For example, if 1 head and 1 tail so far, one should definitely keep tossing. if he gets head the payoff is at least 2/3, and if it is tail then he can get better than 1/3 by keep tossing, thus improving the payoff.
Let f(a,b) be the expectation of the payoff if played optimally, then we are interesting in f(0,0). We can obtain the following recursion:
f(a,b) = max(a/(a+b), (f(a+1,b)+f(a,b+1))/2);
It is not easy to solve it though, since the number of states is infinite. If we fix a finite horizon then we can compute an approximate solution. Another observation is that if one keeps tossing for a long time, with high probability he gets at least 1/2 - \epsilon. Therefore, f(a,b) >= 1/2 for all possible a and b. One good way of getting a good approximate will be setting a horizon T such that f(a,b) = max(a/(a+b), 1/2) if a+b=T, and solve the recursion.
I have no idea how the optimal solution can be computed analytically.
Let f(a,b) be the expectation of the payoff if played optimally, then we are interesting in f(0,0). We can obtain the following recursion:
f(a,b) = max(a/(a+b), (f(a+1,b)+f(a,b+1))/2);
It is not easy to solve it though, since the number of states is infinite. If we fix a finite horizon then we can compute an approximate solution. Another observation is that if one keeps tossing for a long time, with high probability he gets at least 1/2 - \epsilon. Therefore, f(a,b) >= 1/2 for all possible a and b. One good way of getting a good approximate will be setting a horizon T such that f(a,b) = max(a/(a+b), 1/2) if a+b=T, and solve the recursion.
I have no idea how the optimal solution can be computed analytically.
選擇“Disable on www.wenxuecity.com”
選擇“don't run on pages on this domain”
