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Real Odds - Low Content

  • 04-12-2007 11:06pm
    #1
    Registered Users, Registered Users 2 Posts: 2,953 ✭✭✭


    Just something silly i was thinking and im sure theres something simple to it. You know when we calculate odds we divide the number of cards in the deck / outs to get odds of improving our hand.

    Well at a 6 handed table 10 of the 52 cards are dead so for example theres probably going to be at least 3 of each suit that are dead outs on average so why do we not allow for this when making calculations or do we or is it so small its not worth considering, dosent seem that small to me?


Comments

  • Registered Users, Registered Users 2 Posts: 39,902 ✭✭✭✭Mellor


    It is actually considered, and it makes no difference. This is because missing outs decrease our chances and missing non-outs cards increase our chances. A single miss out decreases our chances more than a single missing non-out, but the difference between the increase and decrease is exactly proportional to the chances of that card being removed.


    EDIT: An example might make more sense with the above.

    A bag has 8 balls in it, 2 black balls and 6 white balls.
    If a ball is chosen at random there is a 25% chance that it is black, 2/8

    Now if a unknown ball is removed at random, our chances of chosing black are either 1/7 or 2/7 depending on which was removed.
    6 out of 8 a white is removed and we increase to 2/7, and 2 out of 8 we decrease to 1/7, so
    (6/8*2/7) + (2/8*1/7)
    =12/56 + 2/56
    =14/56
    =2/8......same as before


  • Registered Users, Registered Users 2 Posts: 2,953 ✭✭✭dvdfan


    Thanks Mellor, it took a few minutes to get my head around it but i understand now


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