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Re: The Roulette Conundrum
Allright, here's my stab at the question.
It will be one if the first event gives a negative (non-zero) and the second gives a positive.
Chances are = (probability of negative in first)(probability of positive in second)
= (37/38)(1/38)
Right? now taking two as our hopes,
Chances = (probability of negative in first)(probability of negative in second)(probability of positive in third)
= (37/38)(37/38)(1/38)
Similarly for three,
= (37/38)(37/38)(37/38)(1/38)
and for the n'th,
= ((37/38)^n)(1/38)
Seeing that 37/38 is less than one, I'd put my money on one. If possible, zero, because it gives a probability of 1/38, the highest one.
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ronthepon, capitals avoided. 
And don't ask me why. |
