Mike Plotz: I got the point of Eliezer’s post, and I don’t see why I’m wrong. Could you tell me more specifically than “for the reasons stated” why I’m wrong? And while you’re at it, explain to me your optimal strategy in AnneC’s variation of the game (you’re shot if you get one wrong), assuming you can’t effectively cheat.
In some games, your kind of strategy might work, but in this one it doesn’t. From the problem statement, we are to assume the cards are replaced and reshuffled between each trials so that every trial has a 70% chance of being blue or red.
In every single case, it is more likely that the next card is blue. Even in the game where you are shot if you get one wrong, you should still pick blue every time. The reason is that of all the possible combinations of cards chosen for the whole game, the combination that consists of all blue cards is the most likely one. It is more likely than any particular combination that includes a red card. Because at every step, a blue card is more likely than a red one. Just because you pick a red card, doesn’t give you credit for anywhere a red card might pop up. You have to pick it in the right spot if you want to live. And your chances of doing that in any particular spot are less than the chances of picking the blue card correctly.
There are games where you adopt a strategy with greater variance in order to maximize the possibility of an unlikely win, rather than go for the highest expected value (within the game), because the best expected outcome is a loss. Classic example would be the hail mary pass in football. Expected outcome is worse (in yards) than just running a normal play, or teams would do it all the time. But if there are only 5 seconds on the clock and you need a touchdown, the normal play might win 1 in 1000 games, while the hail mary wins 1 in 50. But there is no difference in variance in choosing red or blue in the game described here, so that kind of strategy doesn’t apply.
Mike Plotz: I got the point of Eliezer’s post, and I don’t see why I’m wrong. Could you tell me more specifically than “for the reasons stated” why I’m wrong? And while you’re at it, explain to me your optimal strategy in AnneC’s variation of the game (you’re shot if you get one wrong), assuming you can’t effectively cheat.
In some games, your kind of strategy might work, but in this one it doesn’t. From the problem statement, we are to assume the cards are replaced and reshuffled between each trials so that every trial has a 70% chance of being blue or red.
In every single case, it is more likely that the next card is blue. Even in the game where you are shot if you get one wrong, you should still pick blue every time. The reason is that of all the possible combinations of cards chosen for the whole game, the combination that consists of all blue cards is the most likely one. It is more likely than any particular combination that includes a red card. Because at every step, a blue card is more likely than a red one. Just because you pick a red card, doesn’t give you credit for anywhere a red card might pop up. You have to pick it in the right spot if you want to live. And your chances of doing that in any particular spot are less than the chances of picking the blue card correctly.
There are games where you adopt a strategy with greater variance in order to maximize the possibility of an unlikely win, rather than go for the highest expected value (within the game), because the best expected outcome is a loss. Classic example would be the hail mary pass in football. Expected outcome is worse (in yards) than just running a normal play, or teams would do it all the time. But if there are only 5 seconds on the clock and you need a touchdown, the normal play might win 1 in 1000 games, while the hail mary wins 1 in 50. But there is no difference in variance in choosing red or blue in the game described here, so that kind of strategy doesn’t apply.