that case B (first card ace) there is no change in the probability of pulling a flush. We also agree that case C (first card ace of spades) there is no change in the probability of pulling a flush. Therefore we can conclude that knowing the suit of the card in question does not change the odds of pulling four more cards to match the suit of the first. Now, in case E (hands w/ the ace of spades) we also agree that the odds of pulling a flush are unchanged. Therefore, we can conclude that the location of the ace of spades is irrelevant, and it is not significant whether we know where the ace of spades is. We have now established that neither the knowledge of suit nor the knowledge of location has any bearing on the odds of pulling a flush. In case d, we now neither the suit nor the location. Tough question: I think I can provide a suitable answer. The difference between the two conditions are that in the case of at least one ace, we know that there is one ace, and the four random cards would be pulled from a pool of 51 cards, as multiple aces would be allowed, which takes us back to # of flushes w/ an Ace = 4*12*11*10*9 =47,520 # of total hands with an ace = 4*51*50*49*48 =23,990,400 In the case of no card higher than a king, we have much different information. We now have no idea what the value of any card is. We also have partial knowledge of what all five cards are not (in this case Aces). Therefore, the # of flushes with no aces allowed is: 12*11*10*9*8=95040 And the total number of hands allowed are: 48*47*46*45*44=205476480 flush probability = 0.046% The difference between the two is that in the first case, your pool of cards remained the same. Yes, you put a constraint on the available permutations by fixing the rank of one of the cards, but this constraint by itself is meaningless. (Akin to pulling all of the clubs out of the deck would not influence the 1:13 chance to pull an ace on your first try). In the second case, you are changing the size of the deck, and this does impact the probability of pulling a flush. (From 0% at only 2 thru 5 remain to 1 in 256 with an infinite amount of cards. One more arguement (and I say arguement in the friendliest most non-confrontational way possible): Take case B. This case is perfectly represented by dealing one card face down from a deck consisting of only the four aces and then combining the remaining three aces with the remainder of the deck (2-K) and shuffling, then dealing four more from the remainder of the deck face down. Now, shuffle the hand that you are dealt. Now the condition of the cards exactly mirror the condition of case d. How did the odds of pulling a flush increase by shuffling our own hand? ------------------ Those who don't learn history are doomed to repeat it.
__________________
Those who don't learn history are doomed to repeat it. |
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12-20-2000, 07:41 AM
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#67 |
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Coordinator
Join Date: Oct 2000
Location: Big Ten Country
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Jumping in here..
Grunion, I don't see how you could say that your equation provides for the cases with more than one ace. When I look at your equation, it seems to me to say, "There is a 1/5 probability that the first card is an ace, and there are no others, There is a 1/5 probability that the second card is an ace, and there are no others, There is a 1/5 probability that the third card is an ace, and there are no others, There is a 1/5 probability that the fourth card is an ace, and there are no others, There is a 1/5 probability that the fifth card is an ace, and there are no others." ------------------ "If we can put a man on the moon, we can grow grass indoors." |
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12-20-2000, 07:45 AM
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#68 |
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Coordinator
Join Date: Oct 2000
Location: Big Ten Country
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I remember always having trouble finding out how many cases of "at least one" of something, until I'd remember to find out how many cases there are of none of them, and subtracting that from the sample set.
So I would say the number of hands with at least one ace is: Number of possible hands - number of hands with no aces Number of possible hands = 52x51x50x49x48 = 311,875,200 Number of hands with no aces = 48x47x46x45x44 = 205,476,480 311,875,200 - 205,476,480 = 106,398,720 ------------------ "If we can put a man on the moon, we can grow grass indoors." |
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12-20-2000, 08:32 AM
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#69 | |
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lolzcat
Join Date: Oct 2000
Location: Annapolis, Md
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Grunion, I think we are making progress now.
Incidentally, the instructions for "quoting" and other formatting are on the faq link at the top right of this page, I believe. Quote:
Okay, we have a problem here. I think you have calculated the number of non-ace flushes incorrectly-- there are quite a number of examples on this thread of a correct calculation, but 12x11x10x9x8 isn't enough-- I believe that your 12 should be a 48-- which rapidly goes to prove my point. My point being-- step by step: -Given: the entire set of hands A has a 0.198% flush probability -We know that the entire set A can be broken into two mutually exclusive (complementary) sets: D = hands with at least one ace F = hands with no card over a king (We know that each and every hand in A is either in D or in F, but not both, and that there are a positive number of each) -If, as you argue, the flush probability of the hands in D is EXACTLY the same as the flush probability in A... -Then, the flush probability in F must be exactly identical to that of both A and D. (Otherwise, adding those hands to D would alter the aggregate set's flush probability) But, as you calculate above (and others have done elsewhere) - this isn't the case. The flush probability of F is clearly not the same as A (or D, for that matter). In fact, it is less than the flush probability of set A (for a variety of intuitite resaons previously stated on this thread). Therefore, the correct balance is this: Add set F (with lower flush probability than A) to set D (with higher flush probability than A) and get set A (with a flush probability in between those two). |
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12-20-2000, 08:49 AM
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#70 | ||
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lolzcat
Join Date: Oct 2000
Location: Annapolis, Md
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Quote:
I appreciate that. Quote:
Good, thought-provoking question. The difference between your two cases is that in case D (the second one you describe), you are not generating the hands randomly. By starting with an ace and then drawing randomly from the rest of the deck, you are (among other things) increasing the chances of getting a hand with multiple aces. This does not map accurately to the same subset of hands generated by a random draw to the entire deck. The shuffling only upsets the order-- but it doesn't magically restore the probability. You could follow that logic ("this technique results in all hands that have at least one ace") to its extreme by starting with an ace (darwm from the 4 aces) and one other randomly-drawn non-ace card (drawn from the 49 non-aces) and then drawing from the remaining shuffled deck of 50 to get the other three. You'll still have some probability of ending up with any of the hands that are within subset D, but you do not have the same probability of drawing each one, since the method of the draw is non-random. So, in essence, I challenge your statement that "Now the condition of the cards exactly mirror the condition of case d." I don't believe that to be so. |
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12-20-2000, 01:41 PM
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#71 | |
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lolzcat
Join Date: Oct 2000
Location: Annapolis, Md
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Grunion, I'm taking a step back here, and making a broader, sinpler appeal to intuition.
Above, your analysis of the puzzle states the following: Quote:
Now, without even getting into any of the main issues upon which we currently disagree... we ought to be able to agree on one thing: A hell of a lot more than 7 or 8 percent of five-card hands are going to have at least one ace. The approximately 8% comes from taking the calculations above of total flush hands versus total hands-- 24 million into 312 million This one you can take to the laboratory, if you like. Go and deal out 100 poker hands, and count up how many have aces. It most certainly won't end up being 7 or 8-- it's going to (intuitively) be 30 or 40 or so, right? If absolutely nothing else I'm arging sticks, this ought to give you a strong hint that there's something fishy with your set of numbers above. |
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12-20-2000, 02:48 PM
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#72 |
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Coordinator
Join Date: Oct 2000
Location: Big Ten Country
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Oddly enough, that 7 or 8 percent figure is 1/13, which is the probability that a ONE card hand will include an ace.
------------------ "If we can put a man on the moon, we can grow grass indoors." |
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12-20-2000, 04:55 PM
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#73 |
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lolzcat
Join Date: Oct 2000
Location: Annapolis, Md
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Passacaglia, it's not odd at all. It's a direct product of the fact that all he's doing (rather than calculating all the hands that have one or more aces) is calculating all the hands that start with an ace. Therefore, the options for the first card are cut from 52 to 4, and therefore the number of hands by a factor of 1/13. It's no mystery-- it's just another product of the errant translation of the conditions described into the count used.
Correctly counting the total number of hands that conatin one or more aces is either pretty easily done using the "subtraction" method that you and I have used above... or a little tougher using the "direct" method that I did above... but either one yields the correct result, which is 106,398,720. Incidentally, this number also passes the "smell test" when compared to what we'd expect. It represents about a third of the total possible hands (311,875,200) -- which is in keeping with what you'd see if you started dealing yourself random hands of cards: about one in three have at least one ace. (You get 5 shots at a 1/13 probability... that's pretty intuitive, but someone could work out the math) |
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12-20-2000, 11:42 PM
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#74 | ||
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Mascot
Join Date: Oct 2000
Location: Richmond, VA
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Hi Guys,
In response to various comments: From Pass. Quote:
1/5(4*51*50*49*48) means that with a guaranteed ace, there is a 1/5 chance that the guaranteed ace will fall in the first position, and it can be one of four aces. The following card can be one of 51 remaining cards, which does include the three remaining aces, with the process continued until all five cards are dealt. The 1/5 factors are meaningless, and will cancel each other out. A flush is a flush is a flush no matter what order the cards were dealt in, so the full equation simplifies to: 4*51*50*49*48. That is how I originally stated it, but broke it out into 1/5's to illustrate a point to QuikSand. The equation above accounts for aces, as there are all 51 remaining cards to choose from. Also, I must apologize, some of my number crunching has been sloppy, and it has been getting in the way drawing this discussion to a conclusion. I have calculated that there is a 34.1% chance of pulling an ace, which appears to agree with QuikSand's figures. From QuikSand: Quote:
I stand by my assessment that this example accurately represents case D. Of course there is an element of unrandomness in the process, but that is because there is a loss of randomness due the the constraint of at least one ace. In my example, the bottom line is we know there is at least one ace, that the remaining four cards were drawn randomly from the remaining 51 cards, and that both the suit of the ace and position of the ace are unknown. I think we agree that we know exactly the same amount of information about my example as your defined case D. It appears that your contention is that I have not undergone a random process to get there, thereby skewing the results. I "start" with an ace because an ace is a constraint of the condition. We know that an ace is present. Pulling a random ace in the beginning of my exersize is done to meet that condition. Your contention that my method increases the probability of pulling multiple aces is entirely correct. It mirrors your constraint. Case D will result in more multiple ace hands than case A. I also completely agree that shuffling will only upset the order, not the probability. That has been one of my major points throughout this discussion, and is the reason that knowing you have a first card ace is no more relevant than knowing you have at least one ace. In both cases, the rank of exactly one card is known. The location of the ace (or the fact that the ace is known to be in the hand, but its precise location is unknown) is irrelevant to the probability of pulling a flush. I fully agree with your contention that drawing a non ace as a second card would provide the same subset of results (since a 5 of a kind is impossible) and yet not be random. The fact that this is true does not mean that it is impossible to accurately model a scenario which is. Your discussion on mutually exclusive sets is interesting. I'm going to sleep on it. ------------------ Those who don't learn history are doomed to repeat it.
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12-21-2000, 08:07 AM
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#75 | |||||
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lolzcat
Join Date: Oct 2000
Location: Annapolis, Md
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Grunion,
Your recently posted: Quote:
But, after some commentary on the inappropriate conclusions this leads to, you have posted this: Quote:
Okay, I presume that your newfound calculation of 34.1% as the likelihood of getting one or more aces from a draw of five cards must reflect that you have revised your counts of either: -the total number of hands possible (a pretty simple calculation, upon which we have agreed); or -the number of hands with an ace (a tougher calculation, upon which we have disagreed). [This presumes that we are in agreement that the flush probability of a random hand is equal to the ratio of these two numbers-- I cannot imagine that we have disagreeemnt there] I strongly sspect the latter is the case. In that case, I now assume that you have joined me in my calculation of 106,398,720 as the correct total of hands that include at least one ace. The laborious version of this calculation is above, stamped posted 12-19-2000 07:08 AM. I'm assuming at this point that you now agree with that calculation. This would, I think you will also agree, invalidate your calculations of the likihood of getting a flush within those 106,398,720 hands-- as your calculations are now invalidated by having a dramatically different denominator. We're not yet to the point of agreeing on the numerator for that calculation, as we have used diffrent methods to calculate the number of flushes that include at least one ace. Mine is contained in the same post I referenced just above. If we can agree to both parts of that fraction, then we may be past this thing. - - - Quote:
Your discussion on this topic misses my main point. Of course case D will result in more mutiple ace hands than case A. My point is that the case you described (starting with an ace, and then randomly drawing from the deck) will give you the same set of hands as those in case D, but those hands will not be distributed in the same manner as in case D. The non-random distribution skews the probability toward an increased number of hands with more than one ace than were represented in the randomly-drawn case D. By doing so, this reduces the flush probability of that set-- and mathematically, it does so back to the flush probability of the original set A. Again, we are stuck on the issue of the substantive difference between set B and set D. Same old song. I'm buoyed by the notion that you agree with one observation of mine-- intuitively, we are standing ont he same ground, it seems: Quote:
So, you'll buy the argument that the "two-card headstart" (selecting two cards by a nonrandom manner, then drawing randomly) will generate exactly the same group of hands but in a different, nonrandom distribution. It doesn't seem a huge leap of faith to me to than recognize that your method-- the "one-card" headstart (selecting one card by a nonrandom manner, then drawing randomly)-- is also likely to yield the same thing: the same domain of possibilitites, but not the same internal probabilities. - - - Quote:
We're in the same time zone... you must lead a far more interesting life than I do if you're just getting around to sleeping on anything at 1am. I'd been cuddled up with my bottle of cough medicine for 2 hours by then. Nighty-nite. |
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12-21-2000, 11:22 AM
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#76 | |
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Mascot
Join Date: Oct 2000
Location: Richmond, VA
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Quote:
I completely disagree with the above statement, for the following reason. A concept that everyone has agreed upon is that the probability of pulling a pair is inversely proportional to the probability of pulling a flush, i.e. as conditions are set which make the odds of pulling a pair greater, the odds of pulling a flush are lessened. In the one card scenario, the odds of pulling any pair do not change. The rank of the first card has no bearing on the overall chance for a pair. Naturally, upon discovery of the first card, the pulling a pair of the same rank as the first card increase, but the odds of pulling a pair of other cards decrease. The net effect is a cancellation. The drawing of a second card changes the scenario dramatically. You've created a condition where we know that the first and second cards do not pair up. This is useful information. We now can eliminate all non-flush hands in which the first two cards match rank, thereby reducing the denominator. As there is no reduction in the numerator, because there are no flush hands where the first two cards pair up, the odds of pulling a flush are greater than the default odds. My selection of a random ace is not to manipulate the percentages, it is merely a mechanism to assure compliance to the stated constraint of case D. The overall compsition of the hand is random as it pertains to the constraint. To me this is as evident as removing the aces from a deck prior to dealing to get a random hand which contains no aces. The mechanism to get there is different, due to the constraint being an inclusion rather than an exclusion. I believe we have successfully isolated our disagreement, I don't know if we are any closer to resolution. I think your prior reduction arguements are invalid, because the addition of a second card changes the situation. I think I have an answer to your mutually exclusive arguement as well. I've edited this message and deleted my response, because it was incorrect. See following posts for discussion. [This message has been edited by Grunion (edited 12-21-2000).]
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Those who don't learn history are doomed to repeat it. |
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12-21-2000, 11:26 AM
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#77 |
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Mascot
Join Date: Oct 2000
Location: Richmond, VA
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Dolaposting to correct a flaw in my logic.
I am missing a subset, cards that contain at least one spade, but less than five spades, which would have a 0% probability of pulling a flush. Disregard my mutually exclusive arguement for now. ------------------ Those who don't learn history are doomed to repeat it.
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12-21-2000, 11:53 AM
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#78 |
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Mascot
Join Date: Oct 2000
Location: Richmond, VA
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QuikSand,
How about this: Set A = all possible hands Subset 1 = all hands w/ at least one club Subset 2 = all hands w/o clubs P flush(1) = (12/51)*(11/50)*(10/49)*(9/48) =0.198% This represents the possibility of each subsequent randomly selected card being a club. Using your logic, subset 2 would have to have P flush(2) = 0.198%, which is cerainly not the case. ------------------ Those who don't learn history are doomed to repeat it.
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Those who don't learn history are doomed to repeat it. |
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12-21-2000, 01:40 PM
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#79 | |
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lolzcat
Join Date: Oct 2000
Location: Annapolis, Md
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Quote:
I'm not sure to what you refer by "my logic" but I would make no such statement. First, I would assert that you are making the same error here that you are making in our general puzzle-- when a condition is states that is general among the five-card hand, you are assuming that the condition may be fulfilled by using nonrandom means to select the first card, and then randomly selecting the others. This is a/the fundamental flaw in your calculation here, and several other places throughout your analyses, as I see them. By this error, you have incorrectly calculated above the probability of "hands with at least one club" of being a flush. Without even attempting the math, I can tell you that the probability of getting a flush among the hands you label above as "subset 1" will be less than that of the entire set of hands, A-- i.e. something less then 0.198%. Intuitively, this is because (as you suggest) we have two mutually exclusive sets of hands, and the intuitive flush probability of the group you label subset 2 will be higher than the whole set A. I'd rather not work through the entire set of calculations to prove all this, but the flaw in your argument lies not in the logic I'm using (about mutually exclusive sets), but in the calculations that you are using to calculate P flush(1). |
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12-21-2000, 02:19 PM
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#80 | |||
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lolzcat
Join Date: Oct 2000
Location: Annapolis, Md
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Grunion, I think I need a recap-- you seem to feel that we have clarified our differences somehow, but I'm not so sure which of your statements you continue to support.
The puzzle, at this point, boils down to this: Which of the two sets of hands has a greater probability of being a flush? A) All five card hands D) All five card hands that contain at least one ace It appears to me that our dueling "conceptual" arguments are not successfully persuading "the other side" in this discussion. So, I propose that we try to isolate this into the simple mathematics of the puzzle. We both seem to agree that one acceptable method to calculate the flush probability of a given set of hands is to calculate: d (denominator) = the total number of hands in that set; n (numerator) = the total number of flush hands in that set; and the flush probability of that set may be calculated as (n/d). I also believe that we have agreed to the entire calculation (using this method) for set A. I will quote your calculations from mid-page 3 of this thread: Quote:
And thus you calculate that the {approximate) flush probability for set A is equal to 0.198%. I agree with these calculations completely. After this point, as I understand it, we seemingly part company. I have made (on page three of this thread) a similar calculation for case D above, which includes an explanation of each of my calculations, hopefully making it a bit easier to follow. This I will excerpt below, for ease of reference: Quote:
While calculating these four numbers and their resulting ratios is sufficient to answer our original question, it also stands to reason that the numbers ought to pass a sort of "double check." Using two of the numbers that we have had to calculate, we can also determine the probability that a random hand will contain one or more aces. Doing so requires that we simply divide the number of hands that have at least one ace (which we have calculated already) into the total number of hands (which we also have already). Doing so, using my figures, works out as follows: Total hands: 311,875,200 Hands with one or more aces: 106,398,720 Probability that a hand will contain one or more aces: 106,398,720 / 311,875,200 = 34.1% This calculation-- again, using my numbers-- comports with theresult that we have already agreed upon, as evidenced by your comment from above: Quote:
- - - So, with that long-winded introduction, I make my argument that: -From the original problem, a hand from set D is more likely to be a flush than is set A. -The flush probability of set A is approximately 0.198% -The flush probability of set B is approximately 0.223% -These results are calculated using methods undisputed by all involved parties -The probability of a random hand containing one or more aces is approximately 34.1% -This result is calculated using the same data and calculations as above - - - - - Grunion, it would be helpful to me if you could lay out (with calculations, if necessary) what you now assert to be: -The total number of hands that contain at least one ace -The total number of flush hands that contain at least one ace -The probability of a hand that contains at least one ace of being a flush It might be useful to then use the first of those three numbers to employ the same "double-check" that I did, and show how it supports the fact that 34.1% of hands will contain one or more aces. In particular, if you assert that the answer to the second item above is different than that shown by my calculations above, it would be very helpful to have a set of annotated calculations that might serve to guide me/us through your logic. My suspicion, however, is that you won't be able to construct a set of numbers that resolve all these issues. Doing so clearly requires you to abandon your previously posted calculations both of the number of hands with one or more aces, and of the number of flush hands with at least one ace. If you actually can get this far, then we'll have really clarified our differences-- and then you may be a lot closer to showing me the error of my ways. Or, I suppose, vice versa. [This message has been edited by QuikSand (edited 12-21-2000).] |
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12-21-2000, 04:25 PM
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#81 |
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Mascot
Join Date: Oct 2000
Location: Richmond, VA
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Ok,
Our discussion has boiled down to whether or not my methods reflect true randomness. If my methods do indeed reflect true randomness, it would stand to reason that my analysis is correct. Consider this: Fred has two red pills and two blue pills. If he takes a red pill and a blue pill together he will die. He must select two at random. Consider the random possibilities: R1R2 R1B1 R1B2 R2R1 R2B1 R2B2 B1R1 B1R2 B1B2 B2B1 B2R1 B2R2 It should be pretty apparent that he has a 2 in 3 chance of croaking. Now lets add a constraint, at least one of the pills is a red pill. R1R2 R1B1 R1B2 R2B1 R2B2 R2R1 B1R1 B1R2 B2R1 B2R2 8 out of the 10 possible outcomes will kill him. Fred has lost the double blue combos which will let him live. Now the question is, are all of the above permutations equally likely to occur? I do not believe that this the case. Permutations are more appropriate when order matters. Since order is irrelevant (as it is in our flush problem), lets look at combinations. Total combinations: R1R2 R1B1 R1B2 R2B1 R2B2 B1B2 There is still a 2/3 chance of Fred dying. For combos w/ the red pill constraint just remove B1B2. Now four out of five possible outcomes will kill Fred, same as above. However, I contend that it is twice as likely for Fred to pull two red pills, given that at least one of them is red, than it is the non conditional scenario. (This should be fairly intuitive, and is fundamentally the same as our agreement that there is a higher probability of pulling an ace given that one is already present.) I think the method you are using in the flush problem applies the condition after the randomization, not before. (Sorry, I have no better way to explain the concept.) Anyway, what I believe your method does is allow Fred to draw four pills completely at random, disregarding any constraints. Then if he is fortunate to pull double blues telling him "Oh, sorry pal, but at least one of those have to be red. Drop those two blue ones back in the pot and try again." Just as in your flush example, if you draw a hand with no aces, it is null and void. The set you are using for selection differs from the one you are using as a condition. This makes a big difference. If your methodology is to pull five random cards from a 52 card deck, and throw out the result if an ace is not present, then I agree with your calculated probability completely. I still disagree with its appication on this constraint. By the way, what's your occupation (just curious, I'm a civil engineer) ------------------ Those who don't learn history are doomed to repeat it.
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Those who don't learn history are doomed to repeat it. |
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12-21-2000, 05:25 PM
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#82 | |
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Mascot
Join Date: Oct 2000
Location: Richmond, VA
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Quote:
Total hands = 52*51*50*49*48=311,875,200 Total hands w/o aces = 48*47*46*45*44 =205,476,480 Total hands w/ at least one ace = 106,398,720 Ace % = 34.12% Total flush hands = 4*13*12*11*10*9= 617,760 Flush % = 0.198% Flush hands which contain an ace =4*12*11*10*9*5 (since this is a permutation) =237,600 Flush w/ ace:Flush = 38.46% Now, if I were to divide 237,600 into 617,760 I would arrive at your answer for D. I can see one conflict with doing so. It is evident that the percentage of flushes with an ace versus flushes is greater than the percentage of hands with an ace versus total hands. This is logical, the reason being that there is no possibility of there being more than one ace in the flush hands, while there is a possiblity of there being duplicate aces when looking at total hands. From this we can conclude: If I have a flush, then it is more likely to have an ace in it than if I have 5 random, non constrained cards. We can also conclude: If I have an ace, then it is more likely that I have a flush than if I do not have an ace. (not directly from above, but we are in agreement about this) However, we can not conclude: If I have an ace, then it is more likely that I have a flush than if I may or may not have an ace. My calculation for case D is as follows: One of four cards must be in the hand (the aces). In addition, the remaining cards all have equal probability of being in the hand. Total combinations equal # of flushes w/ an ace: 4*12*11*10*9=47,520 # of hands w/ an ace: 4*51*50*49*48=23,990,400 % flush = 0.198% To obtain permutations, multiply both calc's by five, which accounts for the location of the ace, and will cancel out. Which brings us back to: # of flushes w/ an ace = 237,600 # of hands with an ace = 119,952,000 Which brings us to our disagreement. I believe that the # of hands with an ace in your arguement is not correct. I think the problem has to do with multiple aces not being accounted for. I'll see what I can do to account for the difference. ------------------ Those who don't learn history are doomed to repeat it.
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12-21-2000, 06:43 PM
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#83 | ||
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lolzcat
Join Date: Oct 2000
Location: Annapolis, Md
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Grunion,
On first blush, I agree that your red/blue pill construct is an appropriate mirror for our disagreement. As you may not expect, I do, in fact, believe that poor Fred's chances of dying increase to 80% if we add the constraint you suggest. Your comment: Quote:
Yes, you are right-- that's exactly how I would suggest that we play the game. And yes, it makes a big difference-- you're darned right it does. My method versus your method is the difference between random (with each outcome having the same chance) and non-random (where some outconmes have different chances). Mine is random, yours isn't. And that is, indeed, a mighty big difference. As for question two... I'm a lobbyist. I deal with tax policy and local government issues. As for your numbers... I have to run right now, but I'll get to them fairly soon. I'm intrigued...I see you get to my answer, but then talk your way out of it. I haven't yet digested the rationale, but you'll have my undivided attention a little later. For now, I'll make this one observation: Quote:
I see these two statements as being mutually inconsistent. Back to my argument about set theory that we had a little trouble with earlier. Later... |
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12-21-2000, 08:25 PM
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#84 | |
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Mascot
Join Date: Oct 2000
Location: Richmond, VA
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Quote:
I think you may be combining permutations and combinations. The probability of pulling exactly one ace is equal to: the probability that the first card is an ace and all of the other cards are not aces plus the probability that the second card is an ace and all of the other cards are not aces, plus.......the fifth card is an ace and all of the other cards are not aces. or (4/52)*(48/51)*(47/50)*(46/49)*(45/48) + (48/52)*(4/51)*(47/50)*(46/49)*(45/48) + ..... + (48/52)*(47/51)*(46/50)*(45/49)*(4/48). This reduces to: 5(4*48*47*46*45)/(52*51*50*49*48) = 29.95% In like fashion, Probability of two cards being aces is: 10(4*3*48*47*46)/(52*51*50*49*48) = 3.99% Three aces: 9(4*3*2*48*47)/(52*51*50*49*48) = 0.16% Four aces: 5(4*3*2*1*48)/(52*51*50*49*48) is essentially 0.00%. Adding these up we get: 29.95+3.99+0.16=34.1%, which we both agree is the correct percentage of hands that contain at least one ace. So my process should be valid. If you agree with the above, it should be intuitive that there is no difference between knowing that your first card is an ace and knowing that there is at least one ace. What happens is that with known cards in unknown locations, the equations are going to collapse on themselves. The probability of having a first card ace equals the probability of having a second card ace equals the probability of having exactly one ace of unknown location. This can be applied to case d: as the number of aces are not known, but we know there is one. We can assign that one ace a "slot" and pull from a 51 card deck for the remainder insuring true randomness. or(4*12*11*10*9)/(4*51*50*49*48). =0.198% ------------------ Those who don't learn history are doomed to repeat it.
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12-21-2000, 08:41 PM
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#85 |
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Mascot
Join Date: Oct 2000
Location: Richmond, VA
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Going back to the pill:
Resolve this paradox: Chance of Fred dying =66.7% Chance of Fred dying if he has at least one red pill = 80% Chance of Fred dying if he has at least one blue pill =80% In all cases Fred is assured of having either a red pill or a blue pill. We are dealing with dependant events: Bayes Theorem states: p(A/B) = probability that A will occur given that B has already occured, where the two events are dependant. p(A/B) = p(A and B)/p(B) Let p(A) = Fred gets a blue pill Let p(B) = Fred gets a red pill p(A and B) = Fred gets one of each (death) If we know that Fred has at least one red pill p(B) = 1 We also know that p(A and B) = 0.67 Therefore p(A/B) which represents the probability of Fred getting a blue pill is equal to 0.67/1 = 0.67 I can assure you that the pill scenario (as well as the flush scenario) deal with dependant events. In order for the events to be independant We would have to have to select pill 1 from a separate set of pills than pill 2. ------------------ Those who don't learn history are doomed to repeat it.
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12-21-2000, 08:55 PM
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#86 | |
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Mascot
Join Date: Oct 2000
Location: Richmond, VA
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Quote:
They are consistent. It goes back to the type of information we receive. Knowing that we have no aces reduces our potential card pool to 48, and we both agree that this makes flushes more difficult. Knowing that we have an ace does not help us at all. We are still working with a pool of 52 cards. Our odds of pulling a flush have not changed (neither have our odds of pulling a pair, three of a kind or full house. However our chance of pulling a staraight would decrease, unless you play with wrap around straights.) And yes, the knowledge that you have an ace eliminates an undesirable subset, which is the one that you know that you do not have an ace. It also enhances an undesirable subset, which is the one where you pull a pair of aces. Unfortunately, even though you have eliminated an undesirable subset, you have not created an advantage. Not having to try construct a flush out of only 48 cards as opposed to 52 does not provide a flush making bonus. It merely assures you that your odds have not diminished. ------------------ Those who don't learn history are doomed to repeat it.
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12-21-2000, 09:30 PM
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#87 | |
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lolzcat
Join Date: Oct 2000
Location: Annapolis, Md
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Trying to reply to several things at once, I'll start at the end...
I find you agreeing with more and more of my supporting statements, without yet agreeing with my conclusions. Go figure. Quote:
Nonsense. Start with a finite set of anything (poker hands, pills, anything) of which a given proportion are of a certain character (flushes, blue). Remove from that set a finite subset of elements which are, on the whole proportionally less dense with the specified character. What remains from the original set, will absolutely be more dense with the character than the original set was. Here's a more concrete example. Take 100 of your blue/red pills, of which exactly 70% are blue. Now remove any number you like-- as long as of the ones you remove, there are fewer than 70% blue. Now, examine what remains from your original 100. Are they 70% blue? Less? Of course not-- they are in every case more than 70% blue. (my notation below is xx/yy where xx=number of blue pills, and yy=number of total pills) Start with 70/100 blue. Subtract 10/20 (50% of removed are blue). Remaining are 60/80 blue = 75%. Start with 70/100 blue. Subtract 40/60 (66% of removed are blue). Remaining are 30/40 blue = 75%. Start with 70/100 blue. Subtract 0/1 (0% of removed are blue). Remaining are 70/99 blue = 70.7% blue. - - - The only case in which the odds in the remaining pool do not change is when the amoung "taken out" approaches zero... and those kinds of measures are obviously meaningless in this exercise with the whole flushes/aces thing. So, back to the original problem, since you've agreed that it does, in fact, work this way (the exclusive sets concept): Start with 311,875,200 hands, of which we agree that 0.198% are flushes. Remove the 205,476,480 hands without aces, whic we now agree have a lower proportion of flushes than the entire set. (Just like my extractions from the 100 pills each contained less than 70% blue) What remains from the set of hands will, mathematically, have a higher density of flushes then the original set. There is, of course, algebra behind all this-- but I simply don't see how it's needed. |
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12-21-2000, 09:53 PM
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#88 | |
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Mascot
Join Date: Oct 2000
Location: Richmond, VA
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Quote:
The identical point can be made for case B. We both agree that P flush for case B equals P flush for case A. There is an inconsistancy somewhere with your logic. ------------------ Those who don't learn history are doomed to repeat it.
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Those who don't learn history are doomed to repeat it. |
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12-21-2000, 09:54 PM
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#89 | |
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lolzcat
Join Date: Oct 2000
Location: Annapolis, Md
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I think I buy everything in here:
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And this also means that you could calculate the precise chances of receiving exactly one ace, given the condition that you have one of these 34.1% of hands which does, indeed, have an ace. Obviously, the chances of this are (using your numbers, which look right to me): 29.25% / 34.1% = 87.8% That is, of all the hand with at least one ace, 87.8% will have exactly one ace. We shoudl agree there. Now, as you propose to set this up-- we'll just draw the ace first, and then revert to a random selection for the rest of the hand. If you're right, that will undoubtedly generate the exact same likelihood of getting exactly one ace. Well, this will be easy, right? I'll just use your preferree method of using the incremental probabilities... Chance of only one ace, given that we start with an ace is: chance that second card is a non-ace x chance that third card is a non-ace x chance that fourth card is a non-ace x chance that fifth card is a non-ace. Again this is easy: 48/51 times 47/50 times 46/49 times 45/48. Which equals... 4,669,920 / 5,997,600 = 77.8%! Geez, sure enough-- "seeding" the first card of the draw by using a non-random method does alter the likelihood of the various possible hands that will result. It makes a rather large difference-- now, instead of single-ace hands representing a full 87.8% of the hands (which is what happend when we allow the hands to be selected randomly, with every combination having exactly the same chance as any other), they now only represent 77.8% chance-- since we (in this example) used a non-random process which inherently makes some hands more likely than others. - - - Our original problem simply said "their likelihood of being a flush," not anything about the method of dealing or any other knowledge that might affect the likelihood of any one hand over another. (Which seems to be what you're trying to read into the puzzle) This puzzle clearly states that we are looking at the entire set, and assessing the likelihood of those hands being a flush. We cimply cannot arbitrarily assign some sort of method to their distribution, and accordingly weigh certain hands as being more likely then others-- based on the way that we might have liked a dealer to have prepared this hand for us. The puzzle simply doesn't allow it. We must work with the concept that each hand is equally likely-- which sends us invariably to the answer that Vaj and Passacaglia provided, and which I and others have demonstrated forwards and backwards. [This message has been edited by QuikSand (edited 12-21-2000).] |
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12-21-2000, 10:00 PM
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#90 | |
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lolzcat
Join Date: Oct 2000
Location: Annapolis, Md
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Quote:
Not at all. Case B also breaks down the set B into two groups: Set B = hands with the first card an ace Set (A=B) = hands with the first card a non-ace I would solidly assert that all three sets A, B, and (A-B) have precisely the same flush percentage. That is because case B and case D are fundamentally different-- the basic point on which you and I disagree. |
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12-21-2000, 10:10 PM
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#91 | ||
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lolzcat
Join Date: Oct 2000
Location: Annapolis, Md
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As for your Bayes pill setup:
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Your error here is clearly in your after-the-fact assignation of the probability of Fred's getting at least one red pill. You state: Quote:
No, no, no. We can't take our ex post facto knowledge of the problem and go back and monkey with the percentages. The percentages are fixed as we set up the problem, and they do not change (that's the idea behind Bayes' Theorem). So, we stick with: Let p(A) = Fred gets a blue pill = 10/12 Let p(B) = Fred gets a red pill = 10/12 p(A and B) = Fred gets one of each (death) = 8/12 And, properly using Bayes' Theorem, we calculate it as you suggest: p(A/B) = p(A and B)/p(B) p(A/B) = (8/12) / (10/12) P(A/B) = 80% Which is exactly what we expact to see-- you rule out the two options that didn't meet the revised criteria, and look back and see that 8 out of 10 equally likely outcomes would have reulted in his death-- therefore he has an 80% chance of buying it. No combining bundles of outcomes or shifting orders necessary, Bayes and all. [This message has been edited by QuikSand (edited 12-21-2000).] |
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12-21-2000, 10:12 PM
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#92 |
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H.S. Freshman Team
Join Date: Oct 2000
Location: Lawn Gisland, NY
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Pardon my butting in.
Grunion, let's deal strictly with combinations. The number of combinations of flushes with an ace is C(4,1)*C(12,4). The possible number of combinations with at least one ace is the sum of the possible number of combinations with one ace, two aces, three aces, and four aces. This is C(4,1)*C(48,4) + C(4,2)*C(48,3) + C(4,3)*C(48,2) + C(4,4)*C(48,1). The quotient of these is 1980/(778320+103776+4512+48) = .0022331.
__________________
"To all of those here who work in marketing or advertising: kill yourselves." -- Bill Hicks "Christianity's such an odd religion. The whole image is that eternal suffering awaits anyone who questions God's infinite love" -- Bill Hicks |
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12-21-2000, 10:12 PM
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#93 | |
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Mascot
Join Date: Oct 2000
Location: Richmond, VA
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Quote:
This is a fairly clear example of you incorrect reasoning. You have five cards which include one given and five unknowns. If one is given, only four can be unknown, this is precisely our discrepancy. I noticed you skipped over Bayes Theorem. ------------------ Those who don't learn history are doomed to repeat it.
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Those who don't learn history are doomed to repeat it. |
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12-21-2000, 10:19 PM
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#94 |
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lolzcat
Join Date: Oct 2000
Location: Annapolis, Md
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You are right that I made an error above - but it is immaterial to everything else I argue, since adjusting it to the proper four terms (which I'll do momentarily) will leave the likelihood at 77.8% - still well below what is correct for an unbiased sampling.
My apologies for murkying the water with my too-hasty typing and multiplying... |
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12-21-2000, 10:30 PM
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#95 |
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lolzcat
Join Date: Oct 2000
Location: Annapolis, Md
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I noticed you skipped over Bayes Theorem.
Nope, just posted that one last. It showed up just before yours - please don't miss it. I'm off to bed... enjoy. |
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12-21-2000, 10:36 PM
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#96 | |
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Mascot
Join Date: Oct 2000
Location: Richmond, VA
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Quote:
It is not ex post facto knowledge. We have used logic to make a correct determination regarding the possible outcome of the event: Given: Fred has at least one red pill. The probability of Fred having a red pill is 100%. We know that before hand, that is the entire purpose of the given. Any time a condition is provided as a constraint, it is always valid to assume that the condition exists. I can not understand what you disagree with: 1. Whether or not we "know" that Fred has at least one red pill. (We obviously do, it was stated as a condition of the problem. We do not need to deduce or infer, it is stated.) 2. That the probability of Fred having a red pill if we know that Fred has at least one red pill is 100%. ------------------ Those who don't learn history are doomed to repeat it.
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Those who don't learn history are doomed to repeat it. |
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12-22-2000, 07:18 AM
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#97 | |
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lolzcat
Join Date: Oct 2000
Location: Annapolis, Md
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What I disagree with is your bastardization of Bayes Theorem, which is a tool to calculate posterior probability by using prior probabilities.
You've created a very elegant little Bayesian problem here, whoch lends itself quite nicely to a Bayesian analysis. However, in doing so, you are not properly using prior probabilities... specifically the prior probability that p(B) = Fred gets a red pill = 10/12. In the Bayesian analysis, we properly look at the subset of outcomes that conform to the limiting condition, in this case that he too at least one red pill. There are ten such outcomes, which were equally likely to have occurred to cause the observed condition. Of the ten equally likely outcomes, eight of them involved the deadly combination. Therefore, the likelihood of Fred's dying is 8/10 - 80%. If we genuinely disagree on this problem, and you honestly believe that the posterior condition allows us to revisit the probabilites of each outcome and start re-weighting some more than others... then we've probably been wasting our time with the needlessly difficult flush puzzle, as our differences are much more elemental than anything required for that. As for your biting question: Quote:
What I disagree with is your claim that either of these tautologies have anything to do with the prior probability of his getting a red pill... which is what this solution calls for using. a commentary here... It seems as though (in both the pill puzzle and the flush puzzle) you are reading an extra step into the puzzle-- something along the lines of: "the setup that we are to analyze was generated by a method that was created for the purpose of generating this particular outcome." Specifically, when you look at the set of outcomes in which Fred has at laest one red pill, you seem to be saying "well, we have to have had some method of getting to this guaranteed outcome, so I'll just advance him a red pill on his first draw." Similarly with the flush puzzle-- in original case D, you seem to be saying "well, we know the hand has to have at least one ace, so we'll just start it off with one and go from there." In each case, this results in a distortion of the probabilities of each possible outcome. In the pill puzzle, it skews the probabilities toward an increased likelihood that Fred will draw a blue pill, since you "assumed" that one of the red ones was already out of circulation. In the flush puzzle, you increase the likelihood of any given multiple-ace hand over any given single-ace hand-- again, distorting the inherent probabilties from the unbiased subset that we're asked for in the puzzle. I think your pill puzzle has served a useful purpose to simplify, or at least illustrate, our differences here. Hopefully, that one is transportable... maybe there is someone else (here or elsewhere) who can do a more articulate job than I in demonstrating the proper applicatin of Bayes' Theorem (or just inductive use of the mutiplicative property of probability) to solve it. [This message has been edited by QuikSand (edited 12-22-2000).] |
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12-22-2000, 07:45 AM
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#98 |
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H.S. Freshman Team
Join Date: Oct 2000
Location: Lawn Gisland, NY
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I suppose now wouldn't be the best time to bring up the Monty Hall problem
__________________
"To all of those here who work in marketing or advertising: kill yourselves." -- Bill Hicks "Christianity's such an odd religion. The whole image is that eternal suffering awaits anyone who questions God's infinite love" -- Bill Hicks |
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12-22-2000, 08:02 AM
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#99 | |
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lolzcat
Join Date: Oct 2000
Location: Annapolis, Md
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Quote:
Boy howdy! |
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12-22-2000, 08:58 AM
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#100 |
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Mascot
Join Date: Oct 2000
Location: Richmond, VA
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QuikSand,
I agree, and we have completely isolated our difference of opinion. It is very evident in the pill problem, and I am fairly certain that our flush disagreement is for the same reasons as our pill arguement. Alernative Solution for the pill problem. Q. What are the odds of Fred dying given that he has at least one red pill. A. If Fred has at least one red pill, then at least one of the following must be true: Pill 1 is R1 Pill 1 is R2 Pill 2 is R1 Pill 2 is R2 If pill 1 is R1 then pill 2 is R2 1/3 of the time, B1 1/3 of the time and B2 1/3 of the time. Fred will die 2/3 of the time If pill 1 is R2 then pill 2 is R1 1/3 of the time, B1 1/3 of the time and B2 1/3 of the time. Fred will die 2/3 of the time If pill 2 is R1 then pill 1 is R2 1/3 of the time, B1 1/3 of the time and B2 1/3 of the time. Fred will die 2/3 of the time. If pill 2 is R2 then pill 1 is R1 1/3 of the time, B1 1/3 of the time and B2 1/3 of the time. Out of twelve possible outcomes, Fred has a red and a blue eight times. He has double reds four times, not two. Also, going back to your contention: If Fred has one red pill he has an 80% chance of dying. If this is so, we can safely deduce: If Fred has one blue pill he has an 80% chance of dying. The original calculated chance of Fred dying is 67%. The statement that Fred with always have either at least one red pill or at least one blue pill is true. We are still left with the paradox stated earlier: In all possible cases Fred has an 80% chance of dying, yet overall, he has a 67% chance of dying.
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