# Problem of the week #3 - April 16th, 2012

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#### Jameson

Staff member
Thank you to everyone who participated last week or submitted a problem for us to use! We hope you guys are enjoying these. This problem is for our young poker players .

Problem:

In Texas Hold 'Em a flush (5 cards of the same suit) beats a straight (5 cards not of the same suit in sequential order), although to many it seems like having a straight is more difficult than a flush. Calculate the probability of a flush (excluding straight flushes and royal flushes) and a straight (excluding straight flushes and royal flushes) to demonstrate why it's correct to say that a flush beats a straight. Show all work and explain each step. Copying the answers from other websites without any explanation is not a solution!

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• Chris L T521

#### Jameson

Staff member

Solution:

First we must establish how many distinct poker hands are possible. Since order doesn't matter, it is equal to $$\displaystyle {52\choose5}=2,598,960$$

We need to exclude all straight flushes from our calculations (a royal flush is a kind of straight flush). The number of distinct straight flushes is 10*4=40. This is derived from the fact that a straight flush is a pair of 5 cards of the same suit in sequential order, the lowest being A2345 and the highest being 10JQKA. The starting card ranges from an A to a 10, giving a 10 total of 10 possible straight flushes for a given suit, which is then multiplied by 4 to account for the 4 suits.

The number of flush combinations is calculated by $$\displaystyle {13\choose5}*4=5,148$$. There are 13 cards in a each suit and we need any combination of 5 of them for a flush and then to account for all 4 suits. We also need to subtract the number of straight flushes from this total, bringing the number of flushes that are not straight flushes as well to 5,148-40=5,108.

The number of straight combinations is calculated as follows. As stated above, the lowest straight is A2345 and the highest one is 10JQKA, giving a total of 10. Each of these cards can be any of 4 suits, so the number of straights is $$\displaystyle 10*4^5=10,240$$. Again we need to subtract the number of straight flushes from the total, bringing the total to 10,240-40=10,200.

So putting this all together, the probability of a straight is $$\displaystyle \frac{10,200}{2,598,960}$$ and the probability of a flush is $$\displaystyle \frac{5,108}{2,598,960}$$. This is consistent with the poker hand rankings although may be counter intuitive at first.

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