Well done indeed.Long run probability of hitting any number is 2/7 = 28.6%. My solution is similar but much simpler. If you're at a given position, in the next round, on the average, you'll skip 5/2 numbers [= 1/6*(0+1+2+3+4+5)] and land into the next one. Therefore you hit ratio is 1/(1+5/2) = 2/7.Hope you enjoyed.fatih

"too complex or intractable"really, all you have to see is that recursivelyp(N) = 1/6.(1 + sum_k<N [ p(k) ] for N<=6andp(N) = 1/6.(sum_k=1..6 [ p(N-k) ] for N>6So, for N<=6, the hitprob is increasing, and after that, it is averaging. Once it's averaging, it's not going to beat the past maximum, right? Same logic holds for j-sided dice... choose j, and long-run avg is 2/(j+1) by the reasoning below. j=1 is a dumb case but the formulas still work.

kr, i personally solved it in the way u described. By the way for N<=6 you could get a closed end formula p(N)=7^(N-1)/6^N

I prefer a more formal solution.p(1) = 1/6 since the only way to win is to get 1 on the first throw.p(i = 2 to 6) = [1 plus the sum from j=1 to i-1 of p(j)]/6. The 1/6 represents the first throw being i, the remaining terms the probabilities that the first throw is i-j.p(i > 6) = [the sum from j=i-6 to i-1 of p(j)]/6 (as kr points out). The first throw is equally likely to be 1 to 6, each term represents the probability of reaching i given a first throw of i-j.This is a full interative definition of p. Before solving anything, it's obvious that the worst number is 1 and the best is 6. p is strictly increasing from 1 to 6. All subsequent numbers are averages of previous values, so they will be between p(1) and p(6).To get the long-term average, we can solve for the eigenvector of the 6x6 transition matrix. It is [6/21,5/21,4/21,3/21,2/21,1/21]. The long term average is the product of this and the transpose of [p(6),p(5),p(4),p(3),p(2),p(1)].p(i<7) = 7^(i-1)*6^(-i) (as zerdna points out).So we need the sum from i = 1 to 6 of [i*7^(i-1)*6^(-i)]/21.Define f(x) = sum from i = 1 to 6 of x^i/126 = (x^7 - x)/[126*(x-1)].The derivative of f(x) with respect to x = sum from i = 1 to 6 of i*x^(i-1)/126 which, when evaluated for x = 7/6, gives the long term average we seek.The derivative is [6*x^7 - 6*x^6 + 1]/[126*(x - 1)^2]. Evaluated at x = 7/6 gives 36/126 = 2/7.1, 2, and 3 are bad; 6 is good. All other numbers have p within 10% of the average value.

QuoteOriginally posted by: AaronI prefer a more formal solution.To get the long-term average, we can solve for the eigenvector of the 6x6 transition matrix. It is [6/21,5/21,4/21,3/21,2/21,1/21]. The long term average is the product of this and the transpose of [p(6),p(5),p(4),p(3),p(2),p(1)].Aaron, I didn't follow what ur trans matrix was. I would use T1/6 1/6 1/6 1/6 1/6 1/6 1 0 0 0 0 0 0 1 0 0 1... 0 0 0 0 1 0which would put on the top of vector next value. Or T^6 if u want a 6 new values. However, it's smth different from what u used, since [6/21,5/21.. 1/21] is not clearly an e-vector of T. It has e-vec [1,..1] though.

I'm confused on this too... Actually I spent about 60 seconds looking at the TM and then realized that I didn't recall the shortcut for computing the determinant in this configuration. But still, your ev doesn't look right... if the thing is approaching a limit, then the correct ev is (1 1 1 1 1 1), no? (or a normalized version thereof)

To unify our approaches, take your transition matrix to the limit (that is, multiply it by itself until it stops changing). Then the first row will be my eigenvector: [6/21, 5/21, 4/21, 3/21, 2/21, 1/21]. So the long-term average will be this vector times the transpose of the first six values, just as in my solution.

got itwithout working it out, I see that in fact the whole matrix is composed of identical rows... the point is that yes, the matrix does the averaging and the shift, but you have to do it this way otherwise you can't get the initial conditions into the problem - and how else are you going to get the right answer without the initial conditions in this case?good one