<t>Maybe another way to answer to this question.Before the clue of left-hander was discovered, we have:60% of being guilty for the main suspect and 40% for anybody else (as we suppose someone is guilty).With the clue, no change for our main suspect (still 60%) and now 2.8%= 40% * 7% for the rest of ...

If you note f(i) the number of such sequences where i is the number of "+1" and f(0)=1, you have the following:f(n)=Sum (j=0 to n-1) f(j)*f(n-1-j)for i=50, the result is close to 2*10^27

A good approximation is 0.5 + mu/(2*pi*(1+sigma*sigma))^0.5Easy to find by using N(x+mu)=N(x)+Int(x-->x+mu)exp(...), the first term gives 0.5 by integrating and the second could be approximated by mu*exp(-x*x/2) to give the solution.

Repoman,Just a little remark,I think you need all E(m,0)=m/p & E(0,n)=n/q to make the double recurrence.Not so easy to calculate all ways to reach E(n,n) but I will try to have a closed form with this approach.

I think it works even if it's not a random walk.If you have p(i) the long term probability to be at vertex i, the average time to return from i to i is 1/p(i).

<t>My first answer on this forum.I think it's greater than 1/5.If you take how many numbers you pick between 1 and 5, you have the following case:* if you take 1,2,3 or 4 numbers, their sum is 0,1,2,3 or 4 (mod 5) with equal probabilities* if you take 0 or 5 numbers, their sum is 0 (mod 5) and you c...