Push your luck?

Let $E_k$ be the expected additional gain you make from a single round of the game, if you have already have a total of $k$ , for any $0 \le k \le 9$ . Clearly $E_k \; = \; \frac{1}{2^{10-k}}\left[\sum_{j=1}^{10-k}\binom{10-k}{j}j - k\right] \; = \; \tfrac12(10-k) - \frac{k}{2^{10-k}} \hspace{1cm} 0 \le k \le 9$ Since $E_k > 0$ for $0 \le k \le 7$ , while $E_k < 0$ for $k = 8,9$ , it follows that the optimum strategy is to keep playing so long as your total gain is $7$ or less, but to take the money and run if your total gain is $8$ or more.

Let $P_j$ , for $0 \le j \le 10$ , be the expected final profit you will make if you are in the middle of a game, playing the optimal strategy, and currently have a total of $j$ . Thus $P_8 = 8$ , $P_9 = 9$ , $P_{10} = 10$ , while, for $0 \le j \le 7$ , $P_j \; = \; \frac{1}{2^{10-j}}\sum_{k=1}^{10-j}\binom{10-j}{k}P_{j+k} \; = \; \frac{1}{2^{10-j}} \sum_{k=j+1}^{10} \binom{10-j}{k-j}P_k$ It is a simple calculation that $P_0 =7.7985122$ , making the answer $\boxed{77985}$ .

Suppose $\$1$ face is heads and $\text{GAME OVER}$ face is tails.

The idea is a simple recurrence system.

Suppose $E(n)$ is the expected value that you can get with $n$ coins remaining, right before you make your choice. Thus $E(0) = 10$ ; if you survive with zero coins left, clearly you should quit, since if you "play" there will be no heads coming up and thus you lose. Also, you're looking for $E(10)$ . Clearly you shouldn't quit before you play the game, so giving the possibility of quitting at the beginning won't change the result.

Now, $E(n)$ is the maximum of the two choices: either you quit, pocketing $10-n$ dollars, or you push on. The latter is simple: $\displaystyle \sum_{i=1}^n \binom{n}{i} E(n-i)$ , corresponding to getting $i$ dollars: there are $\binom{n}{i}$ to get $i$ heads, and after getting $i$ heads, you're left with $n-i$ coins, thus the expected value is $E(n-i)$ . Note that it's not $i+E(n-i)$ or anything; the $i$ heads we have won't necessarily cash out, if it turns out that we lose later, and they are counted in $E(n-i)$ already. Also note that the summation starts from $1$ , since with $0$ heads you lose already. So $E(n) = \displaystyle \max \left( 10-n, \sum_{i=1}^n \binom{n}{i} E(n-i) \right)$ for all $1 \le n \le 10$ . By computing carefully, we obtain that $E(10) = \frac{274385754976485}{35184372088832} \approx 7.79851$ , so $\lfloor 10^4E \rfloor \approx \lfloor 77985.1 \rfloor = \boxed{77985}$ .

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This problem is inspired by Tiny Dice Dungeon , although it has a completely different setup (no fixed number of "coins", any single $\text{GAME OVER}$ loses).