$1000 NLHE Full Ring: flop comes 5,5,2. How do I calculate the odds of another player having a five?

I know the way to work this out. If there are 8 other people at the table therefore 16 cards the chances of everybody not having a 5 are (45/47)*(44/46)*...(30/32). That calculation is (45!/30!)/(47!/32!) = 0.4588
so the chances of somebody having a 5 are 1-0.4588 = 0.5412 = 54%

To test this works you can workout that if there were 23 people sitting at the table therefore 46 cards somebody must have a 5. So it would be (45/47)*(44/46)*...(0/2) which clearly = 0 since 0/2 = 0

If there were 22 other people at the table the chances of everybody not having a 5 would be (45/47)*(44/46)*...(2/4) = (45!/2!)/(47!/4!) = 0.0056 so the chances of someone have a 5 would be 1-0.0056 = 0.9944 = 99.44%

js520 said:

I know the way to work this out. If there are 8 other people at the table therefore 16 cards the chances of everybody not having a 5 are (45/47)*(44/46)*...(30/32). That calculation is (45!/30!)/(47!/32!) = 0.4588
so the chances of somebody having a 5 are 1-0.4588 = 0.5412 = 54%

To test this works you can workout that if there were 23 people sitting at the table therefore 46 cards somebody must have a 5. So it would be (45/47)*(44/46)*...(0/2) which clearly = 0 since 0/2 = 0

If there were 22 other people at the table the chances of everybody not having a 5 would be (45/47)*(44/46)*...(2/4) = (45!/2!)/(47!/4!) = 0.0056 so the chances of someone have a 5 would be 1-0.0056 = 0.9944 = 99.44%

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I took the question to mean the chance of 1 other person having a 5, your way doesn't take into account the times when two people have a 5.

BlueNowhere said:

I took the question to mean the chance of 1 other person having a 5, your way doesn't take into account the times when two people have a 5.

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OP said he has an overpair so the only thing he cares about is if one of the 8 players has a 5

js520 is extremely close, tough he made a very slight mistake (when introducing the factorials) that alters the result in ~2%. He also considered only 8 villains when the OP asks for 9 villains. His calculations do handle the case where the two fives are out there in the villain's hands. I'll reexplain, with the calculations fixed:

First, out of the 52 cards of the deck, we know 5 of them, and 2 of them are fives. This means that there's 47 unseen cards, and 2 of them are fives. We're going to deal a total of 18 cards (9 villains, 2 cards each), and we want to know the chance that one or more of them are fives.

Logically, what we want to calculate is: (Note how if two of the cards are fives, the logical condition still holds true).


To simplify the calculation, we apply a double negation and distribute one of the negations according to De Morgan's Laws, the other just as a normal probability negation. After doing this, we obtain the equivalent expression:


Now this is easier to calculate. The probability that Card1 is not a five is 45/47. The probability that Card2 is not a five is 44/46. Etc Etc. The probability that Card18 is not a five is 28/30. The probability that a series of mutually exclusive events all happen is done by multiplying the chance of all of them. So the final expression to calculate is:


Here's the calculation on WolframAlpha.. The chance that one or more cards are fives is 62,44%. With 8 villains the chance is 56,98% (js520 missed the last fraction of this expression).

BTW, this calculation is useless anyway because if you have a high pair, you should raise, and any non-donk opponent will almost always fold hands with a 5 except pocket pairs, connectors and gappers, and which ones of those he does fold depends on his style, skill and the size of your raise, so it is a completely different problem.

PS: You can't use tricks like the rule of 2 or approximating ORs by doing sums in this calculation, because the calculation is long and too many error accumulates.

Also here's a script which does a simulation and I used to check my calculations:

http://pastebin.com/VXxV0i6Y

It seems some people actually enjoy maths then.

PLEASE TELL ME I DONT NEED TO LEARN ALL THIS TO BE A GOOD POKER PLAYER?????????????????????????????????????

Number of cards remaining = N = 47 here
Number of cards we are interested in (the 5s) = m = 2 here
Number of "spots" (villains times 2) = s = 9 villains x 2 = 18 here

Probability that the cards we are interested are NOT in the s spots =

(N - m)!/(N - m - s)!
_________________
(N)!/(N - s)!

EZ game. So probability is just 1 - that.

For 9 villains (a 10-handed table) this is 62.44%, and for 8 villains, this is 56.98%.

Edited: Had a slight error.

alyoung said:

PLEASE TELL ME I DONT NEED TO LEARN ALL THIS TO BE A GOOD POKER PLAYER?????????????????????????????????????

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You don't. This is actually quite irrelevant to playing poker as the chance of villain having a 5 is nowhere near this as any hand Involving a 5 is highly unlikely in almost everyone's range

Yea my calculation was slightly wrong, should have been 1- (45!/29!)/(47!/31!) which does indeed = 56.98% for 8 villains

Scourrge said:

For 9 villains (a 10-handed table) this is 62.44%, and for 8 villains, this is 56.98%.

Edited: Had a slight error.

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So it looks like about a 7% chance for each villain still left in the hand to have flopped a set when the flop contains a pair. One villain 7%, two villains 14%, on up to 9 villains 63%, give or take. Thanks for doing the math on this question.

Wow... Thanks Lenstra, I would have never been able to figure this out

@jchoop And yeah, when I was asking this question it was more for the purpose of figuring the math out than it was to get a poker strategy answer.

i know. obv you know your shit playign 1k NL lol

haha.. actually, I just chose a limit randomly