You can play a game with me. I have $$10$$ coins, each showing $$\1$$ on one side and $$\text{GAME OVER}$$ on the other. All coins are fair, trust me. The game is as follows: each turn, I will toss all the coins still in the game. If all of them shows $$\text{GAME OVER}$$, then game over, you lose, go home with nothing. If any of them shows $$\1$$, you score a dollar for each of them, I will put them aside, and you get a choice: do you want to go home with what you have, or push your luck, try to get more money, let me toss the remaining coins (those with $$\text{GAME OVER}$$ face side up)?

Here's an example game. I toss the coins, you get $$\1$$ on six coins, so the prize is now at $$\6$$. If you quit now, you collect $$\6$$; if you continue, I will toss the remaining four coins. Suppose that you continue; among the four coins, two more come up $$\1$$, adding to your prize pool to $$\8$$. You could have quit with $$\8$$, but no, you decide to continue again, with only two coins left. I toss them, and they come out $$\text{GAME OVER}$$ for both. Game over, you lose, you don't get the eight dollars.

Let $$E$$ be the expected amount of money you can get by playing optimally. What is $$\lfloor 10^4E \rfloor$$?

(You may use a calculator or similar to do calculations. You may also try simulating it, but I'm pretty sure you won't get it simply with simulation.)

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