What is the probability of having all of one card play in a hand?

[digdeep]

I recently held 95o in the big blind, and checked it to see a flop that came K95. As I didn't suspect anyone to have be holding pocket 9's or 5's I agressively bet the hand, only to have someone turnover 55. Just curious what the odds are that all of one card will be in play in a hand. In this case, what is the probability that player 1 has 55, player two has 5X, and the flop comes XX5? And simply out of curiosity, what are the odds that player one holds 5X, player 2 holds 5X, and the flop comes 55X?

Thanks

[catalinuzzu]

[scottjack]

I don't think this is correct.
these 3 events' combined probability cannot be determined simply by multiplying their individual probabilities.

This only applies if the events are independent.

Clearly, the probability of one player holding 5X is affected by the fact that another player holds 55. These events are therefore not independent.

Am i way off base here?

[BruceZ]

[ QUOTE ]
I don't think this is correct.
these 3 events' combined probability cannot be determined simply by multiplying their individual probabilities.

This only applies if the events are independent.

Clearly, the probability of one player holding 5X is affected by the fact that another player holds 55. These events are therefore not independent.

Am i way off base here?

[/ QUOTE ]

These calculations are correct. It is true that P(A and B) = P(A)*P(B) if and only if A and B are independent; however, it is always true that P(A and B) = P(A)*P(B | A), where P(B | A) means the probability that B occurs given that A has occurred. This is true even when A and B are not independent. You can see that when they are independent, P(B | A) = P(B), which can be taken as a definition of independence. In the calculation above, you can see that we are multiplying by conditional probabilities, since the computation of each probability is computed assuming that the previous event has occurred.

[Dubious]

I'm not sure what the original reply (which was deleted) said. However, as I see the question, the probability of all 4 fives (or all of any card) being in either players hands or on the board, for a ten-player hand with all 5 board cards dealt, is equal to P(a 5 is not in remaining 37 cards not dealt) = (48/52*47/51*...*9/13) = (9*10*11*12)/(51*51*50*49) = 0.18% or only 1 in 547 hands (for 5's or any specific rank of card).

The probability that all four of any rank of card will be in play is 13*0.018% = 2.4% or 1 in 42.

If you know that exactly 3 of a rank has been dealt (one player has a 5, one has two 5's), the probability that the other card will come up on the flop is 1-(41/42*40/41*39/40) = 1-39/42 = 7.1% or 1 in 14. If you, instead, know that exactly 2 cards are out, the probability is (9*10)/(42*41) = 5.2% or 1 in 19 that the other two will show up on the flop).

[Precision1C]

For a 10 handed Hold'em game I get 4.673% for any particular rank to have been dealt out by the river. I admit that is fact is of virtually of no use in any poker calculations since this a priori fact isn't of any use preflop and postflop this a priori calculation isn't worth much.

I derived this percentage by dividing the number of combinations of AAAA in a 25 card draw (20 cards in people's hands and the 5 community cards) and dividing by the number of 25 card draws from 52. 1* C(48, 21) divided by C(52,25). By the by, you can't just multiply 4.673% by 13 to deduce that there is over half a chance of all four card being out of some rank since there is signicant double+ counting.

[Dubious]

37 cards remaining in the deck? What was I thinking? I guess it was just too late at night.

I'm glad you caught that so quickly. Thanks.

Yeah, so it's much more likely. I actually get 6.5% now (or 1 in 15). I'm not sure why you have 48Ch21/52Ch25 instead of 48Ch25/52Ch25 (which simplifies to (27*26*25*24)/(52*51*50*49) if you're calculator doesn't do well dividing numbers that are on the order of 10^14). You're looking at sets of 25 cards no matter whether those sets include or don't include the 4 fives.

And I'm also wrong about multiplying by 13. The correct calculation is 1-(1-P(4 aces))*(1-P(4 twos))*...*(1-P(4 Kings)) = 1-(1-.0648)^13 = 58% chance that all 4 of at least one rank will be in play in a 10-person hand where all 5 board cards are dealt.

And, for completeness, the odds that, given there are two 5's in players' hands, the other two will come on the flop is (3*2)/(32*31)=0.6% (the 3 comes from 3 orders of getting the cards). Given there are three 5's in players' hands, the odds are 3/32=9.4%.

I guess I shouldn't be allowed to post after 11:30 at night.

[Shawsy]

You should end up with the same probability whether you look at the cards in play, or the cards out of play. I think sticking to the assumption of 25 cards in play and 27 cards out of play is a good starting point.

There has to be 4 cards from one rank from the 25 cards in play. This would happen in C(4,4)xC(48,21) ways out of C(52,25) possible combinations. I get 4.67%.

If you look at the cards out of play, there can be zero cards of one rank in the 27 cards. This can happen in C(4,0)xC(48,27) ways out of C(52,27). I get 4.67%.

The problem I have with this is that there are many possible situations where other ranks are entirely in play. I looked at narrowing the problem to having one, and only one, rank entirely in play. The problem got too complex for me pretty quickly. I don't think the problem has much application at the poker table though, so it may not be worth the effort to refine the solution.

[BruceZ]

[ QUOTE ]
I looked at narrowing the problem to having one, and only one, rank entirely in play. The problem got too complex for me pretty quickly. I don't think the problem has much application at the poker table though, so it may not be worth the effort to refine the solution.

[/ QUOTE ]

It is worth knowing the general method as it comes up quite frequently, it is very powerful, and in this case it allows the exact answer to be arrived at very quickly. First, let's answer a slightly easier question, which is "what is the probability of one or more ranks entirely in play?" This is answered straightforwardly by the inclusion-exclusion principle as exactly:

[ 13*C(48,21) -
C(13,2)*C(44,17) +
C(13,3)*C(40,13) -
C(13,4)*C(36,9) +
C(13,5)*C(32,5) -
C(13,6)*C(28,1) ] / C(52,25)

=~ 50.24%.

In the inclusion-exclusion principle, the coefficients of each term alternate between +/- 1. This has the overall effect of counting each eligible deal exactly 1 time no matter how many ranks are actually in play. Now to answer your question, "what is the probability of one and only one rank entirely in play?", we must modify these coefficients so that all of the deals with more than one rank entirely in play are counted 0 times, rather than 1 time, and only the deals with exactly 1 rank entirely in play are counted 1 time by the first term.

For example, since the first term counts deals with exactly 2 ranks in play twice, the coefficient of the second term becomes -2 in order to make these go to 0. But now the first two terms will count deals with exactly 3 ranks in play 3 - 2*C(3,2) = -3 times, so the coefficient of the 3rd term must be +3. In general, the coefficient of the nth term a(n) = -sum{k = 1 to n-1}a(k)C(n,k), with a(1) = 1. This recursive equation is simply a(n) = +/- n, where the positive term is taken for odd coefficients, and the minus sign is used for even coefficients. So our probability of one and only one rank entirely in play is exactly:

[ 13*C(48,21) -
2*C(13,2)*C(44,17) +
3*C(13,3)*C(40,13) -
4*C(13,4)*C(36,9) +
5*C(13,5)*C(32,5) -
6*C(13,6)*C(28,1) ] / C(52,25)

=~ 40.43%.

Note that it is possible to continue generalizing in this way, so that we can answer all questions of the form "what is the probability of exactly N?" or "what is the probability of N or more?". Each solution will have the same form as the inclusion-exclusion principle starting with the Nth term; however, each value of N requires that we derive a new set of coefficients as we have done above for the case N = 1. When we desire the probability of exactly N, the coefficients will always be given by a(n) = -sum{k = 1 to n-1}a(k)C(n,k), with a(N) = 1. When we desire the probability of N or more, the coefficients will be given by a(n) = 1 - sum{k = 1 to n-1}a(k)C(n,k), with a(N) = 1.

[donkeyradish]

Would the probability of all fives being in play not be significantly lower than say, all the Aces?

(Because there are much fewer playable starting hands that contain a five than those that contain an Ace)