[Math] What will be the count of “4cardstraight” in a Poker Game

combinatoricspokerprobability

Let's modify the poker hand(fake): "4cardstraight" — i.e. a straight, but with only 4 cards in a row instead of 5.

Rules of the game:

  1. A hand is counted as a "4cardstraight" if it includes 4 cards in a row, but not 5 (so it's not a regular 5-card straight).

  2. When a hand qualifies as multiple types, it should be counted as the highest-value hand; for this purpose, the value of "4cardstraight" will be just below a straight.

In such game what will be count of "4cardstraight" ?

I found an answer for the regular poker game to count the straight at wikipedia page and explanation in this answer.
So my approach is to count all the number of sequences(having a straight of 4 and 5) and then subtract 10*4^5 from the answer to get count of "4cardstraight".

But I am not able to figure out that how can I calculate the count for all straight of size >=4.

Thanks

Best Answer

It seems straightforward to count directly. From your count of straights, it looks as if you are allowing Aces high or low.

Viewing by bottom card, there are $11$ types of $4$-card straight. But $2$ of the types are special, the Ace low and the Ace high. For each of these $2$ types, the cards that make up the straight can be chosen in $4^4$ ways, and the odd card in $44$ ways. (if we are building a $4$-card straight of type Ace, 2, 3, 4, we will not be using any alredy chosen card, and we must avoid 5's). This gives a total of $(2)(4^4)(44)$.

For each of the $9$ other types, the odd card can have any of $40$ values. For example, if we are building a $4$-card straight that goes 8, 9, 10, J, we must avoid the 7's and the Queens. This gives a total of $(9)(4^4)(40)$.

Add up the two numbers obtained.

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