March 19th, 2017 at 6:08:27 PM
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This may have been answered before, but I can't seem to find a definitive answer anywhere. (Even outside of this site.)

Assuming I have 2 suited cards dealt to me; what I would like to know is the probability someone else in a 10 player hold em game will be dealt 2 suited cards of the same suit that I have. I assume the probability is lower than the standard 23% chance of me getting suited cards, but this may be incorrect. I have taken this stat from Sun Tzu poker odds site.

Also, what effect would that have on me drawing to a flush by the river? (I have seen more than a few postings saying it is unlikely that 2 people will have a flush with only 3 suited cards on the board, but I am looking for something a little more precise, if possible.)

Thanks

Assuming I have 2 suited cards dealt to me; what I would like to know is the probability someone else in a 10 player hold em game will be dealt 2 suited cards of the same suit that I have. I assume the probability is lower than the standard 23% chance of me getting suited cards, but this may be incorrect. I have taken this stat from Sun Tzu poker odds site.

Also, what effect would that have on me drawing to a flush by the river? (I have seen more than a few postings saying it is unlikely that 2 people will have a flush with only 3 suited cards on the board, but I am looking for something a little more precise, if possible.)

Thanks

March 20th, 2017 at 8:31:29 AM
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There are 10 different cases where both of your hole cards are suited and nobody else has two cards of the same suit as yours:

1. Nobody else has a card of the suit

2. One other player has one card of that suit

3. Two other players have one card of that suit

...

10. All nine other players have one card of that suit

Note that (A)C(B) = combin(A,B)

Case 10 has (39)C(2) x (37)C(2) x ... x (23)C(2) possible cases

For case 1, there are (9)C(1) = 9 possible hands that have the one card of that suit, 11 cards of that suit, 39 cards not of that suit that can be the other card, and (38)C(2) x (36)C(2) x ... x (24)C(2) possibilities for the other eight hands

Note this equals the case 10 value x 9 / 1 x 11 / (22 x 2)

For case 2, there are (9)C(2) = 36 possible hands that have one card of that suit, 11 x 39 possible "first hand" pairs, 10 x 38 possible "second hand" pairs, and (37)C(2) x (35)C(2) x ... x (25)C(2) possibilities for the other seven hands

This equals the case 9 value x 8 / 2 x 10 / (23 x 2)

Similarly, for cases 3 through 9, the case N value = the case (N - 1) value x (10 - N) / N x (12 - N) / ((21 + N) x 2).

Add these up, and divide by the total number of deals of the other 9 players, which is (50)C(2) x (48)C(2) x ... x (32)C(2), and you get a probability of 64.51% that no one else has 2 cards of your suit, which means the probability that at least one other player does is (100 - 64.51) = 35.49%.

Note that this is the probability that no one else has 2 cards of your suit after assuming that you have 2 suited cards.

In other words, if you have two suited cards, there is a 35.49% chance that at least one other player has 2 cards of the same suit as yours as well.

This does not take into account the probability of you having two suited cards in the first place.

Yes, I did run a simulation to check that value this time...

March 20th, 2017 at 9:21:56 AM
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Just because I'm dealt two suited cards doesn't mean I'm playing them, so keep that in mind.

March 20th, 2017 at 10:00:26 AM
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Hiding the answers in case others want to work them out I suppose? =P

So for part 2 you're asking what effect someone else having your suited cards would have on you drawing at a flush draw? Pretty simple... There would be 2 less outs for each of you to hit your flush (not caring whose is higher). Say you have 2 suited cards and the player next to you had 2 suited cards... ignoring if other players at the table have your suit or not (like BJ treating unseen cards as undealt cards), then you'd simply have 2 less outs to hit.

Typically when you have 2 suited cards and don't know anyone else's cards you assume you have 9 outs (2 in your had, 2 somewhere on the board, and 9 remaining of that suit in the deck). Going to the river there are 4 community cards, and your 2 hole cards out of the deck, thus you have 9/(52-6) chance of hitting your flush... 9/46 =19.57% chance.

IF someone else also had 2 of your suit (again, regardless of whose is higher) then that would simply take 2 outs away form both of you... Now your chance of catching the flush on the river would be 7/44 = 15.91% chance to hit a flush. So the net change of someone else having 2 of the same suit is "about" 3.5%.

So for part 2 you're asking what effect someone else having your suited cards would have on you drawing at a flush draw? Pretty simple... There would be 2 less outs for each of you to hit your flush (not caring whose is higher). Say you have 2 suited cards and the player next to you had 2 suited cards... ignoring if other players at the table have your suit or not (like BJ treating unseen cards as undealt cards), then you'd simply have 2 less outs to hit.

Typically when you have 2 suited cards and don't know anyone else's cards you assume you have 9 outs (2 in your had, 2 somewhere on the board, and 9 remaining of that suit in the deck). Going to the river there are 4 community cards, and your 2 hole cards out of the deck, thus you have 9/(52-6) chance of hitting your flush... 9/46 =19.57% chance.

IF someone else also had 2 of your suit (again, regardless of whose is higher) then that would simply take 2 outs away form both of you... Now your chance of catching the flush on the river would be 7/44 = 15.91% chance to hit a flush. So the net change of someone else having 2 of the same suit is "about" 3.5%.

Playing it correctly means you've already won.

March 20th, 2017 at 10:34:30 AM
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Quote:RomesHiding the answers in case others want to work them out I suppose? =P

I do it to shrink the amount of space the answers take up. Not everybody is as interested in math and statistics as we are.

March 20th, 2017 at 11:49:44 AM
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...but they should be, and we should force them to read our long responses! =PQuote:ThatDonGuyI do it to shrink the amount of space the answers take up. Not everybody is as interested in math and statistics as we are.

Playing it correctly means you've already won.