Optimal stopping in a card game

A card game is played with eleven cards: an ace, two, three, ... , nine, ten, and a joker. These cards are
thoroughly shuffled and laid face down. You can flip over as many cards as you wish, but they must be flipped over one by one. Once you flip over the joker card, the game is over and you win nothing. If you stop before the joker card is flipped over, you win as many dollars as the sum of the values of the cards you have flipped over (the ace counts for $1). What is an optimal stopping rule? What is the maximal expected gain?

Should stop when the value of the banked money * the likelihood of flipping the joker exceeds the median value of all the non joker cards.

That might be a little conservative as it doesn't take into account building equity by not stopping

Wouldn't know an optimal rule to follow, but I think this might be somewhat close: flip four times then stop unless you are under $21, in which case keep going.

(I didn't do any math, just extrapolated from another similar problem I've heard)

What do you pay to play?

Let n be the number of cards still in the deck.
Let b be the amount of money you have banked.
Let s be the sum of the values of the cards still in the deck.

Then your expected return on the flip is:
(s – b) / n

Optimal play would be to take the flip if this (the expected return) is positive.
The expected return is positive when s > b

Since, s + b = 55

This means you should take the flip if you have $27 or less banked.

PeeMcGee, this is indeed the optimal strategy for the game. Under this strategy the player's average gain is about $15.45 in the long run. Therefore, the payoff odds of a casino offering this game should be no more than 15:1. The game is discussed in greater generality in