Sorry, bad answer for 4

I guess as McCarreira the answer for 2 is 11 (the difference between the 2 preceding terms)an the answer for 4 is 90 (the sum of the 3 preceding terms)

Damn I got it at last... Finally it is all just as if the balls were undistinguishable since exchanging their respective places is not allowed. So my answer was the correct one

<t>QuoteOriginally posted by: bilbo1408Tochiro,What you have said here is correct, but only if the label on the balls is disregarded. I suggested the simple answer of n^m because the original question says that the balls are numbered, which gives me the impression that a difference between which spe...

<t>QuoteOriginally posted by: TochiroQuoteOriginally posted by: MarsdenD'oh! It should be C(M+1,n-1). I'd miss-handled the zero-balls-in-bucket case. I'd better give you at least the framework of a proof, in case I screwed something else up.First, lower case c(M,1)=1, which is also upper case C(M,0)...

<t>QuoteOriginally posted by: MarsdenD'oh! It should be C(M+1,n-1). I'd miss-handled the zero-balls-in-bucket case. I'd better give you at least the framework of a proof, in case I screwed something else up.First, lower case c(M,1)=1, which is also upper case C(M,0). Next, for any (M,n), we can divi...

<t>It is obvious as soon as you watch the final result : the probability of picking y among x,y is 0.5(1+p(y)-p(x)), which is strictly above 0.5 if and only if p(y)>p(x) (and the question is how you get a right guess probability strictly above 0.5 in any case). So the function p has to be strictly i...

The difference is that the p function must be strictly increasing so that p(y) > p(x) [bold]for all[bold] y>x. In your case it happens only ix x and y are of different sign

Couldn't it be said that, Y_n = X_(min(n,100) - min(n,100)/2 being a bounded martingale and tau = inf (n tq X_n=0)) being a stopping time, The expectancy of Y_tau is Y_0? As far as I know the Doob theorem can be applied to any bounded martingale...

<t>The last one is VERY tricky. Anyway I don't think the Arcsin function is known by A-level students, at least in France. But would it be, there would be hardly 1% of correct answers... among the science students! But maybe there are 2 millions candidates each year, so that 1% is more than enough t...

Not exactly, because the payoff date is stochastic as for any american option...