OT - Five card flush puzzle

[QuikSand]

Warning, this puzzle might require some slightly more advanced mathematics than most of my previously posted puzzles, which have tended more toward more intuition than real math. I think this one is solvable with just intuition, but a true proof requires some bigger number-crunching.

This puzzle involves a standard, 52-card deck of playing cards, in four suits and ranks from 2 through ace. A flush is a hand that includes all five cards from the same suit, though they may be of any rank. If the ranks happen to fall in sequence, the hand is designated as a straight flush (4-5-6-7-8 for example) or a royal flush (10-J-Q-K-A only) but in either case, the hand still counts as a flush.

For purposes of this puzzle, all the probabilities are as they seem-- the hands described are the product of some purely random selection process-- there is no trickery involved, just pure probability.

Here's the puzzle:

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

I'll give you a (hopefully) helpful hint-- the ranking will not have five different tiers. There is at least one instance where two or more of the groups listed above have an identical likelihood of being a flush.

Enjoy!

[This message has been edited by QuikSand (edited 12-14-2000).]

[Passacaglia]

I think I know, but I don't want to post it, in case by some freak coincidence, I'm actually right.

------------------
"If we can put a man on the moon, we can grow grass indoors."

[Scarecrow]

Just by first glances, and not actually trying to figure it out statisticaly, I woud say:

A is the most likely
B & D are next with the same likelyhood
C & E are next with the same likelyhood


------------------
If I only had a brain

[Passacaglia]

Okay, now that someone has put up a guess different from mine, I'll say mine: I think they all have the same probablity.

------------------
"If we can put a man on the moon, we can grow grass indoors."

[QuikSand]

I can set your fears to rest, Passacaglia-- you didn't get the correct answer and spoil it for everyone. The correct answer is still out there...

[Passacaglia]

Okay...after a little more thought, and a little more math, my new guess is:

D has the highest porbability.
All others are less likely.

------------------
"If we can put a man on the moon, we can grow grass indoors."

[Vaj]

I'll go with D having the lowest probability, with A,B,C,and E having the same probability

[Celeval]

Looks to me like A, B, and C are equal probablility; while D is less likely than any of those, and E is least likely.

Yes?

Kevin

[WebEwbank]

I think that (D) has the lowest flush probability; the others are all equal and higher.

Reasoning is: presence of any one card has NO effect on "filling" (completing) your flush in the following four cards, since EVERY hand has a first card and it always has a suit.

The reason that (D) has a lower flush probability is that any hand with two aces, or more aces, is de facto not a flush. So within the set (D) you have many hands which can't flush and have eliminated many hands which do flush (any one without an ace). This is the one that needs math, but I don't have time right now to elaborate the proof.

Well, Q ?

[Ctown-Fan]

First (A) is the most probable becasue it includes all flush hands

Second (B) and (D) are equal and the next most probable because they could be the same hand but only include 4 posibilities of receiving an ace.

Third, (c) and (e) are equal becasue it could be the same flush hand with only one posibility for receiving the correct ace.

My stats book is locked away or I would attempt to do the math . . .actually, i think i'll dig it out and give it a shot

[This message has been edited by Ctown-Fan (edited 12-14-2000).]

[Dutch]

Originally posted by QS
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

Possible outcomes:

A: All flush combinations are still available
B: All flush combination are still available depending on the ace.
C: Only Spade Flush are still available
D: All Flush Combination are still available (depending on the ace).
E: Only Spade Flush are still available

Most Likely are A, B, and D.
Least Likely are C and E.



[This message has been edited by Dutch (edited 12-14-2000).]

[Passacaglia]

I think people are on the right track in finding out how many combinations of flushes are available. However, we also need to find out how many hands are available, to get the "denominator" in the probability. I doubt that I did that right with my most recent guess, since I'm tired, and in a terrible, bored, mood, but combinations of hands possible as well as combinations of flushes need to be found.

------------------
"If we can put a man on the moon, we can grow grass indoors."

[DukeRulesMAB]

Hmm...don't have time to give a well-researched answer, but here goes anyway.

The odds of getting a flush in general are:
12/51 * 11/50 * 10/49 * 9/48 = 33/16660

The reason that there are only 4 numbers in that equation is that it doesn't matter what the first card is; the odds are still the same.

In fact, because card order is random anyway, you can reorganize the last two hands such that the ace or ace of spades is picked up first.

Therefore, I'm going with the answer that all the situations are equally likely to land a flush.

Michael

[Bad-example]

Dutch's answer looks pretty convincing to me.

[Celeval]

Okay, here's how to do it:

A, B, and C are identical because the first card you draw has no effect on the probability. Take a deck, draw any card. Let's say it's the 3 of Hearts. The chance of getting a flush is exactly the same as it is before you drew that card; since no matter what card it is (any, any ace, or the ace of spades) the chance of the flush is the chance of drawing your next four cards to match that suit.

For D and E, split it into two pieces: that where you draw the required card first (any Ace, Ace of Spades), and that which you don't. For D and E, the probability is the same if you draw the card first; however if you do not draw the required card first (which you will not for 80% of the flushes you get), then the probability decreases.

D: If you draw a 2-K first, then you HAVE to draw the Ace of that suit later in the hand. Therefore, you have the same chance of remaining in flush until that last card you draw, where you have to draw a single card out of 48 (the ace) rather than any of the 8 cards you could have drawn were you going for a simple straight.

E: Same situation as D, except that you must draw a Spade as your first card; which drops the probability significantly.

The math to go with it:

            Drawn card in hand

          (Prob. of staying in suit)



CARD     1      2      3       4      5



A      52/52  11/51  10/50    9/49   8/48

B       4/4*  11/51  10/50    9/49   8/48

C       1/1^  11/51  10/50    9/49   8/48 

D

 a      4/52  11/51  10/50    9/49   8/48

 b     48/52   1/51  10/50    9/49   8/48

 c     48/52  10/51   1/50    9/49   8/48

 d     48/52  10/51   9/50    1/49   8/48

 e     48/52  10/51   9/50    8/49   1/48



E

 a      1/52  11/51  10/50    9/49   8/48

 b     11/52   1/51  10/50    9/49   8/48

 c     11/52  10/51   1/50    9/49   8/48

 d     11/52  10/51   9/50    1/49   8/48

 e     11/52  10/51   9/50    8/49   1/48

* - It's a given that the first card is an
ace, therefore there's a 4/4 probability
of getting that ace.

^ - It's a given that the first card is the
Ace of Spades, therefore there's a 1/1
probability of getting that card.

Calculation is left to the reader (I'm lazy), but I think this is right.

Kevin

[Celeval]

Should have looked at it more closely first. Bleah, I'm tired.

D 

a      4/52  11/51  10/50    9/49   8/48 

b     48/48   1/4   10/50    9/49   8/48 

c     48/48  10/47   1/4     9/49   8/48

d     48/48  10/47   9/46    1/4    8/48

e     48/48  10/47   9/46    8/45   1/4



E 

a      1/1   11/51  10/50    9/49   8/48

b     11/51   1/1   10/50    9/49   8/48

c     11/51  10/50   1/1     9/49   8/48

d     11/51  10/50   9/49    1/1    8/48

e     11/51  10/50   9/49    8/48   1/1

I think I'm on the right track this time, but I'm leaning towards Dutch now... all have the same chance.

Darnit, I've got to learn to post more so I don't look so stupid when my only posts are wrong.

P.S. I've also got to work my tables out so I don't have to edit 12 times.

[This message has been edited by Celeval (edited 12-14-2000).]

[This message has been edited by Celeval (edited 12-14-2000).]

[This message has been edited by Celeval (edited 12-14-2000).]

[This message has been edited by Celeval (edited 12-14-2000).]

[Ctown-Fan]

Lets try thing again . . .

the probability for A remains constant
(1/4)*(11/51)*(10/50)*(9/49)*(8/48)=
this will never change for a flush hand

(B) the probability remains constant
(1/13)*(11/51)*(10/50)*(9/49)*(8/48)=

(C) remains constant
(1/52)*(11/51)*(10/50)*(9/49)*(8/48)=

(D) does not remain constant becasue the ace can be drawn at any time
if ace is first
(1/13)*(11/51)*(10/50)*(9/49)*(8/48)=
second
(1/4)* (1/51)*(10/50)*(9/49)*(8/48)
third
(1/4)*(11/51)*(1/50)*(9/48)*(8/48)=
and so on for the furth and fifth cards
(E) does to remain constant becasue the ace of spade can be drawn at any time. but the odds only chagne from (d) on the first drawing from the deck. For two reasons, first the odds of drawing the ace of spades first are lower (b and c are not mutually exclusive from d and e) and the odds of drawing a spade with the first card are lower than drawing a card from any suit (52/52)
Ace is first
(1/52)*(11/51)*(10/50)*(9/49)*(8/48)=
Ace is second
(1/4)*(1/51)*(10/50)*(9/48)*(8/28)'
and so on

if I could remember the math then I would try to solve it, but I can. sorry.

Am I at least on the right track?

This is a great question, even though I had to dig out my stats book

[This message has been edited by Ctown-Fan (edited 12-14-2000).]

[QuikSand]

I'm personally having trouble following the probability matrices above... part of it is the spacing, which doesn't line up properly from my current computer.

That said... three comments:

Dutch doesn't have it quite right (ABD / CE).
They are not all equally likely.
Celeval is correct - AB&C are all on the same level.

When I get home, I'l look again at the numbers above, and see if I can follow more clearly looking through IE.

[QuikSand]

I'm still working my way through Celeval's math above... it may be accurate, but it is more complicated than it needs to be. (That happens a lot in combination calculations)

I'll start with intuition, because I feel the whole thing can be solved with just that tool.

Celeval said it very well above:

A, B, and C are identical because the first card you draw has no effect on the probability. Take a deck, draw any card. Let's say it's the 3 of Hearts. The chance of getting a flush is exactly the same as it is before you drew that card; since no matter what card it is (any, any ace, or the ace of spades) the chance of the flush is the chance of drawing your next four cards to match that suit.

I buy this thinking hook, line, and sinker. Seeing one card tells us *absolutely nothing* about the probability of getting a flush... as someone else noted, it will reduce both the numerator and denominator of the fraction that represents the chance... but by proportional amounts, so the likelihood remains identical. Agreed. So A, B, and C will all be on the same level in the correct answer.

That said, can't we assess option E with a similar intuitive approach? E narrows us to the subset of possibilities where one of the cards is the ace of spades-- yet, we know nothing of the other cards. Isn't this essentially the same case as B or C? Does it matter which of the cards we gain some knowledge about, first or otherwise? As long as we only have knowledge about one of the five cards, then (intuitively) we've learned nothing about the hand's chances of being a flush.

So, I would suggest that just by intuition and a basic grasp of the probability involved, we shoudl get to the point where there are only three possible answers:

ABCDE

ABCE
D

or

D
ABCE

I've already (above) ruled out the first one, so we're now stuck with the case that D is either less likely or more likely to be a flush (or else I lied to you).

(Incidentally, both of these were thrown out as possible answers in back-to-back posts earlier in the thread)

That said, who will make the case first here... The intuitive? Or the mathematical?

[QuikSand]

Ctown-fan, I first need to apologize for my memory failure about Vermilion's playing with the "big boys" in high school athletics. I had somehow unfairly lumped you in with the also-rans of Erie County-- Wakeman, New London, that lot. Unfair of me, and I'm sorry.

Now, on to the math. Your calculations above (as of the time I copied them, they seem to be moving a bit):

Lets try thing again . . .
the probability for A remains constant
(1/4)*(11/51)*(10/50)*(9/49)*(8/48)=
this will never change for a flush hand

(B) the probability remains constant
(1/13)*(11/51)*(10/50)*(9/49)*(8/48)=

(C) remains constant
(1/52)*(11/51)*(10/50)*(9/49)*(8/48)=

It looks to me like what you will get with each of these calculations is the probability that a hand meeting those conditions will be drawn. So, your calculation for B would be an accurate representation of the probability that a given hand will have an ace as its first card and will be a flush.

However, this is not on point to the question posed. The question is: given the conditions of each lettered case, what is the probability of the hand being a flush? I trust that you can see the difference between the two constructs.

For any of A, B, or C, I think that the probabilities laid out by Celeval (appropriately labeled "probability of staying in suit") are the correct ones. If you multiply across each row of his layout, you ought to get the 33/16660 number that was calculated earlier by Duke/Michael (I'm taking his number on faith).

The forst probability is by definition 1-- this is the "given" in the setup. Form there, you calculate the incremental likelihood of "staying in suit" and maintaining the flush.

Now, time to crack case D.

[Scarecrow]

Here's what I came up with. According to QS its wrong, but I'm not sure where.

1st 2nd 3rd 4th 5th TOTAL
A: (52/52) (12/51) (11/50) (10/49) (9/48) 0.1981%
B: (4/52) (12/51) (11/50) (10/49) (9/48) 0.0152%
C: (1/52) (12/51) (11/50) (10/49) (9/48) 0.0038%
D: (52/52) (1/51) (11/50) (10/49) (9/48) 0.0165%
E: (13/52) (1/51) (11/50) (10/49) (9/48) 0.0041%

A: 1st card anything, all others same suit.
B: 1st card ace, all others same suit.
C: 1st card Ace of Spades, all others same suit.
D: 1st card anything, 2nd card ace of that suit, all others same suit.
E: 1st card a spade, 2nd card ace of spade, all others same suit.

Therefore, it should be A,D,B,E,C.

------------------
If I only had a brain

[QuikSand]

Scarecrow, see my comment above to Ctown-fan. I believe that you made the same errant translation from the narrative to the mathematical.

[QuikSand]

I'll take a stab at my own set of numbers, but my approach is slightly different. For each case, I'll just calculate two things:

D = total number of cases fitting the criteria stated (flushes or otherwise)

N = total number of cases fitting the criteria that are flushes

Then, I'll just calculate N/D = probability of a flush, given that set of criteria.

(Personally, I find this simpler than the incremental multiplicative probability approach, though they are equally valid)

- - -

Case A - any 5 card hand

D = 52x51x50x49x48
N = 52x12x11x10x9

N / D = 0.1981% (I agree with Scarecrow here)

Case B - start with an ace

D = 4x51x50x49x48
N = 4x12x11x10x9

N / D = 0.1981%

(You can see here that the exact same transfer took place in both cases-- a 52 was reduced to a 4. When we use the ratio, it washes out completely)


Case C - start with a spade

D = 13x51x50x49x48
N = 13x12x11x10x9

N / D = 0.1981%

(Same logic, just 13 different starting points here)


Case E - hand has ace of spades

You can break this into five parts if you insist, but the order is unimportant-- this breaks down into just another run of the same logic...

D = 1x51x50x49x48
N = 1x12x11x10x9

N / D = 0.1981%

- - -

This, we are left with Case D.

I'll leave the rest as an exercise for the readers... for now, at least. Brute force (shown here) will get you to the correct answer, but intuition is still the easier path, I think.

[This message has been edited by QuikSand (edited 12-14-2000).]

[jason_tobias]

I think its D.

[QuikSand]

Once again, our friend J_T cuts through all the nonsense and gets to the heart of the issue. Well done, friend!

[Vaj]

At the risk of replacing my earlier error with another, I now think the probability of a flush for Case D is 0.2233%. I'm sure Passacaglia can confirm this.

[Ctown-Fan]

1. No problem quicksand for the misunderstanding. Since I have moved back to Vermilion, I have learned that conference has dispanded and vermilion plays a much weaker schedule (Oberlin, Firelands, and the like).
2. I think I understand now, my stats was never my strong suit. I only took it after I graduated when I learned the UMaine required it for grad school . . . which i found to be really silly because my M.A. is in rhetoric . . . but back to the point. I think this is really funny. I took stats at Kent State. Thirty people in my class and 27 seven were Fashion Design Majors!! No offense but it was not the most intellectually challenging course I have taken. But, I do see where my assumptions were off.

Thanks, this was a great challenge

[WebEwbank]

Q: was I right or wrong ?

[Passacaglia]

Sorry, Vaj -- i had 0.00223311

------------------
"If we can put a man on the moon, we can grow grass indoors."

[QuikSand]

Web, et al -- the answer is that (D) has the HIGHEST flush probability of the five, while all the others are even.

I'll offer my intuitive explanation... it looks liek Vaj has already crunched the numbers (though I haven't checked fior accuracy).

Intuitively, the main issue with flushes is (as you have pointed out) that they cannot have multiple cards of the same rank.

Group (D) includes all the hands that have at least one ace. The intuitive trick here is to reverse that thinking-- then the math gets much easier.

If the entire set of hands has a certain chance of baing a flush, then than entire set is composed of two discrete subsets-- hands including one or more aces (Group D), and hands without any aces (the rest). By definition, the number of these two subsets of hands must add up to the total number of hands.

So, instead of looking at the hands in Group D, let's look at what remains after we peel them off-- the hands with no aces. This, essentially, means that we're looking at a new deck of cards-- only 48 cards, 4 suits but ranks of only 2 through K. Well, what are our chances of getting a flush from this kind of deck?

Again rather than use math, I'll use intuition. As we shrink the size of each suit, the proportions become dimmer that we can "remain in suit" after each card. Take this toits logical conclusion-- a 40suite deck with only 5 cards in each suit. Clearly, in that case our chances of drawing a 5-card flush become extremely remote-- much more so than with the original 52-card deck, right? So, it stands to reason that as we compress each suit, the likelihood of a flush gets lesser.

So, the subset of hands that contain no aces-- being the equivalent of a hand deal from a 48-card ace-free deck-- is less likely to be a flush than the hand from Group A (or any of the others).

Since the group we just described is the complement of Group D and they combine to form Group A... then by deduction, Group D must have a greater chance of being a flush.

That's the intuitive proof (with a little math) and it works just fine for me. However, if you prefer the math, here's the recipe:

(again, I prefer to use my N and D calculations as at the bottom of page 1 of this thread)

To calculate the chances of Group D being a flush, you must calculate:

D = total number of hands meeting the conditions;

N = total number of qualifying hands that are flushes;

and then calculate N / D.

To get D, I think it is (again) easier to calculate all the hands from a 52-card deck, which is:

52x51x50x49x48

...and then subtract out all the hands from a 48-card ace-free deck, which is:

48x47x46x45x44

The difference should be the number of hands that include at least one ace = D (quickly, I got 106,398,720 on my $1.00 calculator).

To get N is a bit more tedious, but the number of flushes that include an ace (of course they cannot include two of them) would involve assuming the ace, then calculating the number of combinations of the other 12 cards in the suit, and then multiplying by 4 for the different suits, and then by 5 for the five different positions where the ace could appear.

So, N = (12 x 11 x 10 x 9) x 4 x 5
and I calculate N = 237,600.

So, we then calcuate N / D, and I get a number like 0.2233% - which appears to be exactly what my friend Vaj got above and Passacaglia got below.

So, the correct order is:

D
ABCE

q.e.d.

My latest edits corrected my calculator and human errors, and are in italics just above



[This message has been edited by QuikSand (edited 12-15-2000).]

[Passacaglia]

Dola-post.

For the numerator, here's what I did, and probably Vaj as well:

I took all the flushes possible.

52x12x11x10x9 = 617,760

Then subtracted all the flushes with NO aces. Probably a more difficult way of doing it, but it made sense, given how we found the denominator.

Flushes with no aces:
48x11x10x9x8 = 380,160

617,760 - 380,160 = 237,600

237,600/106,398,720 = 0.00223311

That was my reasoning, at least.

------------------
"If we can put a man on the moon, we can grow grass indoors."

[Passacaglia]

Double-Dola.

Hey, QS! You can't be editing your message while I'm trying to correct you! :P

------------------
"If we can put a man on the moon, we can grow grass indoors."

[Passacaglia]

QS! You're making me look like a *real* schmuck here, with me replying to everything, then you editing or deleting it!

------------------
"If we can put a man on the moon, we can grow grass indoors."

[QuikSand]

I got it, Pass. My original calculation was off by x2 from yours, because I made two errors.

I mis-read my overflow of digits on my el cheapo calculator, and was off by a factor of 10 on the number of with-ace hands.

Then, I neglected to multiply my number of flushes by 5, to reflect the five different positions where the ace could appear.

The x10 but /5 errors overlapped to give me double your answer. Now they are untangled above.

Thanks for your help.

[Passacaglia]

QS, I think the flaw might have been in my reasoning.

I wrote that the number of flushes with no aces is:

48x11x10x9x8 = 380,160

Wouldn't it be:

48x12x11x10x9 = 570,240 ?

Then 617,760-570,240 = 47,250 ?

------------------
"If we can put a man on the moon, we can grow grass indoors."

[QuikSand]

And geez, Passacaglia, will you please quit Dolaposting? We're trying to do some math here...

[QuikSand]

Wouldn't it be:

48x12x11x10x9 = 570,240 ?

the above-quoted selection is from a response that Passacgalia has since sheepishly deleted, after seeing the error of his ways

Nope, that's a 13-card suit. You were right the first time, though that is a clever way to get to my original incorrect answer.

[This message has been edited by QuikSand (edited 12-15-2000).]

[Passacaglia]

Ha! Now I'm making *you* look like the idiot, since I'm deleting all my dola-posts!

Hey, if I post something, then delete it, do I still get "credit" for the post?

------------------
"If we can put a man on the moon, we can grow grass indoors."

[This message has been edited by Passacaglia (edited 12-15-2000).]

[Passacaglia]

What do you mean, "that's a 13-card suit?"

------------------
"If we can put a man on the moon, we can grow grass indoors."

[QuikSand]

Curse you, Red Baron!

[Passacaglia]

So here I am, waiting for a response from QS to my question, only to find that he replied to it by editing a post from above! Very sneaky, QS, but it only fooled me for..well, never mind how long it fooled me for! The point is that I know now!

Anyway, I still don't know what you mean by "that's a 13-card suit," but I do see where the x5 comes in, for the different placement possibilities.

AND, I did delete some of my dola-posts, but those were the ones that were covered by your editing of your posts, not the mistakes I made, as you asserted!

------------------
"If we can put a man on the moon, we can grow grass indoors."

[QuikSand]

Pass, my comment about a "13-card suit" is in response to your now-deleted post where you postulated an alternative calculation for how many flushed include no ace. You came up with:

Wouldn't it be:
48x12x11x10x9 = 570,240 ?

And my response is that your 48 is correct, because you may not start with an ace (so only 48, not 52 to start with). However, by using multipliers starting with 12, you are allowing the ace to be included-- which is forbidden by the restrictions. In my parlance above, when you must exclude the ace, you are essentially working with a deck that only has 12-card suits. Your calculation (above) excluded the ace at first, but then re-introduced it in the later stages.

Curiously, that resulted in the exact same wrong answer for N (by subtraction) as mine did (by neglecting to multiply by five).

[Passacaglia]

Yep, that's exactly right. It's weird, I thought I would have been better quipped to do this kind of work today, but it seems I was dead on yesterday. It is interesting how those two incorrect solutions yielded the same answer, though.

------------------
"If we can put a man on the moon, we can grow grass indoors."

[Ctown-Fan]

The only question I have about the final number crunching is this (and maybe I missed it somewhere through all the numbers talk) is, isn't (A) all flush hands? That would mean, by virtue of the question, a set which contains hands with or without aces, increasing the pool of posibilities when determining the answer. I guess when I was responding to this before, I was under the assumption the (A) actually contained (b-E) in it. Sort of the whole ven diagram thing.

[Passacaglia]

If you look at it as a Venn diagram, imagine a circle, including all possible flush hands, inside the box, all possible hands. As you change states, you reduce the circle, since it's not the case that all flushes are possible, and the box, since it's not the case that all hands are possible!

------------------
"If we can put a man on the moon, we can grow grass indoors."

[jason_tobias]

Please, in the name of of an american icon, Rudy Ray Moore (Dolamite), please refer to a dolapost as anything but a dolapost.

I've tried my very best to maintain myself, but I can't further view the abuse constantly perpetrated on the 'Avenging Disco Godfather,'aka,'DOLAMITE!'

Call it the 'Hamburger King.'

[Passacaglia]

Yeah, but that was Dolemite, not Dolamite. And if we don't refer to a dolapost as a dolapost, what WOULD we refer to as a dolapost?

------------------
"If we can put a man on the moon, we can grow grass indoors."

[ez]

yeah, we went through this before. the rock/mineral, mountains, and movie character are dolEmite. the namesake of the dolapost (and dolanuts ) explained that his last name was dola or dola-something, and that his friends(?) therefore tagged him as dolAmite...

[Morgado]

Originally posted by QuikSand:
The intuitive trick here is to reverse that thinking-- then the math gets much easier.

Just wondering, Quik... have you tried going forwards and doing the counting problem directly? It may be an... interesting... check.

[Grunion]

QuikSand,

I'm not sure I agree with your conclusions. (However I'm quite sure that my math skills are not quite as refined as yours so I'm willing to have the error of my ways pointed out.)

Imagine five independant poker hands being dealt (from separate decks).

Case A:
No cards are known. The first card is exposed and is found to be a four of clubs (for sake of example, in case A it could be any card). The odds of pulling a flush would be the odds of each successive card being a club, or 12/51*11/50*10/49*9/48. (0.198%)

Case B:
The first card is exposed and found to be an ace of spades (for example, in case B it could be any ace). The odds of pulling a flush would be the odds of each successive card being a spade, or 12/51*11/50*10/49*9/48. (0.198% which is identical to case A)

Case C:
The first card is exposed and found to be ace of spades (as a conditional requirement). The probability is identical to cases A & B.

Case D:
Your cards are dealt, and the dealer accidentally deals one face up. It is an ace of hearts (or spades or clubs or diamonds the suit of the first known card is always irrelevant). The computed odds of the next four cards matching suit are 12/51*11/50*10/49*10/48. It does not matter that it is not necessarily the first card, order of cards being dealt is not significant, making the probabilities identical to Case B.

Case E:
This time the dealer deals an ace of spades face up. As the suit of the first known card is irrelevant, the odds of a flush are identical.

Another, much quicker way of looking at it is no matter how many constraints (suit, number, or order in which it was dealt) the odds of matching one given card with four more of the same suit is always the same. In every case other than A only one card has any information associated with it.

Also, If it is more likely to pull a flush in case D, it would stand to reason that the same arguement could be made for every single card value, making every single card value a higher probability than the default, five unknown cards condition.

Hopefully the above made sense. Now who's up for some cards!


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