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#1
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[ QUOTE ]
Of course, there's no answer to this problem. But if there were, it would look like this: It's 1 in 40.833. We get this answer from realizing there are 50 cards left in the deck. There are 4 aces in the deck. (4/50)*(3/49)*5 players remaining = 2.449% of the time, or one out of 40.833 times. This is all theoretical of course. In real life questions like this don't have answers. Take your question, write it out in cursive and put it under your pillow and hope the statistics fairy leaves you a gleaming silver dollar in the morning. [/ QUOTE ] On the Broadcast they said on a 9 handed table its 1 in 24. |
#2
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The question was for a six handed table. Multiple the original .0025 number or whatever it was by eight instead of five and you'll get the right answer.
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#3
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with 50 unknown cards you can have a posiblity of 1225 hands 6 are AA. Against 1 opponent then (6/1225).
Against X number of opponents then aprox: 1 - (1219/1225)^x If x approx. 2 then .00977 3 then .01462 4 then .01945 6 then .02903 9 then .04323 Matt |
#4
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[ QUOTE ]
with 50 unknown cards you can have a posiblity of 1225 hands 6 are AA. Against 1 opponent then (6/1225). Against X number of opponents then aprox: 1 - (1219/1225)^x If x approx. 2 then .00977 3 then .01462 4 then .01945 6 then .02903 9 then .04323 Matt [/ QUOTE ] x*6/1225 is more accurate for any number of players. The error is C(x,2)/C(50,4). Your independence approximation works better for some problems where several players can have a hand, such as the probability that any pair is out. |
#5
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[ QUOTE ]
[ QUOTE ] with 50 unknown cards you can have a posiblity of 1225 hands 6 are AA. Against 1 opponent then (6/1225). Against X number of opponents then aprox: 1 - (1219/1225)^x If x approx. 2 then .00977 3 then .01462 4 then .01945 6 then .02903 9 then .04323 Matt [/ QUOTE ] x*6/1225 is more accurate for any number of players. The error is C(x,2)/C(50,4). Your independence approximation works better for some problems where several players can have a hand, such as the probability that any pair is out. [/ QUOTE ] I would have guessed that the independence assumption would work better than the x*6/1225 that you suggest, but of course I haven't calculated it. The precise calculation would get quite messy, I think. It would require calculating the probability of having a total of zero, one, or two Aces in four of other five hands (assuming none of these four hands contain AA), and then calculating the probability of AA in the 6th hand, and multiplying by five. Right? Could you please explain how you calculated the error here, C(x,2)/C(50,4)? |
#6
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[ QUOTE ]
[ QUOTE ] [ QUOTE ] with 50 unknown cards you can have a posiblity of 1225 hands 6 are AA. Against 1 opponent then (6/1225). Against X number of opponents then aprox: 1 - (1219/1225)^x If x approx. 2 then .00977 3 then .01462 4 then .01945 6 then .02903 9 then .04323 Matt [/ QUOTE ] x*6/1225 is more accurate for any number of players. The error is C(x,2)/C(50,4). Your independence approximation works better for some problems where several players can have a hand, such as the probability that any pair is out. [/ QUOTE ] I would have guessed that the independence assumption would work better than the x*6/1225 that you suggest, but of course I haven't calculated it. The precise calculation would get quite messy, I think. It would require calculating the probability of having a total of zero, one, or two Aces in four of other five hands (assuming none of these four hands contain AA), and then calculating the probability of AA in the 6th hand, and multiplying by five. Right? [/ QUOTE ] Nope, there are only 2 terms in the exact inclusion-exclusion calculation since only at most 2 players can have AA. [ QUOTE ] Could you please explain how you calculated the error here, C(x,2)/C(50,4)? [/ QUOTE ] The exact calculation is x*6/1225 - C(x,2)/C(50,4). See the post about the inclusion-exclusion principle. The first term double counts all cases where 2 players have AA, and the second term subtracts this off. Here is a comparison of the methods. <font class="small">Code:</font><hr /><pre> #opponents N P(AA) exact N*6/1225 1-(1219/1225)^N 1 0.49% 0.49% 0.49% 2 0.98% 0.98% 0.98% 3 1.47% 1.47% 1.46% 4 1.96% 1.96% 1.94% 5 2.44% 2.45% 2.43% 6 2.93% 2.94% 2.90% 7 3.42% 3.43% 3.38% 8 3.91% 3.92% 3.85% 9 4.39% 4.41% 4.32% 10 4.88% 4.90% 4.79% </pre><hr /> |
#7
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Thanks for the explanation!
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#8
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This is quite wrong.
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#9
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So say why? I hate this kind of post.
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#10
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This is quite wrong. [/ QUOTE ] Prove it or shut up. |
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