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ccwhoelse?
08-14-2003, 11:38 AM
if you are dealt suited cards, what would be the probability in a 10-handed game that another card of your suit is dealt to the other 9 people.

is it correct to do:

1 - (39!/21!)*(50!/32!) = .996 or 99.6%

if the above is correct, and you flop a four flush, aren't you almost virtually gauranteed to have 8 outs instead of 9 even though your unseen cards still give you 9?

pudley4
08-14-2003, 11:57 AM
I'm not sure if your calculation is correct or not (I didn't take the time to calculate it out); however, even if you know that one or more of your suit was dealt to your opponents, it doesn't matter...

rjc199
08-14-2003, 04:29 PM
The last guy said something wrong. If somebody showed you their hole cards and you saw that they had one of your suit, then the number of outs changes, as does the odds. Say with one card left you have a four flush, and the guy next to you folds and shows you his cards, he was holding the deuce of spades (spades is what you have the four flush in). Therefore you now know there is one less spade in the deck so you have 8 outs, but you have seen 2 more cards so your odds on the river are 8/44 which would be 4.5 : 1.

I don't know how the probable distribution of suits affects the probability of getting a flush. You could calculate the probability distribution of X spades being dealt on the flop (it would be a bell curve), then knowing you have a four flush you could do another probability distribution of the function probability_of_another_spade_on_river with respect to X spades being dealt on the flop, etc.

Wake up CALL
08-14-2003, 05:59 PM
So what you are saying is that if I had nine outs to make my flush on the flop and one of the players has one of my suit in his hand the odds only change if he shows it to me.

Hmmmmmmmmm........

Bozeman
08-14-2003, 07:07 PM
Yup. Any entirely unknown card can be treat the same whether it happens to reside in the stub or in a player's hand.

M.B.E.
08-14-2003, 07:39 PM
"<font color="purple">So what you are saying is that if I had nine outs to make my flush on the flop and one of the players has one of my suit in his hand the odds only change if he shows it to me.</font>"

There's an important concept lurking behind this question. When you state what the odds are, it's always relative to a certain state of known information.

The short answer to your original question is this: it doesn't affect your calculation of the odds of hitting your flush that some of the other players were probably dealt cards of your suit, because based on your knowledge, the cards of your suit in other players' hands can be assumed to be in the proper ratio to the cards of other suits in other players' hands.

So let's say that on the flop you have a diamond draw and you haven't seen anyone else's cards. The odds against hitting your flush on the turn are 38:9 (4.22:1), since the turn is equally likely to be any one of the 47 unseen cards, of which 38 are non-diamonds and 9 are diamonds. Yes it's true that some of the diamonds are likely in your opponents' hands (or in the muck) but so are some of the non-diamonds.

Now if an opponent had accidentally exposed his hand to you before mucking it, let's say it was 7d-2s, then now the odds of hitting your flush on the turn have worsened to 37:8 (4.63:1).

But if three of your opponents had exposed their hands before mucking, and they were 7d-2s, Ac-5h, and Qh-5s, then you can recalculate your odds of hitting your flush on the turn as 33:8 (4.13:1), which is an improvement from your perspective over the case where no extraneous cards have been exposed.

BruceZ
08-14-2003, 08:57 PM
is it correct to do:

1 - (39!/21!)*(50!/32!) = .996 or 99.6%

It's 1 - (39!/21!) / (50!/32!) = 99.6%

or 1 - P(39,18) / P(50,18) = 99.6%

or 1 - C(39,18) / C(50,18) = 99.6%

You had the right number, just the wrong formula.

if the above is correct, and you flop a four flush, aren't you almost virtually gauranteed to have 8 outs instead of 9 even though your unseen cards still give you 9?

You are virtually guaranteed to have no more than 8 cards left in the deck that will make your flush, but you still have 9 outs because the chance of each card left in the deck being a flush card is 9/47. The fact that your opponents almost certainly hold some of the flush cards does not change anything. The 9 outs assume that each of your opponent's cards is as likely to be a flush card as each card in the deck. For that matter, the cards at the top of the deck have the same chance of being a flush card as the cards at the bottom of the deck. In fact, if you knew that there was exactly 1 flush card in your opponent's hands, you would actually have a much greater than average chance of making your flush, because that would mean that there were 8 flush cards out of the remaining 29 cards in the deck, rather than 9 flush cards out of 47 unseen cards. The chance of each card in the deck being a flush card would then be 8/29 instead of 9/47. That is effectively like having 13 outs. Sometimes your opponents will have more flush cards than they are supposed to, and you will effectively have less than 9 outs. On average, you will have 9 outs.

southerndog
08-15-2003, 12:09 PM
I think M.B.E. hit the nail on the head with this statement: "Yes it's true that some of the diamonds are likely in your opponents' hands (or in the muck) but so are some of the non-diamonds." You can work out the numbers from there, but this gives a good feel.

When doing the probability calculations, I usually calculate: P(an event happens)=1-P(event doesn't happen). I think this is the way to go about a lot of poker probabilities.

I think MBE makes an imortant statement that the probability is relative. For example, the probability
you have this second nut flush won before showdown is X. After showdown, the probability is 1 or 0.

ccwhoelse?
08-15-2003, 06:22 PM
[ QUOTE ]
The last guy said something wrong. If somebody showed you their hole cards and you saw that they had one of your suit, then the number of outs changes, as does the odds. Say with one card left you have a four flush, and the guy next to you folds and shows you his cards, he was holding the deuce of spades (spades is what you have the four flush in). Therefore you now know there is one less spade in the deck so you have 8 outs, but you have seen 2 more cards so your odds on the river are 8/44 which would be 4.5 : 1.

[/ QUOTE ]

no, bruce said out of all your opponents, if only one had a flush card your chances would be greater. that is there would be a higher density of flush cards left in the deck. it would be 8/29 as opposed to 9/47 if you didn't see anyone's hand. so you only have 8 outs but they are out of 29 cards, so the chance of hitting an out increases.