OT - Five card flush puzzle

[Grunion]

QuikSand,

First off, I have considered our exchange nothing more than a friendly debate. I just wanted to take a step back and state that. I've noticed our posts have become much more terse than when we began. I can assure you that on my end it is because I am trying to present a large amount of information, and only have a limited amount of time to take from my busy life to do so. I have much respect for your intellect and your reasoning.

I believe everything you have presented is true.....for independent events.

I'll make this one quick, for I know what we will agree on:

2 Dice, chance to roll a nine = 4/36 or 11.1%

Now, given that at least one die is a 6, the chance to roll a nine becomes 1/6 or 16.6%

It does not become 4/21 or 19.0%. The logic of selecting 19% parallels your contentions on both the flush and the pill problem.

However, in order for it to be true, we must have 6 ways of making a 7, 5 ways of making an 8,....to one way of making a 12. But given one 6, we have exactly 2 ways of making each number (6,1 & 1,6), (6,2 & 2,6) etc. Except for the 12 of course, because there is only one 12 combination. However, you must account for the 6 introduced at the beginning of the problem. The 12 must be counted twice for that reason. This is the logic that you are not following.

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Those who don't learn history are doomed to repeat it.

[QuikSand]

...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.

I agree with that statement, as well as your benevolent use of the term "argument."

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.

And once again, you're playing tricks with information gained after the fact.

There are not 12 possible outcomes represented here... there are 10. You simply may not count the two "double red" outcomes (R1 then R2, and R2 then R1) twice as separate events. They are not separate events.

You set up this problem. Fred is going to draw his pills randomly. We agreed originally that each of the 12 possible outcomes was equally likely. All that happened behind some sort fo curtain-- we don't get to go back and change the method of the draw to re-suit the new evidence that we add as a later observation of the outcomes (that at least one pill was red).

When we add the abserved result (that at least one pill was red) we simly eliminate the outcomes from the original set of 12 which do not comport with this observation-- they are B1/B2 and B2/B1. We're left with 10 outcomes which satisfy the observation, each of which were originally likely to have caused the result.

Your analysis is akin to trying to "go back in time" and create some new non-random selection process with some kind of double-weighting for certain outcomes to make sure that when we re-tell the story, the result is 100% likely to conform with the observation. It just doesn't add up.

Out of twelve possible outcomes, Fred has a red and a blue eight times. He has double reds four times, not two.

False.

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.

Nice semantic trick, which has some surface appeal, but doesn't add up to anything at all.

We know that after the pill-drawing (which occurs in a random fashion as stated in the problem) the outcome will be something we could categorize into various subsets. Some of these subsets will have different probabilities of having results in his death than others-- some will be higher than the original set, ome will be lower.

The logical flaw is that the two sets you describe (that he will draw at least one blue; that he will draw at least one red) are not, using the parlance of probability: "exclusive and exhaustive" subsets.

If they were exclusive (meaning no overlap) and exhaustive (meaning they combine to form the entire whole set) then we could have a discussion very similar to the one we had about hands with aces and without aces.

However, it's very clear that these two subsets are exhaustive - meaning that each of the original 12 outcomes is covered by the union of the two subsets. It's equally clear that the two subsets are not exclusive - in fact, it's quite easy to see that there are fully 8 of the 12 original outcomes which are elements of each subset you describe.

In essence, the statement that you're missing in your discussion above is something like "Since we know that the outcome has to be either Fred draws a blue pill or Fred drawa red pill, but it cannot be both..." Were you able to say that (which you wisely have not) then you'd be describing an exclusive and exhaustive group of subsets, and you'd be able to draw some meaningful conclusions about their probabilities.

But by establishing non-exclusive subsets, you can play all sorts of semantic games with what they mean, but they don't add up to any sound logical statement.

[Grunion]

As for your biting question:


quote:
--------------------------------------------------------------------------------

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%.

--------------------------------------------------------------------------------

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.

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."

QuikSand,

Using information provided in a given to draw a logical conclusion is an appropriate use of logic.

Also, I have worked out the math on several specific issues (the odds of drawing exactly one ace, two aces, etc.) with which you have been in agreement with my methodology. I have also demonstrated through use of that methodology, that when the location of the given is unknown, and order is not an issue (as in the case of a flush, or the pills, or the dice) the equation will simplify from something that looks like this:

(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 practical application, yes, I am advancing him a red pill on his first draw.

In a pure mathmatical application, refer to my above alternate solution to the problem.

The constraints in all examples do not change the possible outcomes. Your calculations on the number of outcomes appear to be correct. However, the probability of certain outcomes occuring does change.



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Those who don't learn history are doomed to repeat it.

[QuikSand]

Incidentally, I also have a great deal of respect or your intellect, and for the power of your beliefs. Not many would be willing to stick around for even a fraction of all this... and I don't mind the tone at all. Glad you agree there.

Now, on to your little dice puzzle:

I'll make this one quick, for I know what we will agree on:

2 Dice, chance to roll a nine = 4/36 or 11.1%

Now, given that at least one die is a 6, the chance to roll a nine becomes 1/6 or 16.6%

It does not become 4/21 or 19.0%. The logic of selecting 19% parallels your contentions
on both the flush and the pill problem.

However, in order for it to be true, we must have 6 ways of making a 7, 5 ways of
making an 8,....to one way of making a 12. But given one 6, we have exactly 2 ways of
making each number (6,1 & 1,6), (6,2 & 2,6) etc. Except for the 12 of course, because
there is only one 12 combination. However, you must account for the 6 introduced at the
beginning of the problem. The 12 must be counted twice for that reason. This is the logic
that you are not following.

There are 11 outcomes in which at least one die shows a six. They are:

1/6
2/6
3/6
4/6
5/6
6/6

6/1
6/2
6/3
6/4
6/5

Of these 11 outcomes, exactly two of them add to nine.

The likelihood of two dice adding to nine, given the fact that at least one is a six is 2/11 = 18.18%.

The examples are getting simpler, and your logical flaw is becoming more red-lettered.

We do not re-visit the random sequential rolling of two dice and re-establish their method to ensure that we'll get the observed outcome. You're making the exact same mistake again-- you;'re just placing the first die down with a six showing, and then rollin the second die. It's a fallacy-- it simply does not comport with the problem as you set it up. And yes, it's the exact same fallacy that has clouded your results in the flush probalem and the pill problem.

[QuikSand]

Back to the flush puzzle for a moment. Since your concept seems to be "we'll revisit the process of selecting the possibe hands, to ensure that we get the subset that comports with the final observed outcoe (even if that corrupts the probabilities of the component elements of that subset)."

I once asked: why not just start off the hand with one ace and one non-ace? You rejected this method as unfairly skewing the outcome. I ask you now: how can you choose between these two different methods-- both of which will generate the same group of hands, but with different probabilities of certain ones happening? How can you say that "staring the hand with an ace" is appropriate, yet "starting the hand with an ace and then a non-ace" is inappropriate?

Once you betray randomness, how do you ever draw any lines? What's wrong with starting the hand with an ace then a non-ace?

[QuikSand]

I openly wonder how we might settle this. The examples you are providing keep getting simpler and simpler... ar we getting close to something that we could actually conduct ourselves, using playing cards or the like?

I fear that we might not agree on how to set it up... but I'm not certain how else you'll "show me the light" here. (Or conceivably vice-versa)

[QuikSand]

Maybe we can use your dice problem as our vehicle. Here is the expanded version of how I interpret the dice problem, with overbearing narrative to draw out our differences.

- - -

Two fair dice are rolled, and placed into separate spots (so we can tell which was the first, and which was the second). This takes place randomly, behind a curtain.

We then open the curtain, and we are asked "does at leat one of the dice show a six." We correctly respond affirmatively.

Given this knowledge, what is the likelihood that the sum of the two dice is nine?

- - -

Do you agree with this re-phrasing of the puzzle?

[QuikSand]

Using information provided in a given to draw a logical conclusion is an appropriate use of logic.

Yes, it is. But that's not a fair characterization of what you are doing. You are using what is defined as posterior knowledge-- knowledge we gain only after the events in question have occurred-- and you are using it to re-define the events themselves.

The statements:

-The hand has at least one ace;
-Fred drew at least one red pill; and
-At least one die shows a six

...all reflect posterior information and by definition are not what you are calling "given." That's exactly the problem.

If you want to have a dice puzzle that includes the presence of a six as a given, then you must state it something like this:

"Two dice are rolled using a certain method that guarantees that at least one of the two dice shows a six. What is the probability that the two dice together add to nine?"

But of course then, you won't have any answer, since there are multiple methods that may have been used to generate the given conditions. If you want a puzzle with the singular answer you suggest, then you have to spell out the method that you want to have as a given:

"Two dice appear before you. The first was manually placed with the six showing. The second was rolled randomly. What is the probability that the two add to nine?"

And then, you'd finally have a puzzle that gives all the information you're taking as a given and places it all right up into the pre-conditions of the puzzle. And if you asked me that puzzle, I'd have no choice but to give you the answer you seek-- 1/6.

Absent the inclusion of that kind of information all contained within the given, it's just a wholly different puzzle. of course, a parallel argument exists for the pills and the flushes-- but I agree that the simpler puzzles help make the thinking more intuitive.

[This message has been edited by QuikSand (edited 12-22-2000).]

[Grunion]

Once you betray randomness, how do you ever draw any lines? What's wrong with starting the hand with an ace then a non-ace?

Pretty clear answer here. It creates an entirely different set of circumstances. Starting off in the above manner is not consistant with the constraint. We know that there is at least one ace. And while we know there is at least one non-ace (due to their being only four aces), the arbitrary assignment of a second card non-ace artifically eliminates potential cominations of cards. As I stated in the past, in determining the probability of pulling a flush, knowledge about one card is meaningless, knowledge about more than one card is important (even partial knowledge, such as no cards are an ace).

Back to the dice:

There are 11 outcomes in which at least one die shows a six. They are:

1/6
2/6
3/6
4/6
5/6
6/6

6/1
6/2
6/3
6/4
6/5

Of these 11 outcomes, exactly two of them add to nine.

Lets rephrase the question:
Given two dice, and at least one of them is a six, what are the odds of the other die being a three?

[Passacaglia]

dola-post

[QuikSand]

Lets rephrase the question:
Given two dice, and at least one of them is a six, what are the odds of the other die
being a three?

Okay, I'll bite.

In the 10/11 likely event that the two dice are different, the chances are 1/5 that the "other die" is a three.

In the 1/11 likely event that the two dice are both sixes, the chances are 0 that the "other die" is a three.

The combined chances of the two dice adding to nine are (10/11 * 1/5) + (1/11 * 0) = 10/55 = 18.18%

q.e.d.

[QuikSand]

Responding to my comment:
Once you betray randomness, how do you ever draw any lines? What's wrong
with starting the hand with an ace then a non-ace?

Grunion wrote:

Pretty clear answer here. It creates an entirely different set of circumstances. Starting off in the above manner is not consistant with the constraint. We know that there is at least one ace. And while we know there is at least one non-ace (due to their being only four aces), the arbitrary assignment of a second card non-ace artifically eliminates potential cominations of cards.

Simply false. There are no hands that satisyfy the requirement "hands that have at least one ace" that may not be generated by the process of selecting one ace, selecting one non ace, and then randomly selecting the remaining three hands.

Of course this process skews the hand selection, making some more probable than other. And of course that's a problem.

But it's the same problem that occurs in the numerous incorrect analyses of problems in this thread's recent history. You simply cannot alter the "selection method" to try to conform with information gained after the fact. The "one card seed" and the "two card seed" is just more clear evidence of the perils of doing so.

[Grunion]

The statements:

-The hand has at least one ace;
-Fred drew at least one red pill; and
-At least one die shows a six

...all reflect posterior information and by definition are not what you are calling "given." That's exactly the problem.

I don't agree. We have posterior confirmation not posterior information.

You can not solve these problems setting a constraint, then ignoring the constraint to obtain results, and then throwing out the results that do not match the defined constraint. This process is not mathmatically sound. The information provided in the constraint must be accounted for in the randomization process.

For your conclusions to be valid, your statement should be:

Draw two pills, and ignore any result that does not have a red pill in it. (Which is fundamentally different than draw two pills with the constraint that at least one of the pills is red.)

Then, 80% death would be a valid assessment.

[QuikSand]

It seems as though we are focusing in on our semantic differences, which might be the only ones we have. I confess I didn't see this coming two pages ago, but it certainly makes it clear why we seem to disagree on such fundamental issues.

Your interpretation of theze puzzle wordings is one that I have yet to hear from anyone, in any context. You defend it artfully and crispy, and the math that we've exchanged has (save for a few haste-induced errors on both sides) been sound.

Our differences are seemingly in the statement and interpretation of the problem, and I don't know how to resolve that to anyone's satisfaction. I can say with relative certainty that problems like the dice and pill problems you have set up will be replicated (or nearly so) in a wide variety of textbooks, and that in virtually no case will any problem be interpreted or presented in the manner you have argued.

I'm certain that were you to take a test from the instructor materials from any common probability textbook, your interpretation would lead you to any number of answers that would, in the eyes of the text and the instructor, be judged to be "wrong."

I'm doing my best here to avoid saying that you are wrong and I am right, because I now see it more as a matter of convention. Your answers have been, at least to a degree, correct for your interpretation of the problems-- no matter how unusual those interpretations may be.

I do think, though, that your interpretation suffers from a lack of precision. Back to the flush puzzle for a moment-- we probably could agree that there are a number of different non-random methods that could generate five-card hands that contain an ace. You seem to suggest "seeding" an ace as the first card. I've offered that we could "seed" the first card as an ace, and the second card as a non-ace. These two different approaches probably are not the only possiblemethods that could be used to genrate the entire domain of hands that contain one ace-- though with admittedly different probabilities of given hands within that subset. So, how can you ever be sure of a single answer with your interpretations of these puzzles? How can you ever be sure that the nonrandom method you chose to comply with the posterior information is the same method as the unknown being who set up our five card hands?

I don't know what else to suggest. Your various challengs to me, while sometimes cleverland challenging, simply dance around these same issues of interpretation. I'm doing my best to field each of your challenges, but I'm doing so with the nearly universally-accepted principles of probability on my side, so it's almost unfair.


[This message has been edited by QuikSand (edited 12-22-2000).]

[Grunion]

Okay, I'll bite.

In the 10/11 likely event that the two dice are different, the chances are 1/5 that the "other die" is a three.

In the 1/11 likely event that the two dice are both sixes, the chances are 0 that the "other die" is a three.

The combined chances of the two dice adding to nine are (10/11 * 1/5) + (1/11 * 0) = 10/55 = 18.18%

Definately not true, the probability of two dice being different will always be 5/6. (unless, of course, we have a given such as both dice are the same or both dice are different). This should be fairly intuitive.

Given that at least one die is a six, we have one known value and one unknown value. We do not know which die is the six, but that is irrelevant, order does not matter in this case. (Which is defined in combinations, which we have been using excusively in our dice discussion.) What you are stating implies that the possibility of rolling a three on a given die increases because a separate die happens to land on a six.

The odds of rolling a nine given that at least one of the dice is a six equals the probabilty of rolling a nine given that the first die is a six which equals the odds of rolling a nine given that the second die is a six.

I think we can agree that the odds of two dice matching is one in six. Your logic would result in that probability being one in 11. (Just replace the nine with a twelve, and then apply it to ones to get two, twos to get four, etc.)

[QuikSand]

You can not solve these problems setting a constraint, then ignoring the constraint to obtain results, and then throwing out the results that do not match the defined constraint. This process is not mathmatically sound.

You can solve these problems that way, and statisticians the world around insist upon it. It is the only mathematically sound way of doing so.

The information provided in the constraint must be accounted for in the randomization process.

By making the "randominzation process" non-random? How does this possibly comply with the concept of these puzzles?

I realize that these questions/answers seem like they splitting hairs, but I'm genuinely on the fence right now between believing that (a) you have a valid, if unusual, interpretation of this kind of probability puzzle which has its own branch of defensible solutions; or (b) you're just flat out wrong, and the source just happens to be in the setup of the problem.

To get me toward (a) instead of (b), I think I'd need you to convince me why I can't use my "two-card seed" assumption in the flush puzzle, when we talk about the hands with at least one ace, but why we must use your "one-card seed" assumption. How on earth can you generalize something like that, once the problem gets more complicated than five cards and one suit? (which it obviously can)

[Grunion]

To get me toward (a) instead of (b), I think I'd need you to convince me why I can't use my "two-card seed" assumption in the flush puzzle, when we talk about the hands with at least one ace, but why we must use your "one-card seed" assumption. How on earth can you generalize something like that, once the problem gets more complicated than five cards and one suit? (which it obviously can)

Without getting into the calculation (if I need to I will), we agree on your calculation on all other subsets. If you use the two seed method on those, you would arrive at a different result than the ones which we unilaterally agree upon.

[wignasty]

Damn. I have read over about 1/4 of this debate, and I have come to a preliminary conclusion.

You are both right.

I promise to finish reading all the info, but as I see it right now, you aren't really debating the answer to the problems. It seems as though you are debating the "correct" way of looking at it. I see one of you looking at it from a mathematical point of view, while the other is looking at it from a logical point of view. I have also noticd that you both seem to flip sides a few times. It's funny how you never seem to be on the same side at any time, though. This intersting exchange has brought up another puzzle.

1. wignasty, QuikSand and Grunion each have a die.
2. They are each behind a seperate curtain, so they can't see eachother.
3. They each roll their die 1 time.
4. wignasty states that he rolled a 5.
5. QuikSand and Grunion explain the probability of getting a total of 13 between the 3 dice, without saying what they each rolled.
6. Neither QuikSand or Grunion can leave or eat until they both come to the same conclusion.

Q1 - What is the probability that they will ever agree on what each of them rolled.
Q2 - What are the odds that one of them will die of starvation, and then will the other one cheat and look behind the curtain.
Q3 - What are the odds that wignasty lied about his die, because he was afraid "The Man" could use that information against him.

wignasty

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Making a better America..... through paranoia

[QuikSand]

Without getting into the calculation (if I need to I will), we agree on your calculation on all other subsets. If you use the two seed method on those, you would arrive at a different result than the ones which we unilaterally agree upon.

I believe that we have agreed on the total number of hands that include at least one ace.

What we have not yet agreed upon is within that total number of hands, what proportion of them are flushes? Or what proportion have more than one ace? or how any have exactly three aces?

I believe that the puzzle, as originally stated, asks about that set of hands, and makes a clear assumption that each of the hands within that set is equally represented. I would answer all the above questions by making the original assumption that we're examinning the randomly-created set, with each hand equally represented, and the calculations are absolutely clear.

You believe that the puzzle asks about that set of hands, and presumes that we know the non-random method that was used to generate them, which (as it turns out) results in some hands being represented more heavily than others. Based on your assumption of the generation method (which is not the only one which complies completely with the observable characteristics of the set of hands) you then weight certan hands more than others, to calculate a different answer to each of the subordinate questions.

Theoretically, someone else could come along, and using your idea that we are enabled to "re-build" the process used to generate the hands, they could use my "two-card seed" method to create the hands. This method, which is every bit as logially valid as yours, would result in a completely different set of answers to those questions. (what is the flush probability? what is the probability that the hand has more than one ace? what is the pobability that the hand has exactly three aces? etc.)

How can you argue that your anwer is "correct" when this hypothetical third person follows your exact logical reasoning, then comes to a fork in the road and chooses the other option (of the two I have thought of... there may be an infinite number of methods) and generates an entirely different distribution of hands, and therefore a completely different answer to the puzzle? (He'll say that the set D is less likely to be a flush, I'm guessing)

---

So, on your point-- we've agreed how many different hands exist that contain at least one ace. This admittedly took some doing.

The essential question upon which we still disagree is: given the fact that we know thos hand has at least one ace (and no other facts - there is nothing else stated in the original problem) what is the likelihood that the hand is a flush?

My Method

Using a neutral calculation of a random distribution of all the possible hands that fulfill the stated observation, I get 0.223%.

I'm comforted by the fact that it is impossible to use my logical reasoning here, do the math correctly, and come up with any other result.

Your Method

Using a biased calculation of a non-random distribution of all the same possible hands, granting more weight to some than others by virtue of a selection process that you decided was appropriate, you get 0.198%

Another guy uses your method

The other guy agrees with us on the count of total hands uses your logic, but chooses a different selection process (the two-card seed) and gets a different non-random distribution than yours, which grants different uneven weights to certain hands than your uneven weights. He then calculates the flush probability to be another number, perhaps 0.15% (I have not done the math)

My question is this: what is the airtight argument that your selection of your preferred (but admittedly non-random) "hand generation method" is correct, and that the other guy's method is incorrect?

And what happens when we have a more complicated problem, where there are an intuitively obvious and much larger number of methods thatcould be used to satisfy the observed results? How do we know how to choose the right one?

[This message has been edited by QuikSand (edited 12-22-2000).]

[QuikSand]

Grunion, I may be confused about one thing here. Do you believe that this non-random selection process:

1) pick card #1 from the 4-card ace stack
2) pick card #2 from the 48-card non-ace stack
3) pick cards 3-5 from the re-shuffled remaining 50 cards

Do you argue that the set of possible hands resulting from this selection process (setting aside their various probabilities of occurring) is in some fashion different from the set of possible hands represented by the statement:

All hands which include at least one ace


If you believe there to be a difference in these two sets, then I understand your difficulty in following my arguments about the "two-card seed" option. If that's the case, I urge you to reconsider that belief-- I don't think you'll find it to be substantiated.

[QuikSand]

Just in case the horse isn't "dead enough" yet...

Back to the "flush probability of hands with one or more aces" puzzle. I have dreamed up a few more things:

- - -

Method #1:

-Draw card #1 from the 4-card pile of aces
-Draw cards 2-5 from the remaining 51 reshuffled cards

Method #2:

-Draw card #1 from the 4-card pile of aces
-Draw card #2 from the 48-card pile of non-aces
-Draw cards 3-5 from the remaining 50 reshuffled cards

Method #3:

-Draw cards 1-4 from the shuffled 52-card deck
Look at cards 1-4
-If they contain at least one ace, draw card #5 from the remaining shuffled 48
-If they don't contain atleast one ace, pull the four aces from the remainign cards, shuffle them, and draw card 5 from that four-card stack

Methods #4, #5, #6

Various combinations along the same lines as #3-- seeding the ace in as card #4, #3, or #2 if a check of the cards in the hand thus far does not reveal an ace, then drawing the rest of the hands randomly from the reshuffled deck.

- - -

Okay, here are six different methods that might have been used to generate any of the possible five card hands that contain at least one ace. None of these methods will generate a five card ghand that does not meet this observed requirement.

When the original puzzle asks:

Rank the following sets of 5-card poker hands in descending order of their likelihood of being a flush:

(A) All 5 card hands
(B) Hands whose first card is an ace
(C) Hands whose first card is the ace of spades
(D) Hands with at least one ace
(E) Hands with the ace of spades

It seems clear that you (Grunion) are "assuming" that the "set of hands" labeled D was generated by a particular method designed to generate only that set of possible outcomes. (I believe I'm articulating your position properly here)

My question is: how did you know which of these six methods was used? Each of them will in every single case generate a hand that satisfies the requirement stated. As an added bonus, it's also true that every single hand that does satisfy that requirement can be generated by each and every one of these six different selection methods.

For your analysis, you chose selection method #1 from above, and you then worked through an internally consistent set of calculations to determine that the likelihood of that set of hands being a flush was exactly and indisputably 0.198%.

How is it, from the statement of the puzzle (quoted verbatim above) that the solver is to know to use "selection method" #1 from above, and to use that particular non-random method to determine the likely characteristics of that set? Why not use selection method number 2, or number 3, or any of the others that I have chosen above? (Which is only a sampling of what may well be an infinite number of selection methods that generate the entire set of hands with aces)

Mathematics demands certain answers to properly stated problems. This puzzle seems to have two different interpretations-- at least that's the synthesis of our collective argument.

My interpretation (that the set is evenly distributed) leads to a clear result, with absolutely no dispute about its veracity. There is no judgment involved, and there is only one correct answer.

Your interpretation (that we need to divine some implicit "selection method") leads to a potentially infinite number (but certainly a multiple) of options, each of which might lead to a different calculated result.

How can this be so? Is there some guideline that a puzzle-solver should employ when deciding what "selection method" might have been used to build this nonrandom group of hands? Why #1 over #2 or #3? How do we expand your solution to generally cover all similar cases? (This is, of course, a snap for my solution)

[Grunion]

QuikSand,

Get back to me on my dice reponse (8:47 AM), I believe its extremely relevant. I'm just starting to wade through your recent posts now.

[Grunion]

I believe that the puzzle, as originally stated, asks about that set of hands, and makes a clear assumption that each of the hands within that set is equally represented. I would answer all the above questions by making the original assumption that we're examinning the randomly-created set, with each hand equally represented, and the calculations are absolutely clear.

No, I do not believe that it does. In order to achieve that scenario, the condition would be stated as follows.

What are the odds of drawing a flush from five random cards, given that all hands not containing an ace are to be discarded.

Which is not identical to:

What are the odds of drawing a flush given that your hand contains at least one ace.

The constraint effects the overall probability of obtaining a flush. Knowing that you have an ace makes the probability of certain outcomes more likely than others, which should be fairly self evident. Your methods to not refine the process enough (which is quite ironic, and fairly non-intuitive, being that I am the one using the less complex equations).

Your method says: either the event happens, or it doesn't, and if the event happens there is an equal probability for each particular combination (of dice, pills, cards) which is possible to occur.

Which I think closely parallels a statement such as this:
2 green balls one yellow ball, one pick. The pick can be green or it can be yellow, therefore both have an equal chance of occuring.

[QuikSand]

Will do.

In response to my comments:

[quote]In the 10/11 likely event that the two dice are different, the chances are 1/5 that the "other die" is a three.

In the 1/11 likely event that the two dice are both sixes, the chances are 0 that the "other die" is a three.

The combined chances of the two dice adding to nine are (10/11 * 1/5) + (1/11 * 0) = 10/55 = 18.18%[quote]

Grunion replies:

Definately not true, the probability of two dice being different will always be 5/6. (unless, of course, we have a given such as both dice are the same or both dice are different). This should be fairly intuitive.

Given that at least one die is a six, we have one known value and one unknown value. We do not know which die is the six, but that is irrelevant, order does not matter in this case. (Which is defined in combinations, which we have been using excusively in our dice discussion.) What you are stating implies that the possibility of rolling a three on a given die increases because a separate die happens to land on a six.

The odds of rolling a nine given that at least one of the dice is a six equals the probabilty of rolling a nine given that the first die is a six which equals the odds of rolling a nine given that the second die is a six.

I think we can agree that the odds of two dice matching is one in six. Your logic would
result in that probability being one in 11. (Just replace the nine with a twelve, and then apply it to ones to get two, twos to get four, etc.)

This is just another outpouring of our "difference of opinion" of how to react to after-the-fact observations about random events.

Your entire discussion rests on the belief that the die-rolling was not a random event, but instead was some sort of non-random event built to satisfy the conditions that were observed.

My statement (which conforms to univerally-held probability concepts) is that your puzzle:

I'll make this one quick, for I know what we will agree on:

2 Dice, chance to roll a nine = 4/36 or 11.1%

Now, given that at least one die is a 6, the chance to roll a nine becomes...

(I snipped out what follows immediately thereafter, because it's your solution)

...this puzzle simply must be understood to have the following meaning:

Part I:
We roll two fair dice.
We inspect the results.
We calculate the likeihood that the sum is nine.

Part II:
If, upon inspection of the dice after the process in Part I, we see that the actual result is within the subset of all originally-possible results described by "at least oen die shows a six" then we can "back out" the calculation of the probability taht the two dies show a sum of nine.

This is nothing more than you re-stating your interpretation of this sort fo puzzle, and me telling you it doesn't make any sense. And, apparently, vice versa.

Just an idle question-- I don't mean anything prejudicial about it. Are you either a non-native American, or did you perhaps study in another country? I'm wondering if there might be some difference in international conventions with this sort of thing... (which I still find hard to buy given the variability within your arguments, but I'm trying to find some benefit of the doubt)

- - -

Regardless, this brings us to your statement:

Definately not true, the probability of two dice being different will always be 5/6.

Well, now we can both agree that isn't true. I can set down two dice with each side showing a six... and there won't be a 5/6 chance that those two dice are different.

It's all a matter of what knowledge we claim to have about these dice.

I claim (again, in keeping with the universally-held conventions of probability study) that we know the following:

-the two dice were rolled randomly

-the result happened to be one of 11 equally likely combinations that show at least one six

With that knowledge about these two dice, it decidedly untrue that the dice have a 5/6 chance of being different. They have an exactly 10/11 chance of being different.

I suggest that you'll dispute my claim to the two bits of knowledge stated above (back to our semantic differences), but you can't claim that if those things are true that my calculation is off.

[Grunion]

Grunion, I may be confused about one thing here. Do you believe that this non-random selection process:
1) pick card #1 from the 4-card ace stack
2) pick card #2 from the 48-card non-ace stack
3) pick cards 3-5 from the re-shuffled remaining 50 cards

Do you argue that the set of possible hands resulting from this selection process (setting aside their various probabilities of occurring) is in some fashion different from the set of possible hands represented by the statement:

All hands which include at least one ace


If you believe there to be a difference in these two sets, then I understand your difficulty in following my arguments about the "two-card seed" option. If that's the case, I urge you to reconsider that belief-- I don't think you'll find it to be substantiated.

In the context of our discussion, no the above is not a random process, for the reason that we have arbitrarily reduced one of our selection pools to 48, when it clearly should be 51.
In a general context, this could be be considered a random process, if it matched the constraint.
Such as:
At least one card is an ace, and one of the remaining four cards can not be considered to be an ace. (And yes I realize we do know this in the original case D, but it is a function of our relatively complex set of four cards of each value for 5 places, not by some additonal constraint placed on the problem.) If you are not buying into the above, consider a similar question, but either only drawing four cards versus five, or introducing a fifth suit. We could attempt to wade through the details of this particular scenario, but I think you'll agree that we'll end up spending a lot of time hashing through the complexity of the problem, and not focusing on our clearly defined difference of opinion.

[QuikSand]

...the condition would be stated as follows.

What are the odds of drawing a flush from five random cards, given that all hands not
containing an ace are to be discarded.

Which is not identical to:

What are the odds of drawing a flush given that your hand contains at least one ace.

And this, again, is where we differ. I strongly believe that an overwhelming majority of any "reliable" sources (textbooks, academics, whomever) in such matters would agree that the two questions you juxtapose above are, indeed, identical.

[QuikSand]

In your 11:11am post, when you answer:

no the above is not a random process

You are not responding to my question. We agree these are not random processes. Both your process (seed an ace, randomly draw the rest) and all the other processes I describe elsewhere are definitely non-random processes. They all skew the distribution within the universe of possible outcomes.

However, my argument is that they all have the exact same universe of possible outcomes. They all can generate each and ever hand with an ace, and they all can generate no other type of hand. I hope you follow my point here.

My poit is since you find it acceptable to assume one particular non-random method, and you justify it on the grounds that it generates only the hands which conform to the condition in the puzzle... why stop with that selection method? I've found a bunch of other ones-- how did you choose, and how can you defend that as a unique correct answer to the exclusion of some other answer?

[QuikSand]

Which I think closely parallels a statement such as this:
2 green balls one yellow ball, one pick. The pick can be green or it can be yellow,
therefore both have an equal chance of occuring.

Give me a break.

[QuikSand]

While I don't have a proof of this, I strongly suspect that using your logic and following it to the fullest extrame, the only possible answer to any of the puzzles we've discussed is "there is insufficient information int he puzzle to answer it."

Since you decide that all of these random events with defined probabilities are going to be re-visited and altered into some sort of non-random systems that are rigged to cpmply with our ex post facto observations... then I submit that there will exist more than one method of generating the results. Since there is ny systematic way to choose one among many (or infinitely many) possible non-ranodm systems, I believe you paint yourself into a corner on any one of these puzzles.

So, when you are confronted with a puzzle like:

Two dice were rolled.
At least one came up a six.
What is the chance that they add to nine?

You cannot reach the correct answer of 2/11, but instead you must respond by saying: "To answer that, I need to know the details of the system that was used to generate this result. There are too many systems that generate two dice with at least one six, and I cannot chose among them, and therefore cannot answer the question."

It's nonsensical, but I think it's the only outcome of your way of thinking here... the more I consider it, the more convinced I become that your entire logic leads to this-- at least in many cases, if not every case.

[Grunion]

Method #1:

-Draw card #1 from the 4-card pile of aces
-Draw cards 2-5 from the remaining 51 reshuffled cards

Method #2:

-Draw card #1 from the 4-card pile of aces
-Draw card #2 from the 48-card pile of non-aces
-Draw cards 3-5 from the remaining 50 reshuffled cards

Method #3:

-Draw cards 1-4 from the shuffled 52-card deck
Look at cards 1-4
-If they contain at least one ace, draw card #5 from the remaining shuffled 48
-If they don't contain atleast one ace, pull the four aces from the remainign cards, shuffle them, and draw card 5 from that four-card stack

Method 1 is random, with respect to case B, and since order doesn't matter to case D as well. Although the equation ends up being the same for cases a, c and e. This is not a suitable method of modeling those scenarios. This would seem to validate my contention that no matter how much or how little we know a particular card, it is of no added significance in calculating flush probabilities. Order, starting suit, starting card value, and location of known card do not change the probabilities, provided that all known information pertains to the same card.

Method 2 is incorrect in all stated cases. The other part of my initial contention is that we need to know something about more than one card for there to be any chance of obtaining significant information. We now have knowledge about two cards, and this knowledge needs to be accounted for in our calculation. I believe that you are presenting method two not because you agree with it, but because you think that I find it a valid concept. I do not and never have.

Method 3 is also non valid, because we also have additional information, this time it is being introduced in the middle of the process. We then make a decision base on the information we know about four cards. And to bounce back to one of my main contentions, one we have information about more than one card, there is additional information which can alter our flush probability (in this case full knowledge of four complete cards). I'm not about to do the math for this, because it involves a decision tree.

I am very perplexed that you are able to (correctly) dismiss methods as invalid
because the result does not match those for cases A, B, C & E; yet have been unwavering in your assessment of D, even though it falls into the same category.

------------------
Those who don't learn history are doomed to repeat it.

[Grunion]

And this, again, is where we differ. I strongly believe that an overwhelming majority of any "reliable" sources (textbooks, academics, whomever) in such matters would agree that the two questions you juxtapose above are, indeed, identical.

Sorry, but I have to call you out on this. Its happened more than once. If you are referring to statistical textbooks, etc. and want to use the info you dig up I strongly encourage you to do so. However, its improper and unfair to make an opinionated statement (I stongly believe....) to draw a conclusion about proper statistical analysis. I stongly believe the reverse is true, because I believe I am correct. However it is still not appropriate for me to argue that the academic community is in my corner.

[QuikSand]

Here is a new, and rather simple, puzzle which might get to my point more clearly.

- - -

Let's go back to Fred, and his potentially deadly pills.

There are now six pills from which Fred will draw.

2 are blue
2 are red
2 are green

Fred will draw and ingest two pills, drawing randomly from a container including those remaining.

If he draws and ingests one blue and one red pill, he will die.

- - -

That is the basic setup. That is all we know in advance. That information is not allowed to change (as much as you would prefer that it did later).

First, calculate the likelihood that Fred will die.

We would agree (I suspect) that this is easy:
-there are 6 x 5 = 30 different pill combinations
-there are 4 x 2 = 8 deadly combiantions

The likelihood of Fred dying is 8 / 30 = ~23%

I'm pretty sure we'll agree here (unless I just made a bonehead mistake in my math - I dont think this reveals any of our differences is my point).

- - -

Now, let's add some posterior information:

Fred did not select both a green and a red pill.

What, now, is the likelihood that Fred will die?

For me, this is fairly easy-- I'll spell it out.

I had 30 possible outcomes, which were equally likely to begin with. Now, I know that 8 of those outcomes did not happen, based on the posterior knowledge that we have gained. Therefore, I am left with the remaining 22 possible outcomes, which remain equally likely (with respect to one another, though Bayes' Theorem properly applied tells us that they each now represent a larger probability of having been the even outcome).

Of the 8 deadly combinations, they all remain as possible among the group of 22 possible outcomes. So, there are 8 of the 22 equally likely outcomes that result in Fred's death.

I calculate Fred's likelihood of dying with this set of knowledge as being 8 / 22 = ~36%.

Now, how do you solve this one? You seem to need a "random" method of ensuring that the final outcome will not contain both a red and a green pill. I built this problem because (at least to me) it isn't intuitively obvious how one might do that. Do rush in before he dtarts drawing pills and flip a coin, and take out both the greens if it's heads and take out both the reds if it's tails?

I'm not sure how you determine the "selection method" here in any ambiguous way.

What's your answer to this puzzle? Can you even reach any answer?

[This message has been edited by QuikSand (edited 12-22-2000).]

[QuikSand]

Sorry, but I have to call you out on this.

Point taken. I have edited another such remark out of my last post (the red/blue/green pill puzzle) and I won't use them again.

[Grunion]

While I don't have a proof of this, I strongly suspect that using your logic and following it to the fullest extrame, the only possible answer to any of the puzzles we've discussed is "there is insufficient information int he puzzle to answer it."

Essentially, your are correct in my assessment. But it can be answered, we just can not draw pertinent conclusions from the question posed. Information on a second card is required to alter the probabilities,
Such as,
Ace of Hearts and four of spades
or ace of clubs and seven of clubs.

Interestingly enough, while knowing that we have the ace and seven of clubs helps us (because we have sucessfully restricted our pool) knowing that we have at least two clubs does not (as every possible hand will have at least two matching cards).

So, we need info pertaining to a minimum of two cards for us to obtain additional conclusions.

Having information on two cards may or may not help us, depending on how this information pertains to the characteristics of our deck and our flush subset.

Knowing we have at least one 4 and one seven does not help, but knowing we have two kings does.

[Grunion]

Also,

You are still referencing ex post facto observations, without addressing my claim that we are merely dealing with posterior confirmation, not posterior information.

I contend that you are dealing with posterior information. You wait until the process is complete to "throw out" results (i.e. two blue pills) rather than correctly applying the constraint of a given at the beginning of the process. In all examples, my only possible results agree with the given. In yours, a decision needs to be made on the result to determine if it agrees with the given. This is what is skewing our results.

[Grunion]

The likelihood of Fred dying is 8 / 30 = ~23%

agreed

Fred did not select both a green and a red pill.

What, now, is the likelihood that Fred will die?

Remembering that order is not important:
P(1) has a 1/3 chance of being red, a 1/3 chance of being green and a 1/3 chance of being blue.

If P(1) is red, P(2) has a 2/3 chance of being blue and a 1/3 chance of being red (as a red/green combo is impossible)

If P(1) is blue, P(2) has a 2/5 chance of being red, a 2/5 chance of being green and a 1/5 chance of being green.

If P(1) is green, P(2) has a 1/3 chance of being green and a 2/3 chance of being blue (as a red/green combo is impossible)

P(red/blue) = .33(.67)+.33(.4)+ .33(0)= 35.5%

[QuikSand]

Grunion, I confess that I have not been consulting a textbook or other resource to guide my terminology. I may well have made loose use of certain terms like ex post facto and posterior information, and may have used the terms in a context that is not in keeping with some convention.

However, I think the context of all my discussions have been clear, and I think that we've successfully isolated our differences-- which revolve around the proper use of the information, and not really its label.

- - -

However, you're right when you characterize my method as being (paraphrasing): we let the random events occer, and just throw out all the outcomes that don't match the
observed condition.

That's exactly how I do it, and I continue to assert that this is the proper way to approach any such problem, unless the "condition" referenced is not an observation of outcomes, but is instead clearly built into the set of givens.

I argue that in each of the puzzles we've discussed, these various observations (the hand has at least one ace, Fred drew at least one red pill, at least one die shows a six, Fred did not draw both a red and a green pill) are all of the exact same character-- they are all observations made after the random event- not conditions around which one should build nonrandom events.

[QuikSand]

With my red/green/blue pill puzzle - well done. I neglected to make the puzzle complex enough to avoid a simple brute force calculation, which would require you to depart into your theory of "selection process." You win there.

(Though, isn't it at least a little troubling to you that my method of "throwing out" the random outcomes that don't mesh with the observed condition generates the correct answer, which you also got by intuitive brute force? You've suggested that my methods are somehow laughable, yet they're right here. Of course, that proves nothing-- it could have been a coincidence, or a by-product of the setup somehow)

Note: I've since revised my position on your answer to this puzzle, and I now disagree with you. I was temporarily fooled that your method (which, of course differs from mine) resulted in an answer which was within a rounding error of mine.

- - -

I'll try again to illustrate my point.

Let's say that my original flush puzzle contained an option F that read:

F - hands that contain at least one ace or at least one queen, have at least two face cards, and do not contain the jack of diamonds

(or some other nonsensically complicated description)

If the puzzle asks you to determine the likeilhood of this set of hands being a flush... how would you go about calculating that?

My method is simple: I calculate the total number of hands that meet the restrictions, and then I calculate the total number of hands that meet the restrictions and happen to be flushes. I divide the latter into the former, and have the correct answer.

Since you reject my notion that each possible hand must be equally-weighted, I presume that you would have a different approach to a problem like this.

Consistent with your solution to the original flush puzzle (specifically the original case D) you would be required to come up with some hand-selection methodology that would be certain to generate the exact subset of hands described by the set F. Am I right-- that's how you would approach such a puzzle?

Can we generalize this flush puzzle to discuss our various aproaches?

My approach is for any given set of conditions x, you calculate the total number of hands that meet x, the total number of flushes that meet x, divide the latter into the former and you have the correct "flush probability" of the set defined by x.

Can you generalize? Do you have any way to solve a problem like the hypothectical F I describe above (or something even more tedious and complicated)?

That puzzle should be solvable, shouldn't it?

[This message has been edited by QuikSand (edited 12-22-2000).]

[This message has been edited by QuikSand (edited 12-22-2000).]

[Grunion]

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).

The hand listed above should be included in both sets. This is also a core difference in our conceptual disagreement.

[QuikSand]

I'm guessing that last post was just a hiccup-- I think we're pretty far beyond that stage, and have more closely zeroed in on our differences.

I think we now both understand the precise differences between the way you're calculating things and the way that I am-- the card-counting stuff was one vehicle that got us this far.

[QuikSand]

Remembering that order is not important:

P(1) has a 1/3 chance of being red, a 1/3 chnce of being green and a 1/3 chance of being blue.

If P(1) is red, P(2) has a 2/3 chance of being blue and a 1/3 chance of being red (as a red/green combo is impossible)

If P(1) is blue, P(2) has a 2/5 chance of being red, a 2/5 chance of being green and a
1/5 chance of being green.

If P(1) is green, P(2) has a 1/3 chance of being green and a 2/3 chance of being blue (as a red/green combo is impossible)

P(red/blue) = .33(.67)+.33(.4)+ .33(0)= 35.5%

Oops. I got fooled by the coincidence that your answer happened to round off to the same percent as mine.

I, of course, disagree with your approach to this puzzle. You have cleverly come up with a "pill-draw generation system" which does result in only the outcomes specified, and you calculated your probabilittes based on that. The nonrandom "system" you use seems to be "draw one random pill, and then if the second pill causes an invalid combo, draw a different second pill."

Your answer breaks down as 48/135 = 35.55%

I would argue again that the correct answer is 8/22 = 36.36%

I disagree with your very first assertion-- that P(1) has a 1/3 chance of being each color. I disagree, and would argue that of the 22 combinations that could have generated the "non red and green" combo we have observed, that 10 begin with a blue pill, 6 with a red, and 6 with a green. I would, of course, argue that P(1) has a 10/22 chance of being blue, a 6/22 chance of being red, and a 6/22 chace of being green.


Regrettably, my puzzle was not sufficiently complex to require you to have much difficulty in coming up with a "pill-draw generation system" that ends up with the same distribution of outcomes. You did, and when you worked through the math, you came up with an answer consistent with your logic on all the other puzzles.

We again disagree on the proper use of the after-the-event observed information. It's just the same story over again.

I think my puzzle above (with the ridiculously complicated information about the hands of cards) is a better example of the meaningful differences between your approach and mine.

Sorry for this too-simple puzzle getting us off-track, and proving to just be redundant.

[This message has been edited by QuikSand (edited 12-22-2000).]

[Grunion]

Unfortunately, I believe our answers are similar because of coinsidence, not because both are correct. I come to this conclusion because our denominators, factor differently. (Yours are 2 and 11 and mine are 3 and 5, with repetitions).

This brings me to an obsevation (as much as I would like to draw a definitive conclusion, I am not well educated enough on the subject to do so).

It would stand to reason, when dealing with 52 cards, we should not have to deal with prime numbers in the numerator or denominator greater than 52. (or 36 for the dice, or 6 for the pills). Yet pretty much all of your equations introduce primes greater than the original set. I've long since thrown out my scratch paper I was working with yesterday, but your case d total combination (106 million and change) broke down into many prime numbers, one of which being 2309 (if I recall correctly). This seems suspect, but as I said, I can not make a rock solid arguement that this should not be.

F - hands that contain at least one ace or at least one queen, have at least two face cards, and do not contain the jack of diamonds

Yes, to calculate an answer using my method would be tedious, but that does not make it inappropriate.

Addressing the first condition, (and I choose this condition not because it is sequentially first, but because it contains an or statement) at least one ace or one queen, I would make the following statement.

There is an equal chance that a hand that has at least one ace in it will have a queen in it as well as that a hand that has at least one queen in it will have an ace in it as well. I could prove this, but since there are the same number of aces and queens in the deck I believe it is unneccesary.

Therefore, there is a 1/2 chance that case one of the or statement will drive the remainder of the problem, and a 1/2 chance that case two will drive the remainder of the problem.

This is valid, as long as my model for one given allows for the inclusion of the other condition down the line. And yes, this will result in particular duplicate answers, which I have contended all along to be necessary.

Now, since the cases are not identical with respect to the constraints (the queen is a subset of the face cards, while the ace is not) The probabilities of both cases would have to be analyzed separately, as in the six pill problem.

There is no way I'm going to crank this thing out, but since case 2 is more complex, lets look at that case closer.

We know we have a queen, and therefore a face card, so we now must model four cards such that at least one of the four is a face card, and none of the four is a jack of diamonds, and that they combine to make a flush (working in probabilities, not total hands). Additionally, we know that if the given queen is the queen of diamonds, we can not have a flush, so only 3/4 of the queens work.

Anyway, I hope you were not expecting me to solve this, but I am confident that I could.
The main point is, with or statements, you can assign one as given as long as your model allows for the inclusion of the other condition (as opposed to either/or statements, which are exclusive)
Then you have to multiply the percentage of case one occuring followed by the percentage of pulling a flush given case 1 and add it to the percentage of case 2 occuring multiplied by the percentage of flush for case 2. And, of course, each case would have subcases which would have to follow the same logic.
And once again, it is OK to have combinations of hands in case one which would also be valid for case two. Each case describes an applied constraint, not the presence of a card. (although, in this particular example, by the nature of the constraint, we know that there is an ace in case 1 and we know there is a queen in case 2.

Or:
We have at least one ace or queen.
In case 1 we know we have an ace, we may have a queen.
In case 2 we know we have a queen we may have an ace.

Also,
We would not need to break the problem down into cases if the two face cards were not an added condition. As far as flushes are concerned, aces are as good as queens.

This leads me to an interesting question:

Case G: Hands that contain at least one ace or one queen.

I would contend that the answer is 0.198%, I assume your answer would be different, most likely slightly less than your answer for case D. The odds would get lower and lower antil you reached hands that contain at least one card greater than a two, as having five cards is impossible, at that point your answer would become 0.198% as well.

[QuikSand]

Responding to my post:

Method #1:

-Draw card #1 from the 4-card pile of aces
-Draw cards 2-5 from the remaining 51 reshuffled cards

Method #2:

-Draw card #1 from the 4-card pile of aces
-Draw card #2 from the 48-card pile of non-aces
-Draw cards 3-5 from the remaining 50 reshuffled cards

Method #3:

-Draw cards 1-4 from the shuffled 52-card deck
Look at cards 1-4
-If they contain at least one ace, draw card #5 from the remaining shuffled 48
-If they don't contain atleast one ace, pull the four aces from the remainign cards, shuffle them, and draw card 5 from that four-card stack

Grunion replied:

Method 1 is random, with respect to case B, and since order doesn't matter to case D as well. Although the equation ends up being the same for cases a, c and e. This is not a suitable method of modeling those scenarios. This would seem to validate my contention that no matter how much or how little we know a particular card, it is of no added significance in calculating flush probabilities. Order, starting suit, starting card value, and location of known card do not change the probabilities, provided that all known information pertains to the same card.

Method 2 is incorrect in all stated cases. The other part of my initial contention is that we need to know something about more than one card for there to be any chance of obtaining significant information. We now have knowledge about two cards, and this knowledge needs to be accounted for in our calculation. I believe that you are presenting method two not because you agree with it, but because you think that I find it a valid concept. I do not and never have.

Method 3 is also non valid, because we also have additional information, this time it is being introduced in the middle of the process. We then make a decision base on the information we know about four cards. And to bounce back to one of my main contentions, one we have information about more than one card, there is additional information which can alter our flush probability (in this case full knowledge of four complete cards). I'm not about to do the math for this, because it involves a decision tree.

I am very perplexed that you are able to (correctly) dismiss methods as invalid
because the result does not match those for cases A, B, C & E; yet have been unwavering in your assessment of D, even though it falls into the same category.

First of all, how is it that you can fairly describe "seeding" the first card of the hand (Method #1) as an ace to be "random?" You've already acknowledged that it will cause there to be a higher density of multi-ace hands than the method I endorse, which has each hand weighted with absolutely equal probability. You skew the likelihood of given hands, making some more likely then others-- that is inerently non-random.

I add Method 2 and the others not because I believe them to be "valid" but because I believe them to be of the same level of validity as Method #1 (which I also deem to be "invalid" using your terms). My point is that any of these selection methods could have been employed by the dealer-- how can you choose which one?

I think you may not follow my use of these "method" descriptions. I am discusing a process that would be used by a third party to prepare the "set of hands" described in the original puzzle. The dealer follows the rules set forth in the various methods in order to prepare the hands that conform to the conditions specified. The fact that at some intermediate point the dealer "knows" what certain cards are does not add anything to our knowledge, which seems to be the principal reason you reject my alternative methods.

---

Again, I point back to my last discussion of the generalized flush puzzle.

If asked the question:

"What is the probability that this set of hands meeting condition x will be a flush?"

I have a simple decision process.

Count D = total number of hands meeting X
Count N = total number of flush hands meeting X
Calculate N / D = probability of flush within that set of hands

- - -

In the generic, the best summary I can put on your solution process is something like this:

Divine or select a hand-creation process (presumably but not necessarily non random) that would result in exactly the hands that meet the description X.

By following the probabilities of the divined/selected process, calculate the chance that the result would be a flush.

- - -

Is that a fair statement of our two approaches?

Or do you put more conditions on the process used to determine the "hand selection" method? You accept what I label Method #1 as appropriate, but reject Method #2 and the others. Why is that?

What if I modified Method 2 to read:

Method #2:

-Draw card #1 from the 4-card pile of aces
-Draw card #2 from the 48-card pile of non-aces
-Draw cards 3-5 from the remaining 50 reshuffled cards
-Before looking, shuffle cards 2-5, to randomize the location of the guaranteed non-ace card


Does this make it better? (It shouldn't, of course) This hand-creation method simly translates to: we know the first card is an ace, and we know that there's one non-ace card somewhere in the hand.

Obviously, we haven't gained any more knowledge from that, right? We weren't going to have 5 aces anyway...

[Grunion]

(Though, isn't it at least a little troubling to you that my method of "throwing out" the random outcomes that don't mesh with the observed condition generates the correct answer, which you also got by intuitive brute force? You've suggested that my methods are somehow laughable, yet they're right here. Of course, that proves nothing-- it could have been a coincidence, or a by-product of the setup somehow)

One minute you essentially accept my calculation as correct (you did add the disclaimer though),and further evidence that your methods are correct because they agree.

I, of course, disagree with your approach to this puzzle. You have cleverly come up with a "pill-draw generation system" which does result in only the outcomes specified, and you calculated your probabilittes based on that. The nonrandom "system" you use seems to be "draw one random pill, and then if the second pill causes an invalid combo, draw a different second pill."

The next minute, you dismiss them completely, on the basis that they actually don't agree with yours, and then state that as further evidence that your process is correct. If you are assessing correctness based on the outcome of my result as compared to yours, we may as well quit right now.

Regrettably, my puzzle was not sufficiently complex to require you to have much difficulty in coming up with a "pill-draw generation system" that ends up with the same distribution of outcomes. You did, and when you worked through the math, you came up with an answer consistent with your logic on all the other puzzles.

Also, I don't think the creation of a scenario which A) has complexity greater than my statistical knowledge B) is so complex as to greatly increase my chance of error or C) would require massive amounts of time and effort to calculate would really help prove your point. We have pinned down our disagreement lets keep it simple.

If I wanted to be a pain in the ass, I would ask that you recalculate your solution for case d using only probabilities, not by calculating total combinations and dividing them into each other. (And please don't take that as a request or challenge, I wouldn't expect you to do it.)

[QuikSand]

(On the red/blue/green pill puzzle)

Sorry if I wasn't clear. On first blush, since I saw your answers was very close to mine, I presumed it to be correct. I then realized that it was not (in my judgemtn) correct. Rather than completely covering my tracks, I added my post-script right after my original commentary on your answer. Sorry if that caused any confusion-- that was the opposite of my intent.

Let me be clear here. I disagree with your method, and your answer for that puzzle. I believe our disagreement to be resulting from the exact same fundamental disagreement abotu how to solve this kinf of problem, and more specifically on how to "use" the information we receive about the outcome of random processes.

[QuikSand]

Also, I don't think the creation of a scenario which A) has complexity greater than my
statistical knowledge B) is so complex as to greatly increase my chance of error or C)
would require massive amounts of time and effort to calculate would really help prove
your point. We have pinned down our disagreement lets keep it simple.

If I wanted to be a pain in the ass, I would ask that you recalculate your solution for
case d using only probabilities, not by calculating total combinations and dividing them
into each other. (And please don't take that as a request or challenge, I wouldn't expect
you to do it.)

My point is not to exhaust your statistical knowledge, nor to "trip you up" into making an arithmetic error. I couldn't care less about either one.

My point with all this is to demonstrate that your process for solving these puzzles (replacing the completely random selection with a new non-random one of your choosing) is frought with critical problems.

I have tried to make the same point using our original problem, by using my "method #1" and "method #2" illustration, but I have failed to sufficiently articulate that point (or so it seems) and you continue to insist that there is something inherently "vaild" about your method and something that makes the other method(s) "invalid."

My point is that trying to divine some process of building the set of hands-- about which we know nothing except teh results-- is unsupportable. It invariably leads to making assumptions, choosing one method over another, and so forth. In my judgment, it leads to every such problem being impossible.

[Grunion]

First of all, how is it that you can fairly describe "seeding" the first card of the hand (Method #1) as an ace to be "random?" You've already acknowledged that it will cause there to be a higher density of multi-ace hands than the method I endorse, which has each hand weighted with absolutely equal probability. You skew the likelihood of given
hands, making some more likely then others-- that is inerently non-random.

Because I am not "seeding" an ace. (although the full calculation that I use will cross cancel itself so that the remaining calculation will mirror the "seeded ace" calculation.
So yes, the reduced calculation is identical to a seeded ace calc, which would seem to support my contention that we need info from more than one card.

The full calc is as follows.

Q. P(flush) given at least one ace.

(Commentary - given the constraint, we know that the hand will contain at least one ace. However that is all we know, so we must identify not only the possible cases in which a known ace can occur, but the probability of each case occuring. To be truly mathmatically appropriate, we should also account for the suit of the known ace, which is also unknown. We could do so either by creating four subcases for each case, or by creating 20 cases instead of five. I can assure you the same result would be achieved as below.)

Case 1 = The known ace is in the first position.
Case 2 = The known ace is in the second position.
Case 3 = The known ace is in the third position.
Case 4 = The known ace is in the fourth position.
Case 5 = The known ace is in the fifth condition.
P(case 1) = 1/5
P(case 2) = 1/5
P(case 3) = 1/5
P(case 4) = 1/5
P(case 5) = 1/5

Case 1:
P(flush) = 4/4*12/51*11/50*10/49*9/48
=0.198%
Case 2:
P(flush) = 48/51*1/4*11/50*10/49*9/48
=0.198%
Case 3:
P(flush) = 48/51*11/50*1/4*10/49*9/48
=0.198%
Case 4:
P(flush) = 48/51*11/50*10/49*1/4*9/48
=0.198%
Case 5:
P(flush) = 48/51*11/50*10/49*9/48*1/4
=0.198%

P(flush)=1/5(0.198)+1/5(0.198)+1/5(0.198)+1/5(0.198)+1/5(0.198)
=0.198%

As you can see, I have not seeded an ace. The above process gives us no better idea where the ace is than the original constraint posed.

Based on your earlier contentions, I do not believe that we have any disagreement on the calculation for each individual case.
I don't know for sure, but I believe that you have no problem with my assigning case 1 through 5 an equal probability of occuring.

And, I do not believe I described method one as merely random. I described it to be random to case B. We should have no difficulty convincing you of this since you agree it is a valid solution for case B. Technically, it should not be considered random for case D. However the full process to determine case D, which is as above but adding cases or subcases to deal with unknown suit, will result in a calculation which will cross cancel until it is the exact of the one used for B. So from a practical standpoint, method 1 does accurately model case d as well.

Which reduces our discussion to this:
It is appropriate to break down the question into separate cases in the first case?

[Grunion]

You skew the likelihood of given hands, making some more likely then others-- that is inerently non-random.

There are 6 ways to roll a seven with two dice, and ony one to roll a 12. Yet the results are fully random.

The equation:

P(x) = C(1)P(1)+C(2)P(2)

Which states the probability of x occuring is equal to the probability of case 1 occuring times the probability of x occuring given case 1 plus the probability of case 2 occuring times the probability of x occuring given case 2. Where C(1)+C(2) must equal 1.0
(Of course cases could be added so long as the addition of the probability of all cases equals 1.0)

The above equation can make some outcomes more likely than others, yet is valid (this equation is presented in my civil engineering handbook, which I paid $140 for, so it better be valid).

[QuikSand]

The equation:

P(x) = C(1)P(1)+C(2)P(2)

Which states the probability of x occuring is equal to the probability of case 1 occuring times the probability of x occuring given case 1 plus the probability of case 2 occuring times the probability of x occuring given case 2. Where C(1)+C(2) must equal 1.0
(Of course cases could be added so long as the addition of the probability of all cases equals 1.0)

The above equation can make some outcomes more likely than others, yet is valid (this equation is presented in my civil engineering handbook, which I paid $140 for, so it better be valid).

Right you are. It is, of course, possible for a system of different outcomes to have different probabilities and still be a "valid" system.

However, when we're given no set of conditions about the distribution of the original set of hands (as we were not in the original flush problem) I argue that it is invalid to use the conditions that are observed, and to then use any process to generate anything other than an even distribution of hands.

So, I'm of course right back to my model-- you take all the hands that exist, and eliminate the ones which don't meet the specificed condition(s). Then, look at how many from that set are flushes... and you have your flush probability. I realize you disagree with this, but it's the exact same model I would use to solve any problem of this general type.

[This message has been edited by QuikSand (edited 12-22-2000).]

[Vaj]

...the condition would be stated as follows.

What are the odds of drawing a flush from five random cards, given that all hands not containing an ace are to be discarded.

Which is not identical to:

What are the odds of drawing a flush given that your hand contains at least one ace.

As I see it, this is the core disagreement that problems involving pills and dice are obscuring. FWIW, my calculation on page 4 assumed the equivalence of these two statements, as I restricted the possible universe of randomly drawn hands to only those including at least one ace.

Perhaps we can attempt to calculate the probability of having 1, 2, 3, or 4 aces in hands generated by these two methods, and see how they compare.