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

    1. Ganzir says:

      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.


    2. Ganzir says:

      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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