<t>Let L(0)=m, R(0)=n, i.e. we start at (m,n).E(X)=Sum_i {i*P(we reach (i,0))*P(we reach out into right pocket)}+Sum_ j {j*P(we reach (0,j))*P(we reach out into left pocket)}Note that the probability of each path from (m,n) to (i,0)*P(we reach out into right pocket) is the same: (the last fraction b...

<t>QuoteOriginally posted by: BramJIs it possible to characterize all N for which N+1 divides N!+1? Edit: easy to verify that necessary condition is N = even, but N=8 shows that this isn't a sufficient conditionSuch N are characterized by the Wilson's theorem: p is prime iff p divides (p-1)!+1So the...

<t>An afterthought:The formula P(correct)=0.5+p2/2 suggests that we can improve the probability of guessing correctly the higher number. But that is not exactly so.Note that p2 depends on the distribution from which we generate the number z, and as stilyo mentioned, that could be any distribution of...

<t>We can probably reformulate the problem as follows: Starting from the point (m,n) on the x-y plane having integral grid, we move either south or west until we reach either x=-1 line or y=-1 line. Find the average value of the x or y intercept when one of the above lines is reached.Solution: Sum[i...

<t>Suppose we define x,y,z as follows:x is the smaller of the two numbers written on paper chits, y is the greater of the two numbers written on paper chits. z = any randomly generated numberLet p1 = P(z<x), p2 = P(x<z<y), p3 = P(z>y); (p1+p2+p3 = 1)P(the proposed strategy gives a correct guess) = p...

<t>I think that we need to replace n by (n+1) in our calculations. This is because we do not have to find “the expected number of JB left when one bag gets empty”, rather, we have to find the “ expected number of JB left when we reach again into an empty bag” (in other words, we have to find the num...