## How to get probabilities wrong

2010-01-30

A couple of days ago, one of my friends asked me to calculate some probability related to Texas hold’em. The probability he wanted to calculate was quite simple, but exactly this is the problem. The easier it looks to calculate, the more opportunities for mistake. This reverse can be also true, if it’s hard to calculate you probably won’t even try.

The problem is the following: given that the 2 cards you have in your hand are of the same suit, what is the probability that you will get 4 cards of the same suit after the flop.

Since people learn from their mistakes, but smart people learn from other people’s mistakes, lets see how to calculate this probability. The way it works is that you know 2 cards so you have 50 left. Now, it can be argued that the other players have each 2 cards too, thus it will be less than 50 cards in the game. But from your point of view it really doesn’t matter if the other players have cards or how many players are there. Imagine the following scenarios:

• You have a pack of 52 cards. You give yourself 2 cards, then you draw some cards and just leave them on the table, then you draw 3 more and put them face up (the flop in Texas hold’em). Here it is obvious that you have 50 cards and you just draw 3 of them randomly
• You have a pack of 52 cards. You give 2 cards to you and a number of other players and then draw 3 more and place them face up. There is no difference in the way the cards are drawn compared to the previous scenario.

Of course if you would knew the cards of the other players you could calculate more accurately the probability that we’re looking for. But based only on the knowledge of your 2 cards you can use the first scenario regardless of the number of players at the table. It’s the same as if you have a dice. If you just know it’s just a dice you can say the probability to get the number 1 is 1/6. But if you would know in detail the mass distribution of the material of the dice, the initial position of the dice, and the initial force and moment you could give a more accurate probability to get number 1. But this is already getting into the philosophy of probability so lets leave this for another post.

So we have 50 cards and we draw 3 of them. This means C(50, 3) is the total number of outcomes. But we already have 2 card of a particular suit and we want 2 more of the same suit. Since we are left with 11 cards of our suit of interest in the rest of the cards, and we want any 2 of them that means C(11, 2) favorable outcomes. And for the third card in the flop we can have anything else, so that means 50 – 2 = 48 outcomes. This gives us:

$p = \frac{C(11, 2) \cdot 49}{C(50, 3)} = 13.47\%$

(Un)fortunately we had the right result, and it wasn’t what we got. The mistake was that for the third card we multiplied by the number of cards remaining in the deck. However there are cards of the same suit after we draw the 4 we were interested in. So what we calculated was the probability to get 4 or 5 cards of the same suit after the flop. The requirement was to calculate the probability to get only 4 cards of the same suit.

So the correct answer if that for the third card we can have any card of another suit. That means 39 outcomes giving the right result of:

$p = \frac{C(11, 2) \cdot 39}{C(50, 3)} = 10.94\%$