Re: Real tough problem.....pls try...

"harry389@xxxxxxxxx" <harry389@xxxxxxxxx> writes:

Well as you said you can bet a portion X everytime so that eventually
you survive the loss and also get a gain. Instead of X being 1 -
2/(N+1) it can be 1/2 everytime for N >=3. The worst case wud be wen u
lose the bet in the Nth round wen u lose 2^N-1 - 2^N-1/2 times the
initial amount eventually leaving u 2^N-2 times the initial amount.

Your maths is decidedly suspect. What you propose is sub-optimal.

David

.

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