Problem Link:
Difficulty:
Easy
Pre-requisites:
Probability
Problem:
In a game between team A and team B, you are given the probability that team A will win. If you bet M rupees on team X, and team X wins, then you gain 2 * (1-PX) * M rupees, whereas if team X loses, then you lose this M rupees. Starting out with 10000 rupees, what is the expected maximum value you can have, if you place your bets optimally?
Explanation:
We are given PA, and PB (= 1-PA). Let us bet M rupees on A and N rupees on B. Then, the expected gain/loss we will have is:
PA * (Payoff for team A winning) + PB * (Payoff for team B winning)
= PA * (2 * PB * M - N) + PB * (2 * PA * N - M)
= M * PB * (2 * PA - 1) + N * PA * (2 * PB - 1).
Note that, M * PB>= 0, and N * PA>= 0. Therefore, if PA> 1/2, the above is monotonically increasing in M, and decreasing in N. Thus, it is best if we bet M = 10000, and N = 0 in this case. If PA< 1/2, we have it monotonically decreasing in M, and increasing in N. Thus, in this case, it is best we bet M = 0, N = 10000.
Thus, final answer is as follows:
f(PA):
if(PA >= 0.5):
return 10000 + 10000 * (1 - PA) * (2 * PA - 1)
else:
return 10000 + 10000 * PA * (1 - 2 * PA)
Setter's Solution:
Will be updated soon.
Tester's Solution:
Can be found here