5 ms·

> If random really means random, then that's presumably true by definition. No it's not. Randomness is always with respect to a given distribution. There is no

> If random really means random, then that's presumably true by definition. No it's not. Randomness is always with respect to a given distribution. There is nothing special about the uniform distribution over a threshold. The space of strategies also include strategies where your opponent redraw only when they hit a number between 0.1 and 0.2, or where they redraw if the second decimal digit of the first draw is even. > The thing is that maximizing the expected value is optimal in the absence of any information about the opponent's strategy. Your statement is meaningless without a prior on the strategy. Under which prior is maximizing the expected value optimal? If you assume the other party randomly picks a threshold between 0 and 1, then the optimal answer is (sqrt(39)-3)/6 not 0.5. Not that this prior makes any sense. It's far more sensible to assume a strategic opponent. Rethrowing below 0.5 is typically the optimal strategy when the opponent has no strategy at all, i.e. always or never rethrows.

foldr 10y ago
If a strategy is any function from the preceding game state to keep/redraw, then I don't have the mathematical chops to specify a uniform prior over all such strategies and compute the optimal strategy that way. (In fact it may not necessarily be possible to do this in a sensible way, as Bertrand's paradox suggests, as another poster pointed out.) However, there is a much simpler argument suggesting that maximizing EV would be the right approach in the absence of any knowledge about the opponent's strategy. Say the opponent chooses to redraw at random. Then in effect they are drawing once. If we redraw when the value is < a, then we just want to maximize the probability of winning, (1-a)(a + 0.5) + 0.5a, giving a = 0.5. Modeling a process that we know nothing about as a random process seems reasonable. And the reasoning above does not actually involve computing EVs, though it happens that the right answer is the answer that maximizes the EV. To justify this a bit further, we can reason as follows. If we know nothing about the opponent's strategy, then their initial draw tells us nothing about whether or not they will choose to redraw. So our estimate of their probability of redrawing, p, must be independent of their initial draw. Then the probability of winning is 0.5ap + (1-a)(1-p)(a + 0.5) + (1-a)p(a + 0.5) + 0.5a(1-p), which is invariant with respect to p, and which is at its maximum when a = 0.5. >It's far more sensible to assume a strategic opponent. Well, this is going round in circles, but I would say that this is sensible only if you know something about the opponent! I don't see how it's sensible to assume anything about your opponent if you know nothing about them. And if we're talking about the "real world", then of course you do know something, but you certainly don't know that they're perfectly rational.

foldr 10y ago

If a strategy is any function from the preceding game state to keep/redraw, then I don't have the mathematical chops to specify a uniform prior over all such strategies and compute the optimal strategy that way. (In fact it may not necessarily be possible to do this in a sensible way, as Bertrand's paradox suggests, as another poster pointed out.) However, there is a much simpler argument suggesting that maximizing EV would be the right approach in the absence of any knowledge about the opponent's strategy. Say the opponent chooses to redraw at random. Then in effect they are drawing once. If we redraw when the value is < a, then we just want to maximize the probability of winning, (1-a)(a + 0.5) + 0.5a, giving a = 0.5. Modeling a process that we know nothing about as a random process seems reasonable. And the reasoning above does not actually involve computing EVs, though it happens that the right answer is the answer that maximizes the EV. To justify this a bit further, we can reason as follows. If we know nothing about the opponent's strategy, then their initial draw tells us nothing about whether or not they will choose to redraw. So our estimate of their probability of redrawing, p, must be independent of their initial draw. Then the probability of winning is 0.5ap + (1-a)(1-p)(a + 0.5) + (1-a)p(a + 0.5) + 0.5a(1-p), which is invariant with respect to p, and which is at its maximum when a = 0.5. >It's far more sensible to assume a strategic opponent. Well, this is going round in circles, but I would say that this is sensible only if you know something about the opponent! I don't see how it's sensible to assume anything about your opponent if you know nothing about them. And if we're talking about the "real world", then of course you do know something, but you certainly don't know that they're perfectly rational.