Dead cards and counting outs.

[OrianasDaad]

I'm likely missing something very simple, but in my quest to understand, I must pose the question.

How come, when you are counting outs, you don't subtract the probability that one or more of your outs are dead in other hands? Numerous books and articles all use the term "unseen cards". Part of me knows that this is correct (because it isn't challenged), but another part of me wonders why this is correct. Lets use an example.

You've got an open-ended straight draw on a rainbow board with unsuited cards. Precisely 8 outs, with 47 unseen cards, or 1-4.875 to make your hand. Let's also say that 9 other players got cards. Without doing math I'm incapable of doing (yet), I'm going to assume that on average, one of the outs is going to be dead. Wouldn't you need 1-5.71 to draw to your straight, instead of 1-4.875?

I feel that my logic is wrong somewhere.

[dansalmo]

If you assume that 1 out is dead amoung the 18 cards that the other players hold, then your odds are 7 out of 29 unseen cards, not 7 out of 47. You see now why it does not matter where the outs are?

[gaming_mouse]

Calculating the probability based on "unseen" cards automatically takes into account the possibility that one of the cards is in another player's hand.

Of course, other player's betting patterns change the odds -- but this is difficult to quantify. For example, if you are on the button with a flush draw on the flop, and 6 people have called a bet before it gets to you, there is a good chance you are up against another flush draw.

However, that issue aside, it should be easy to see that the possibility of an opponent holding one of your outs does not affect the calcualtion. Think of it like this to convince yourself:

Imagine you are dealt your hand first, A[img]/images/graemlins/heart.gif[/img] K[img]/images/graemlins/heart.gif[/img]. No one else has been dealt a hand. Now the flop is dealt: 2[img]/images/graemlins/heart.gif[/img] 5[img]/images/graemlins/heart.gif[/img] 8[img]/images/graemlins/club.gif[/img]. At this point it should be clear that you have 9 outs. And the chance the next card dealt, the turn card, completes your flush in 9 in 47.

Now clearly we can now throw out the rest of the deck with the exception of that next card and it will not affect the probability we just calculated. We can throw the rest of the deck into the fireplace, and there will still be a 9 in 47 chance that the next card in the deck completes your flush.

So instead of throwing the deck out, deal in the remaining 9 players, but deal them in from the bottom of the deck, so that nobody gets the next card. So the chance that you complete your flush still will not change.

But wheather we deal the other players in from the bottom of the deck or the top of it can't matter, because the deck is randomly shuffled. So the chance that someone else holds one of our outs can't affect the calculation either.

Does that help make it clear?

gm

[OrianasDaad]

Not really. I know it is correct, and I completely understand your example, but I've still got that itch. I'll try some math. Here goes.

My 9 opponents have collectively C(47,18) sets of possible cards in their hands. I want to see what percentage of the time I'm drawing to all my outs. To do that, I'm going to define the set of hands my opponents could collectively have that don't have my outs in them, or C(47,39)

C(47, 18) = 4,568,648,125,690
C(47, 39) = 314,457,495

I get an exponential number when I do this percentage, so subtract this from the total possible sets, and get the percentage that I'm NOT drawing to all my outs:

99.993117%

I'm not sure if my math is correct, as I'm just getting into combinatorics. Someone bash me on the head if I butchered this too much.

All this being said, I know (though not intuitively) that it doesn't matter. I just don't get it.

[gaming_mouse]

Well, yes, of course most of the time one or more of your flush cards will be in someone else's hand. The point is that this doesn't affect the calculation. When you calculate odds by counting outs, the method takes this possibility into account.

You might want to try the following exercise:


NOTE: P( A | B) = P(event A occuring given that event B has already occureed)

P(make your flush) = P(make your flush | no oppo has an out)*P(no oppo has an out) +
P(make your flush | 1 oppo holds 1 out)*P(1 oppo holds one out) + ...
+ P(make your flush|opponents hold all outs)*P(opponents hold all outs)

The "long way" of doing the calcualtion (the right hand side of the eqtn) will allow you to explicitly factor in the situations that confuse you. You will see that you come out with the same answer as the counting outs method.

gm

[OrianasDaad]

Whew, that seems like some bit of work, but I'll give it a go... tomorrow.

Just to make sure I'm clear, I multiply the probablity of me making my flush based on how many outs I assume my opponents hold by the probability that my opponents hold exactly that number of outs. I then add all 10 of these probabilities (for outs 0,1,..9) together, and that will equal around .1914 (9/47)?

[gaming_mouse]

correct.