Timeline for Guessing each other's coins
Current License: CC BY-SA 4.0
12 events
| when toggle format | what | by | license | comment | |
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| Apr 25, 2022 at 8:16 | comment | added | Frederik Ravn Klausen | Does anybody have intuition about why one has to make the specific ordering: {010: 2, 011: 2, 001: 1, 110: 0, 100: 0, 101: 1}? In particular, it seems very far from the 1/3 strategy where I just guess on the first place where I myself have a 1. Can anyone prove that I can't have an ordering where I guess on a place where I have a 1 myself? | |
| Apr 2, 2019 at 9:05 | comment | added | mihaild | Mixed strategy can't give better result then any pure strategy - just enumerate all possible results and for them one-by-one replace mix with number that maximizes winning probability. | |
| Apr 1, 2019 at 23:02 | comment | added | John Gunnar Carlsson | For what it's worth, I have just used CPLEX to solve the cases $N\in\{2,3,4\}$ and verified that your proposed solution is optimal for these cases. That is, the optimal ratios are $0.3125$, $0.34375$, and $0.3477$. In fact, it remains optimal even if you allow a mixed strategy (i.e. for a given sequence, your selection of the digit is random) | |
| Mar 31, 2019 at 21:03 | comment | added | Guillaume Aubrun | A formula that matches the known estimates for even $N=2,4,6$ is $p_N = \frac{7}{20} - \frac{3}{5 \cdot 4^N}$ | |
| Mar 31, 2019 at 15:47 | comment | added | Guillaume Aubrun | Note that for even $N$, $p_N \neq \frac{7}{20} - \frac{2}{5\cdot 4^N}$ since the latter fraction is not of the form $k/4^N$. | |
| Mar 31, 2019 at 15:21 | comment | added | Yoav Kallus | @MaxAlekseyev, I found strategies for N=3,5,7,9 satisfying your inequality (with equality, I should note) using a very simple algorithm (fix random f_A and optimize f_B; fix f_B and optimize f_A; and so on until fixed point, repeat with different random initial condition). | |
| Mar 31, 2019 at 15:04 | comment | added | Max Alekseyev | For what $N$, $7/20$ works in reverse direction, i.e. $p_N\geq \frac{7}{20} - \frac{2}{5\cdot 4^N}$ ? | |
| Mar 31, 2019 at 10:59 | history | edited | mihaild | CC BY-SA 4.0 |
edited body
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| Mar 31, 2019 at 1:18 | comment | added | Yoav Kallus | In general, your method gives $p_\text{opt} \ge (4^N p_N-1)/(4^N-4)$. Using Guillaume's $p_5$ in this bound also gets you to $7/20$. | |
| Mar 30, 2019 at 22:08 | history | edited | mihaild | CC BY-SA 4.0 |
added 15 characters in body
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| Mar 30, 2019 at 21:50 | review | First posts | |||
| Mar 30, 2019 at 21:56 | |||||
| Mar 30, 2019 at 21:47 | history | answered | mihaild | CC BY-SA 4.0 |