Push your luck?

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.)

Image Credit: Flickr Coins

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