Probabilities of poker

In five-card draw poken, players are trying to obtain the best possible combination to win. There are \({52}\choose{5}\) possible hands:

\[\begin{split}{{52}\choose{5}} & = \frac{52!}{(52-5)!5!} \\ \\ & = \frac{1\cdot 2\ldots\cdot 52}{(1\cdot 2\ldots\cdot 47)(1\cdot 2\cdot 3\cdot 4\cdot 5)} \\ \\ & = \frac{48\cdot 49\cdot 50\cdot 51\cdot 52}{120} \\ \\ & = \frac{311,874,200}{120} \\ \\ & = 2,598,960\end{split}\]

One fine hand to end with, is four-of-a-kind, e.g., four aces, or four sevens, etc. How many ways are there to get such a hand? The hand is build in successive steps, so we can use the multiplication principle in our calculations.

First there is the rank choice. There are 13 ranks in a playing card deck and we chose one. Formally, we can write \({{13}\choose{1}}\) but this evaluates to \(13!/(12!1!)=13\). The second step involves the suit. There are four suits and our choices are \({{4}\choose{1}}=4\). And finally, we have the last step: the fifth card that can be anything else (except for the rank that we use for the four-of-a-kind). There are \(12\) such choices. All together, we have \(13\cdot 12\cdot 4=624\) ways to pull a four-of-a-kind hand.

And so we can write:

\[\begin{split}\text{Probability of four-of-a-kind} & = \frac{\text{number of ways to pull a four-of-a-kind}}{\text{number of all possible 5-card hands}} \\ \\ & = \frac{624}{2,598,960} \\ \\ & \approx 0.024\%\end{split}\]

Let’s look at another hand, the single pair: two of the five cards are of the same rank. The successive steps, and their corresponding factors in the multiplication principle, are:

\[\begin{split}\underbrace{{{13}\choose{1}}}_{\substack{\text{same}\\\text{rank}}}\ \underbrace{{{4}\choose{2}}}_{\substack{\text{two}\\\text{suits}}}\ \underbrace{{{12}\choose{3}}}_{\substack{\text{any}\\\text{three}\\\text{other}\\\text{cards}}}\ \underbrace{{{4}\choose{1}}{{4}\choose{1}}{{4}\choose{1}}}_{\substack{\text{of any suit}\\\text{for each of}\\\text{3 cards}}} &= 13\cdot 6\cdot 220\cdot 4^3 \\ &=1,098,240\end{split}\]

There are, in other words, 1,098,240 different ways to end with a single pair. And the probability is

\[\begin{split}\text{Probability of single pair} & = \frac{\text{number of ways to pull single pair}}{\text{number of all possible 5-card hands}} \\ \\ & = \frac{1,098,240}{2,598,960} \\ \\ & \approx 42.26\%\end{split}\]