Photo courtesy of Maxime Perron Caissy
In poker, when calculating winning chances, why don't we consider the cards in opponents' hands? They clearly aren't available to us.
What if one of our outs is in the opponent's hand? We can't get it, so it should be discounted, no?
Good question
This is actually a really good question. When we count outs, we actually close our eyes to this difficulty. We pretend it doesn't exist and hope it will go away.
The good news is, there's a reason why we don't have to bother. We can solve it once and for all and then it actually goes away.
So, let's do that.
Stupid answer
Say we have a set of sixes on the turn and are looking for the chances of hitting quads on the river. We have a single out, the remaining six. What are our chances of getting it?
Normally we would say that 46 unseen cards can come on the river and only one of them is good. Our winning chances are 1/46.
But now we want to consider the possibility that our card is in an opponent's hand. How can we include this in our calculations?
Assume that nine players received cards in this hand. Then there are two possibilities:
- The winning card is in the deck. Our chances are 1/34. (There are 34 cards left in the deck.)
- The winning card is in the opponent's hand. Our chances are 0.
But the probability that the winning card was dealt to an opponent is 18/46. The probability that it's still in the deck is 34/46.
Put it together and you find that the probability of getting the six is:
P = 1/34 * 34/46 + 0/44 * 18/46 = 1/46
That is, (probability we win if card is in deck)*(probability card is in deck) + (probability we win if opponent has it)*(probability opponent has it).
As we can see, all "unwanted" factors cancel out and we're back to the normal result, 1/46.
Probability is about information
In most other examples, the calculations would be more labored. But the principle holds. We will always return to the "normal" simple result.
If you think of it, this makes sense. Since we don't know where the various unseen cards are, their position cannot influence our calculation.
On the other hand, if we DO know where they are, our calculation certainly changes.
For example, if we know that villain has a pair of aces, our chances of "quadding" go up to 1/44. A total of 44 cards are unknown, one of them is our six.
/Charlie River
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