After getting into BattlePoker as I mentioned in my previous blog post, I’m now sampling regular poker with fake money tables on PokerStars. Surprisingly, I was allowed to include a period in my username, so I chose this blog’s domain name: If you came here after seeing my name in PokerStars, welcome 🙂


3 thoughts on “PokerStars

    • My initial instinct is (7 choose 5) divided by (52 choose 7), but that’s obviously far too small. I’ll think about it some more and get back to you.


    • 52!/45! = 674,274,182,400 possible 7-card configurations
      Divide that by 7! = 133,784,560 distinct 7-card configurations, all with equal probability

      Now we need to find how many distinct 7-card configurations contain a straight and divide that by 133,784,560 to get our answer.

      There are 9 distinct straights in terms of face value, ranging from (2,3,4,5,6) up to (10,Q,J,K,A). Any of those 5 cards can have 4 different suits, so multiply 9 by 4^5 = 9,216 straights. Given that you have 5 cards which form a straight, there are (52 – 5) * (52 – 6) = 2,162 possibilities for the other two cards, divided by 2! to make them distinct. So there are (9,216 * 2,162)/2! = 9,962,496 distinct 7-card configurations which contain a 5-card straight.

      Final answer: 9,962,496/133,784,560 ~= 7.5%

      I still am not certain this is right, so I’ll think about it some more later.

      Liked by 1 person

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