Question 994021
0.4^10=0.0001
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Probability wins 0=0.6^10=0.006
Probability wins 1=10*(0.4)*(0.6)^9=0.0403
wins 2= 10C2(0.4)^2)(0.6)^8=0.1209
wins 3=10C3(0.4)^3(0.6)^7=0.2150
wins 4=10C4(0.4)^4(0.6)^6=0.2508
These sum to 0.6330  This is what we don't want.  The probability of winning at least 5 is everything else or 1-0.6330=0.3670
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This is the probability of 1-probability of winning no hands, and that is 0.994
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That would be 0.4^4, since this is considered independent=0.0256
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Expected value is 10*-0.4 or 4 hands.