Timeline for 5 people each have 5 equal cards
Current License: CC BY-SA 3.0
17 events
| when toggle format | what | by | license | comment | |
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| Nov 4, 2016 at 11:50 | comment | added | kaine | @2012rcampion Thank you. That was the concept I was aiming for but I didn't want to use that term unless I was certain it was appropriate. I'm an chemical engineer so that isn't quite in my normal vocabulary. | |
| Nov 4, 2016 at 11:04 | comment | added | 2012rcampion | The technical term for the solution you found is a Nash equilibrium. I calculated it using a different method and got the same result, so I'm fairly certain you have the right solution. | |
| Nov 3, 2016 at 18:15 | history | edited | kaine | CC BY-SA 3.0 |
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| Nov 3, 2016 at 18:06 | history | edited | kaine | CC BY-SA 3.0 |
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| Nov 3, 2016 at 17:59 | comment | added | kaine | @Jik just for you (sarcasm) I removed the (P1+P2+P3..) from the equations, calculated the probability of a draw for each one which is $P_d_x = P_x^4+6P_x^2( \sum (P_o^2))+4P_x( \sum (P_o^3))$. multiplied that by .2 and added it to each win. My previous answer did bias it further towards the 4s and 5s. The numbers in the answer will be edited shortly. | |
| Nov 3, 2016 at 17:43 | comment | added | kaine | @JiK I originally using the divide by (P1+P2+P3...) to take care of that. Realized that was wrong later. You are correct. My method aimed to maximize win percentage for each round instead of improving win to loss ratio. | |
| Nov 3, 2016 at 17:36 | comment | added | JiK | It seems to me that your formulas don't take into account the fact that the round is replayed if there is no clear winner. | |
| Nov 3, 2016 at 15:28 | comment | added | corsiKa | "A single idiot at the table changes everything" - just like in real cards. Some idiot being opening with 2 9 suited under the gun and calls a 3 bet... you can't beat stupid players with logic! | |
| S Nov 3, 2016 at 15:25 | history | suggested | Shufflepants | CC BY-SA 3.0 |
spelling: probability
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| Nov 3, 2016 at 15:16 | review | Suggested edits | |||
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| Nov 2, 2016 at 22:35 | comment | added | ffao | I'm surprised we have P5>0, but the math is sound. Upon further thought, it makes sense, because lower numbers are more likely to tie, so it's easier to be a loner with a 5. | |
| Nov 2, 2016 at 19:05 | history | edited | kaine | CC BY-SA 3.0 |
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| Nov 2, 2016 at 18:36 | comment | added | kaine | Mistake corrected and got updated results. Did another type of simulation where actual games were run and results were within 0.1% in value. | |
| Nov 2, 2016 at 18:35 | history | edited | kaine | CC BY-SA 3.0 |
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| Nov 2, 2016 at 16:26 | comment | added | kaine | Note, I have found a mistake in caculations. I did not take into account 3 people having a single number....damn | |
| Nov 2, 2016 at 15:35 | history | edited | kaine | CC BY-SA 3.0 |
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| Nov 2, 2016 at 15:27 | history | answered | kaine | CC BY-SA 3.0 |