4 ms·

There is a difference between being willing to gamble and beating a game. It is obvious to see that each time you play this game you are losing $9901 dollars (.

by Cantdog 17y ago

There is a difference between being willing to gamble and beating a game. It is obvious to see that each time you play this game you are losing $9901 dollars (.99x1+.01x-10000). That is the expected outcome each time you play the game. Probability is often not intuitive. It is easy to think you "beat" the system because you won a couple bucks. This game may better illustrate my point. There is a hat with 100 tickets. 99 of the tickets produce a $1 win. The last ticket requires you kill yourself. Would you play this game once or twice?

splat 17y ago
I'll agree with your semantics. To me, "beating a game" implies that you can consistently come out ahead over a long period of time. Nevertheless, playing an "unbeatable" game over a short period of time is not necessarily irrational. Although your expected value in your first version of the hat game is -$9901, you are not losing $9901 every time you play. If you play long enough it will average out to be this, but every time you play you will either win $1 or lose $10,000. If the chance of a devastating loss is negligible, it's hardly unwise to make a gamble, even if it has a negative expected value--this is why we occasionally do risky things that have a small chance of killing us, like driving or eating sushi. The problem in thinking in terms of an expected value is that an expected value presupposes many trials--given a single experiment, it's a meaningless concept. Indeed, given a single experiment, both the classical and frequency interpretations of probability are meaningless with regards to what the probability of a given outcome really is. In the case of a single experiment, the only interpretation that makes sense is the Bayesian interpretation. I would not play your second game, but the reason I wouldn't has nothing to do with expected values (notwithstanding the fact that it's technically impossible to calculate an expected value for an unquantifiable outcome like death). In your first game, I feel that I would come out ahead, and in the second game I feel like it's not worth it.

by Cantdog 17y ago

splat 17y ago

I'll agree with your semantics. To me, "beating a game" implies that you can consistently come out ahead over a long period of time. Nevertheless, playing an "unbeatable" game over a short period of time is not necessarily irrational. Although your expected value in your first version of the hat game is -$9901, you are not losing $9901 every time you play. If you play long enough it will average out to be this, but every time you play you will either win $1 or lose $10,000. If the chance of a devastating loss is negligible, it's hardly unwise to make a gamble, even if it has a negative expected value--this is why we occasionally do risky things that have a small chance of killing us, like driving or eating sushi. The problem in thinking in terms of an expected value is that an expected value presupposes many trials--given a single experiment, it's a meaningless concept. Indeed, given a single experiment, both the classical and frequency interpretations of probability are meaningless with regards to what the probability of a given outcome really is. In the case of a single experiment, the only interpretation that makes sense is the Bayesian interpretation. I would not play your second game, but the reason I wouldn't has nothing to do with expected values (notwithstanding the fact that it's technically impossible to calculate an expected value for an unquantifiable outcome like death). In your first game, I feel that I would come out ahead, and in the second game I feel like it's not worth it.