Odds for drawing to a flush, consider opponents cards?

[FredBoots]

Sorry if this has been asked and answered.

Table of ten, folded around to you on SB with Ax diamonds. You call and big blind checks. Flop has 2 diamonds, and you count 9 outs to get your flush.

But by my calculations, the chances that everyone who folded DIDN'T have a diamond are about 1% to 3%. So over 90% of the time, at least one of your outs got folded. Shouldn't this fact count when considering my outs. Shouldn't I consider that some of my outs were lost when those people folded?

[Uncle_Billy]

I'll probably embarrass myself here since I'm a newbie, but since I actually had the same question a while back and crunched the numbers...

Short answer is that your actual outs are likely less than 9, but that your odds of hitting the flush remain the same if you calculate it either way. Ratio of (remaining diamonds / remaining cards) equals ratio of (unseen diamonds [9]/ unseen cards[47]).

The numbers work out if you assume that the distribution of diamonds across the folded cards averages 25%.

[Copernicus]

[ QUOTE ]
Sorry if this has been asked and answered.

Table of ten, folded around to you on SB with Ax diamonds. You call and big blind checks. Flop has 2 diamonds, and you count 9 outs to get your flush.

But by my calculations, the chances that everyone who folded DIDN'T have a diamond are about 1% to 3%. So over 90% of the time, at least one of your outs got folded. Shouldn't this fact count when considering my outs. Shouldn't I consider that some of my outs were lost when those people folded?

[/ QUOTE ]

Yes, but when you take them into account it has no effect, and here is a simpler example showing why. Lets say youre playing some kind of game where the deck is dealt down to the last few cards, and to win, you need a 10. Youve been card counting and there are 2 Ts and 1 non-ten left, but before you get your card one is burned. Your question is equivalent to "shouldnt I take into account the high probability that one of my Ts is the burn card?"

Without thinking about the burn card, you think your chances of a T are 2/3.

Now consider the burn card. 2/3 of the time the burn card is a T and you only have 1/2 chance of getting the other T. So in this case you have 2/3*1/2 = 1/3 chance of getting a T.

The other case is 1/3 of the time the burn card is the non-10. Now you have 100% chance of getting a T, or 1/3 chance in this situation.

1/3+1/3 = 2/3...same as your original calculation.

Note: this example demonstrates why people who berate and blame 3rd base in a 21 game for their losses because 3rd base hit or stood inappropriately are out of line. Their action in the long run will not affect the dealers chances of breaking or making his hand.

Edit: btw this is only because the distribution of cards is assumed to be random, and you have no information to refine that. Even in poker you can make deductions about what cards a player has from his actions, and if you deduce some hands that include one or more of your outs, then you do need to reduce your outs to recognize that.

Eg. if a players actions indicate strongly to you that he can only have AA, AK, KK, and QQ, and Aces are your only outs (and you dont have any Aces and you have 1 K) then you cant assume you have 4 outs. There are 27 hands in the range you deduced, and 6 of them have 2 Aces and 12 of them have 1 A. So on average you need to reduce your A outs from 4 to 3.1 (and 3 is close enough for any pot odds evaluations).

[RocketManJames]

[ QUOTE ]
I'll probably embarrass myself here since I'm a newbie, but since I actually had the same question a while back and crunched the numbers...

Short answer is that your actual outs are likely less than 9, but that your odds of hitting the flush remain the same if you calculate it either way. Ratio of (remaining diamonds / remaining cards) equals ratio of (unseen diamonds [9]/ unseen cards[47]).

The numbers work out if you assume that the distribution of diamonds across the folded cards averages 25%.

[/ QUOTE ]

Maybe this is wrong, but I'm going to throw it out here for so the rest of you can tell me what you think.

Say you've got Ax diamonds as in the original post...

Flop is unpaired and contains 2 diamonds, and say you have 4 opponents. If 2 of the 4 opponents fold for a single bet, I think it is reasonable to assume that neither of the folded hands were suited diamonds. So, then can't we assume that we've seen 2 more cards?

So the number of unseen cards goes from 47 to 45. I know it doesn't change much, but wouldn't this be accurate? Or, am I missing something?

-RMJ

[Copernicus]

[ QUOTE ]
[ QUOTE ]
I'll probably embarrass myself here since I'm a newbie, but since I actually had the same question a while back and crunched the numbers...

Short answer is that your actual outs are likely less than 9, but that your odds of hitting the flush remain the same if you calculate it either way. Ratio of (remaining diamonds / remaining cards) equals ratio of (unseen diamonds [9]/ unseen cards[47]).

The numbers work out if you assume that the distribution of diamonds across the folded cards averages 25%.

[/ QUOTE ]

Maybe this is wrong, but I'm going to throw it out here for so the rest of you can tell me what you think.

Say you've got Ax diamonds as in the original post...

Flop is unpaired and contains 2 diamonds, and say you have 4 opponents. If 2 of the 4 opponents fold for a single bet, I think it is reasonable to assume that neither of the folded hands were suited diamonds. So, then can't we assume that we've seen 2 more cards?

So the number of unseen cards goes from 47 to 45. I know it doesn't change much, but wouldn't this be accurate? Or, am I missing something?

-RMJ

[/ QUOTE ]

You can certainly adjust downward, though im not sure 2 cards is the right number. The average number of non-diamonds when neither hand has 2 diamonds would be greater than 2 I would think. (Though thats somewhat offset by some small chance that 2 diamonds might fold since you hold the A)

[FredBoots]

Thanks Uncle_Billy and Copernicus. I think I understand it now.

I was so confused that I wrote a computer program that simulated hitting your 4-to-a-flush on the turn and/or river. I found the results were the same as predicted by the theory: 35%.