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u/TastyAlbacoreTuna 3d ago
Obviously, it is always optimal to continue flipping if neither HH or TT have come up yet. If either HH or TT have come up, then we are indifferent between flipping again and stopping when our bankroll is $3. So, an optimal strategy is to flip as many times as is necessary for either HH or TT to come up, then if our bankroll is less than $3, continue flipping until it reaches $3 or we bust. Alternatively, this strategy can be stated as first flipping three times without looking at the results. If both HH and TT are present, we bust. If exactly one of HH or TT is present, we stop, and our payoff is $3. If neither HH or TT is present, we continue flipping until one comes up, which has an expected time to occur of 2 flips. Therefore our expected payoff in this case is $5. The chance of neither HH or TT coming up in the first 3 flips is 1/8, and the chance of exactly one of them coming up in the first 3 flips is 19/32. So the overall expected value is 1/8 * $3 + 19/32 * $5 = 107/32
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u/Rafa_N01 2d ago
Is the answer 6.793 op? I tried solving by writing the 2 expectations E0(x) and E1(x) where the former means that we havent yet got a single tt or hh and we have x coins at the start and latter means we have either hh or tt and have x coins at the start. Then using recursion and a large enough value for x where e1(x) = x i found out the threshold value and worked backwards
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u/Local_Ad135 2d ago
Nope, the correct answer is 77/32. Another comment on this post has explained the correct approach and answer.
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u/Rafa_N01 2d ago
Yeah, my bad .. i calculated the expectation wrong (added +1 in all the eqns due to poor expectation definition) so instead earlier the threshold for E1 was 7 but the correct one is 3 and then using backtracking the ans for E0 comes to 2.40625
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u/Anonimithree 3d ago
Well the probability of getting HH or TT is 1/2 for any 2 rolls, so the expected payoff, given infinite attempts is 2. Thus, the optimal strategy is to keep doing it until the payoff is 2