[Math] Probability of being dealt a straight in Poker

probability

In poker, what is the probability of being dealt a straight?

Attempt:

First, we find the size of sample space, the set of all possible 5-hand combinations, this is ${52 \choose 5}$. Now, in how many ways can we obtain a straight? So, this is my thinking,

1) For the first card, we have $14$ possibilities. Let's say we pick an ace. So the other cards must be 2,3,4, and 5.

Notice the ace can be chosen in 4 ways. Now, same argument works until first card is J since then we cant have straight. So we only have $11$ choices. Thus, we have that the number of ways to get straight is

$$ 11 \times 4 \times 14 $$

and hence,

$$ Probability \; \; = \frac{ 11 \times 4 \times 14 }{ {52 \choose 5 } } $$

Is this correct argument?

Best Answer

Your argument is not correct.

A straight is a poker hand consisting of five cards of sequential rank, not all of the same suit.

  1. Choose the low card, which may be an ace, 2, 3, 4, 5, 6, 7, 8, 9, 10.
  2. Choose the suits for each of the five cards in the hand, keeping in mind that a flush (all cards of the same suit) is not permitted.

The number of favorable hands is $$10 \cdot (4^5 - 4)$$

Related Question