Date: 12/10/2002 at 13:23:01 From: David Stadler Subject: Poker probability From a standard 52-card deck you pull out a 5-6-7-8 (suit is irrelevent for the problem). You shuffle the remaining 48 cards and deal 3 more cards. What percentage of the time will you complete the 5-card straight (by receiving a 4 or 9 with any one of the three cards dealt). My friends and I disagree on the answer. Here is my math: 1st draw = 16.6667% 2nd draw = 17.0213% 3rd draw = 17.3913% adding the three chances together I get 51.08% you will get a 4 or 9. My reasoning: Ignoring the slight increase in percentage when a non-winning card is dealt, your odds are 8/48 or 1/6, similar to dice. If I select one number from a die and roll 3 times (3 cards), my odds are 50% to roll my number. Please help us settle this dispute - it is amazing how stubborn we all are!
Date: 12/10/2002 at 20:20:07 From: Doctor Ian Subject: Re: Poker probability Hi David, The problem with these 'at least one' situations is that you have to consider _all_ of the possibilities: p(4,-,-) '-' = not 4 or 9 + p(-,4,-) + p(-,-,4) + p(9,-,-) + p(-,9,-) + p(-,-,9) + p(4,9,-) + p(9,4,-) + ... It's a headache, to say the least. Fortunately, there is a trick you can use that works very nicely: p(at least once) = 1 - p(never) Does that make sense? There are four 4's and four 9's remaining in the deck, which has 48 cards. So the probability of drawing something other than a 4 or a 9 would be (48 - 8)/48 The probability of doing it again would be (47 - 8)/47 And again, (46 - 8)/46 So the probability of missing on all three draws would be 40 39 38 -- * -- * -- 48 47 46 5 39 19 = -- * -- * -- 6 47 23 5 13 19 = -- * -- * -- 2 47 23 = 1235/2162 = 0.57 So the probability of hitting at least one would be p = 1 - 0.57 = 0.43 Does this help? - Doctor Ian, The Math Forum http://mathforum.org/dr.math/
Date: 12/12/2002 at 10:16:39 From: David Stadler Subject: Thank you (Poker probability) Dr. Ian - Thank you very much for explaining the poker probability question to me. I made a program to simulate and came up with the same answer before I sent the question, but still couldn't see what I was missing. Thanks again. D. Stadler
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