Timeline for One flip coin game
Current License: CC BY-SA 4.0
13 events
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Nov 6 at 21:03 | comment | added | Fedor Petrov | The maximal coefficient is at $n/2$ because the sequence of the coefficients of this polynomial is symmetric (obviously) and log-concave by en.m.wikipedia.org/wiki/Newton%27s_inequalities | |
Oct 26 at 7:05 | comment | added | Nate River | Yes, the framework of maximising simple expectation is appropriate when the amount that you get is small enough i suppose. @Snared | |
Oct 26 at 6:16 | comment | added | Snared | It's akin to saying that if I give you a $(1/2+\epsilon)$ coin-flip opportunity to double-or-nothing any amount of money, obviously the best expected value from that would be to flip as much money as you can. But you wouldn't flip your entire net worth because your expected utility falls off a cliff - you "die" $1/2-\epsilon$ amount of the time and are "twice as wealthy" $1/2+\epsilon$ of the time, but it would have been better to just bet say 5% of your net worth so you could still be in the same general band of wealth even if you lose. | |
Oct 26 at 6:13 | comment | added | Snared | @NateRiver The reason why that should be obvious as well. It is optimal in regards to maximizing your expectation, not maximizing your utility. The utility often involves something like the expectation minus a constant times the standard deviation of your strategy. Setting that constant to zero gives your question, but if this game were about real money that is meaningful to someone, then simply maximizing the expectation would be a losing strategy with respect to maximizing their utility from the game. But the constant chosen for risk is a personal choice, though $0$ is never the best. | |
Oct 26 at 6:11 | comment | added | Nate River | Well yes, but it is still unclear why betting your entire fortune is always optimal. Of course the computations say so! @Snared | |
Oct 26 at 6:07 | comment | added | Snared | @NateRiver It should be obvious why. When $n$ is odd, you get 1 coin which is indifferent to choosing between heads and tails $(p=0.5)$. But when $n$ is even, every coin prefers exactly one of heads and tails $(p \neq 0.5)$. | |
Oct 25 at 19:31 | history | edited | SmileyCraft | CC BY-SA 4.0 |
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Oct 25 at 15:07 | history | edited | SmileyCraft | CC BY-SA 4.0 |
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Oct 25 at 14:23 | vote | accept | Nate River | ||
Oct 25 at 14:23 | comment | added | Nate River | Wow, thanks for the really thorough answer! I agree with your proof of the claim that you should always bet your entire fortune to maximise the expected value. It is interesting that the optimal value changes between $n$ odd and even instead of being strictly monotone. The computation of the expected value, even knowing the optimal strategy seems really tough. | |
Oct 25 at 14:08 | history | edited | SmileyCraft | CC BY-SA 4.0 |
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S Oct 25 at 14:02 | review | First answers | |||
Oct 25 at 17:21 | |||||
S Oct 25 at 14:02 | history | answered | SmileyCraft | CC BY-SA 4.0 |