Originally posted by wolfgang59
So we are in Tibet at the Holy Temple of Gangwolf and two novices are eagerly awaiting their final test before entering the order. Unfortunately the monastery is almost fully booked; there being only one vacancy.
In time honoured tradition they are given a challenge. Who can get highest up the temple stairway without stumbling?
The novices can stop ...[text shortened]... how their opponent has done.
Can you formulate a solution for P% chance of steps crumbling?
To a Tibetan, what is winning?
But, seriously. You need to know how many steps (N) there are, to make your chances 50/50 against an equally gifted opponent, which is the best you can expect to do.
Then, saying there are N steps, give a weight of .9 to step 1, .9*.9 to step 2, .9*.9*.9 to step 3, and so on, up to .9*.9... N times, to step N. Increment the sum as you go along, identifying each increment with its step number (e.g., step 1 is all numbers >0 to .9, step 2 is all numbers >.9 to 1.71, etc.) Add the N results together to obtain M. Randomly select a number R, >0 to M inclusive of M. Find at which step R is contained, in the range of sums arrived at between that step and the next step. Try to get to that step, and if you do, stop.
This way, against your equally rational opponent, your predicted chances of stopping while still in the game, are the same as your actual predicted chances of getting to that step. The optimal strategy will be the same for your opponent.
If you have no idea how many steps there are, look around and think about it and guess at N.
Do not count on your opponent knowing anything about Nash equilibria.