Timeline for Is there a better way of calculating the chance of winning than $\mu * (1 - (\sigma * f)) * 100$ for the card game schnapsen?
Current License: CC BY-SA 4.0
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| Jan 2, 2022 at 12:10 | history | edited | nbro | CC BY-SA 4.0 |
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| Nov 15, 2018 at 14:02 | vote | accept | Aura Lee | ||
| Nov 12, 2018 at 17:00 | answer | added | Dennis Soemers♦ | timeline score: 2 | |
| Nov 11, 2018 at 20:45 | history | edited | Aura Lee | CC BY-SA 4.0 |
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| Nov 11, 2018 at 20:25 | history | edited | Aura Lee | CC BY-SA 4.0 |
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| Nov 11, 2018 at 20:13 | comment | added | Aura Lee | The standard deviation and mean is calculated with all different plays. If one game state has 5 different ways to continue it calculates the percentage of winning (each continuation calculated again) for each one and then uses the formula and returns the value for the previous tree node to calculate. A win returns 100 percent, a loss returns 0. (My AI returns a higher percentage for a 3 point loss and a lower one for a 1 point win but that is not that important) | |
| Nov 11, 2018 at 18:19 | comment | added | Neil Slater | What is the standard deviation taken of - the three different plays? So you have calculated means and stds for each branch point in the tree? Presumably these are calculated with even weights on each choice? | |
| Nov 11, 2018 at 14:40 | review | First posts | |||
| Nov 11, 2018 at 21:35 | |||||
| Nov 11, 2018 at 14:39 | history | asked | Aura Lee | CC BY-SA 4.0 |