Timeline for Expected Value : Poker
Current License: CC BY-SA 4.0
8 events
| when toggle format | what | by | license | comment | |
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| S Jul 15, 2022 at 20:32 | history | suggested | jerryjrxie | CC BY-SA 4.0 |
fixed typo on equation for (2, 1)
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| Jul 15, 2022 at 19:44 | review | Suggested edits | |||
| S Jul 15, 2022 at 20:32 | |||||
| Nov 25, 2020 at 19:50 | history | edited | Pontus Hultkrantz | CC BY-SA 4.0 |
added 251 characters in body
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| Nov 25, 2020 at 19:47 | comment | added | Pontus Hultkrantz | @oskarszarowicz. I added a bigger update.So previously, my final expected bankroll was 267. The thing is that in some cases no matter what you bet, it won't change the expectation, just like you showed by cancelling of $x$. However, we can consider higher moments, such as variance, and find the bet size that minimizes it. So the "optimal" unique strategy, assuming we don't like variance, has expectation 267, zero variance, and is accomplished by only betting on guaranteed events, except on the second card, where you bet 1/3 of your bankroll on the color not on the first card. | |
| Nov 25, 2020 at 19:40 | history | edited | Pontus Hultkrantz | CC BY-SA 4.0 |
Major walkthrough of strategy, consider variance, and recursive solution in python.
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| Nov 22, 2020 at 23:52 | comment | added | user524813 | strategy is $\frac{1}{2} \cdot (200+x) + \frac{1}{2} \cdot (200-x) = \frac{200+x+200-x}{2} = \frac{400}{2} = 200$ Same as before. Just a curious find I really liked your approach though! | |
| Nov 22, 2020 at 23:51 | comment | added | user524813 | Nice solution but for what its worth if we only care about expected value (and not variance) we are very welcome to bet on the third card in your second case. Let us say we see {red, black}, what is the expected value of a bet of $x$ on a red card? $\frac{1}{2} $ we are correct and in which case we end up with $100+x$ after this third card and then end up with $200+2x$ after I double my money on the certain third card. In $\frac{1}{2}$ the case that I was wrong I then have $100-x$ after loosing $x$ which I double at the end to end with $200-x$ Hence the expected value of this | |
| Nov 22, 2020 at 17:03 | history | answered | Pontus Hultkrantz | CC BY-SA 4.0 |