The Tao of Gaming

Boardgames and lesser pursuits

President’s Day Web Walking

  • Michael Chwe’s manuscript on “Folk Game Theory,” which analyzes Jane Austen, Oklahoma!, Brer Rabbit, and other popular stories from a game theoretic perspective, is certainly interesting. Although I’m stunned that the opening sentence of his manuscript isn’t “It is a truth universally acknowledged, that a single man in possession of a good fortune, is not in a Nash Equilibrium.”  (or some variant). [Hattip to Cheeptalk]
  • By my calculation, the odds of one partnership in bridge having NO high card points between them happens roughly once in every 21,700 hands. So that’s why it was the talk of the bridge club last week. Update — You know, there are 16 honor cards, not 12. So by my corrected calculation it’s once every 1,950,000 hands. Update Squared — Half that. Once every 975,000 hands. See below
  • I do enjoy strategy articles based off of massive data analysis of online games. Even for a trifling like 7 Wonders.
  • My pointless quest up the total fan count leaderboard for Rock Band 3 continues apace. I was going to pre-order the Squier pro-guitar from Amazon, but apparently Best Buy has an exclusive. What are ya’ll doing for that? (If anything).

So, how’s your weekend going?

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Written by taogaming

February 20, 2011 at 10:42 am

3 Responses

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  1. Wouldn’t it be half that, since either partnership could have no high card points?

    frunk

    February 20, 2011 at 10:33 pm

    • Well, those were the odds of the combination. You do the combination twice (once for each partnership) but the second partnership’s cards are totally dependant on the first.

      So, yes, those were the odds of dealing out 26 cards without an Ace-Jack. I guess I should have added it with the odds of dealing all 12. I’m not sure if that would be half, or smoe other number. OK, let’s try it a 3rd time.

      We want # of possible NO honor hands + # of all possible ALL honor hands divided by # of possible hands. And, in this case, a hand is 26 cards.

      T1 = # of 2-hands with no honors = 36 choose 26.
      T2 = # of 2-hands with all honors = 16 choose 16 * 36 choose 10 (which reduces to 36 choose 10)
      T3 = # of 2-hands = 52 choose 26
      T4 = (T1 + T2) / T3
      Via Google Calc
      T1 = 254 186 856
      T2 = 254 186 856 (which makes sense)
      T3 = 4.95918533 × 10^14
      T4 = 1.025 x 10^(-6)
      1/T4 = about 1 in 975,450

      So, half of that it is. Sigh.

      taogaming

      February 21, 2011 at 10:48 am

      • Math is hard. I’m going shopping. For a game.

        taogaming

        February 21, 2011 at 10:48 am


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