#1
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Quick question
Sorry if asked a million times. What are the odds you'll have the same hand as another player in a full (10) ring game?
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#2
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Re: Quick question
Zero.
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#3
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Re: Quick question
[ QUOTE ]
Sorry if asked a million times. What are the odds you'll have the same hand as another player in a full (10) ring game? [/ QUOTE ] P(pair)*P(matching pair) + P(XYs)*P(other XYs) + P(XYo)*P(other XYo) = (1/17)*P(matching pair) + (4/17)*P(other XYs) + (12/17)*P(other XYo) From the inclusion-exclusion principle: 1/17*[9*1/C(50,2)] + 4/17*[9*3/C(50,2) - C(9,2)*3*2/C(50,2)/C(48,2) + C(9,3)*3*2*1/C(50,2)/C(48,2)/C(46,2)] + 12/17*[9*7/C(50,2) - C(9,2)*(4*3 + 2*2 + 1*2)/C(50,2)/C(48,2) + C(9,3)*(4*3*1 + 2*2*1 + 1*2*1)/C(50,2)/C(48,2)/C(46,2)] =~ 4.1%. |
#4
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Thanks Bruce ......n/m
n/m
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#5
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Re: Thanks Bruce ......n/m
Yeah thanks a lot for the help.
Like Iv'e said before in another post, I'm only 17 and my math knowledge goes up to only a first semester of AP Calculus AB, so I really have to learn most of this stuff. But thanks for helping me, I'll get right to reading your post (we learn something new every day [img]/images/graemlins/smile.gif[/img]!) From now on I won't psot any math unless I know it's right, so I don't make myself look like an idiot. |
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