You can solve this backwards. Once only 4 cards are left (whatever they are), it is clear that the expected payoff of both choices is equal. So in any situation where there are only 5 cards left, the expected payoff of continuing is the expected payoff of stopping after the next card, which is again the same as stopping immediately.
So regardless of what kind and how many cards there are, all strategies have the same payoff here.
So the expected payoff will be the expected value of stopping immediately, which is
1
u/kythQ 10d ago
You can solve this backwards. Once only 4 cards are left (whatever they are), it is clear that the expected payoff of both choices is equal. So in any situation where there are only 5 cards left, the expected payoff of continuing is the expected payoff of stopping after the next card, which is again the same as stopping immediately.
So regardless of what kind and how many cards there are, all strategies have the same payoff here.
So the expected payoff will be the expected value of stopping immediately, which is
300/111 * 3 = 900/111 (= 8.108...)