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#1
10-18-2007, 01:50 AM
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Re: Prob of not drawing card in x trials...help
Wow,
You guys here in probability are wicked smart. I had a strong feeling that this solution would be significantly more complex than what I'm used to dealing with in hold 'em (creation of my own odds and outs charts and such). After about ten minutes of mulling this over I actually grasp this, sort of. I'm lost with the Pascal's triangle coefficient reference, but all is good. I can make this work to solve for other values of n. Thank you very much for the help Bruce. Take care. 
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#2
10-18-2007, 02:42 AM
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Re: Prob of not drawing card in x trials...help
Note that I recomputed the 2nd result.
[ QUOTE ] I'm lost with the Pascal's triangle coefficient reference, but all is good. I can make this work to solve for other values of n. [/ QUOTE ] You need the coefficients from Pascal's triangle for other values of N. If you refer to the Pascal's triangle given on this page, the coefficients are obtained from the Nth column, except that their alignment is a little off. They should be: N=1: 1,1,1,1,1,1... N=2: 1,2,3,4,5,6,... N=3: 1,3,6,10,15,... N=4: 1,4,10,20,... N=5: 1,5,15,... Just extend the triangle downward as far as you need. Each number is a sum of the number directly above it and the one diagonally above to the left. If you use the other form of the triangle shown later on that page, the coefficients appear on the diagonals. Another way to look at it is that the values for each N form a running sum of the numbers from the previous value of N.
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