# Aces in the pack

Today’s big question is about poker. For some reason, statistics books shy away from gambling, reasoning that it’s somehow harmful or evil. It’s true, it can be addictive (although so can, for instance, using the computer or reading books) but it’s actually the whole reason statistics exists. (This Italian lad, Cardano, wanted to figure out how to win more money playing dice, so he figured out the basics and took everyone to the cleaners. He lost it again, though, but them’s the breaks.)

In any case, I like gambling. Or rather, I hate actually putting money on stuff, because I’m a skinflint and don’t like losing stuff, but I like the maths of gambling. Which is almost the same thing, right?

So, the problem I want to attack today is, what’s the probability of getting four aces in a Texas Hold’em poker hand? If you don’t know the rules, look them up. It’s a good game. Here are the basics:

- everyone gets two cards for themselves.
- there’s a round of betting.
- three community cards (the flop) are dealt into the middle - these can be part of anyone’s hand.
- there’s another round of betting.
- another community card (the turn) is dealt, followed by a round of betting.
- a final community card (the river) is dealt, followed by a final round of betting.
- anyone who hasn’t given up takes part in a showdown to see who has the best hand - which wins the money everyone has bet.

The problem of your hand being four aces boils down to four cards out of a random seven cards all being aces. There are a couple of ways of working this out, at least:

1) The probability tree way. If you do it ‘properly’ thinking of every possible card, you get $52 \times 51 \times 50 \times 49 \times 48 \times 47 \times 46$ branches. That’s about a trillion, so don’t do it that way. You could also, more sensibly, do it as ‘ace/not ace’, which trims the tree down to 128 branches - but still too many for a quick problem.

2) The combinations way. This way works out the probability of getting one version of the hand you want and seeing how many ways there are to shuffle it around.

So, let’s say the first four cards are aces. The chances of that are $\frac{4}{52} \times \frac{3}{51} \times \frac{2}{50} \times {1}{49} = \frac{1}{13} \times {1}{ 17} \times{ 1}{25} \times {1}{49} = \times{1}{ 270,725}$ - about four in a million. We don’t care about the probabilities of the other three cards right now.

How many ways are there to arrange four aces and three goats? It turns out that Pascal’s triangle is your friend. Because the aces are interchangeable (and so are the goats), the number of ways you can arrange the cards is $^7C_3$, or 35. To get the probability of four aces in your hold ‘em hand, you simply multiply 35 by the number from before - it’s $\frac{1}{7,735}$. A lot smaller than I thought!

Here’s a challenge question for you: what’s the probability of you being dealt two aces? And what’s the probability of you getting a hand of four aces, given that you have two aces already?

(Image by Merydith used under a Creative Commons CC-by-sa licence.)