# Unbiased coin game [closed]

Suppose Alice and Bob want to play a game sharing an unbiased coin.

In her turn, Alice flips the coin only once. If she gets heads she sums one point, none otherwise. Then she hands the coin to Bob.

To make things more interesting, each turn he chooses how many times he'll flip the coin beforehand. If he decides to flip $$k$$ times and he gets heads $$k$$ times in a row he sums $$2^{k-1}$$ points.

$$\begin{array}{|c|c|} \hline \text{Times he flips the coin} & \text{Probability of getting all heads in a row}& \text{Points} \\ \hline 1 & \frac{1}{2} & 1 \\ \hline 2 & \frac{1}{4} & 2\\ \hline 3 & \frac{1}{8} & 4\\ \hline \vdots & \vdots & \vdots \\ \hline k & \frac{1}{2^{k}} & 2^{k-1} \\ \hline \end{array}$$

If he gets tails anytime, his turn is over and he doesn't sum points.

The player who reachs 1000 points first is the winner.

Bob knows his maths, so he always chooses k to be the number that maximises his chances of winning.

What is the probability of Bob winning the game?

## closed as off-topic by Saad, ancientmathematician, Gibbs, Jose Arnaldo Bebita-Dris, Mark S.Sep 22 '18 at 11:58

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• Can Bob choose a different $k$ each turn? – Mike Earnest Sep 21 '18 at 17:34
• Yes. I've missed that detail. – Luciano Arlandis Sep 21 '18 at 17:41
• Project Euler asks that you not solicit help for their problems. – Ross Millikan Sep 21 '18 at 20:54

Going all-out $$k=11$$ every time gives B around $$1/e$$ chance of losing, compared with $$3/e^2$$ for $$k=10$$, and $$71/3e^4$$ for $$k=9$$. So choose $$k=11$$.
Hint: Make a $$1000\times 1000$$-table containing at the index (i, j) the probability that Bob will win when Alice has i points and Bob has j points, and it is Bob's turn. And maybe another table storing what is the best move for Bob in each situation. Then start to calculate the table from the end, so you can utilize the previously calculated probabilities in the calculations for smaller $$i$$ and $$j$$. Also, find the $$k$$ ($$T$$ in ProjectEuler's notation) that maximizes the probability.