Timeline for Bayesian rank of game characters
Current License: CC BY-SA 3.0
12 events
| when toggle format | what | by | license | comment | |
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| Jun 18, 2012 at 4:34 | history | tweeted | twitter.com/#!/StackStats/status/214576761730646017 | ||
| Jun 17, 2012 at 2:54 | answer | added | Zen | timeline score: 2 | |
| Mar 18, 2012 at 16:18 | answer | added | charles.y.zheng | timeline score: 4 | |
| Mar 18, 2012 at 15:52 | history | edited | charles.y.zheng | CC BY-SA 3.0 |
oops, I don't have power to delete
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| Nov 19, 2011 at 8:42 | comment | added | alan | @whuber, so if you take q(i) and q(j) as the Beta distributions linking to each character, assume p(i, j) < 1/2 and p(i, j) = f(q(j) - q(i)). So, the q values are estimated. So there are a number of biases without predetermining an outcome, i.e. p(i, j) < 1/2 to represent the size bias. The actual size difference size(Aj - Ai) is available also. On each round, an inference can be made in some way to capture the paired outcome, increase the probability of Aj, decrease the probability of Ai, and size(Aj - Ai) can be used to infer a variable representing size consistency. Does this make sense, an | |
| Nov 18, 2011 at 17:27 | comment | added | whuber♦ | It seems rather too strong to assume you know $q(i)-q(j)$. This is tantamount to saying, with no data at all, that you already know some things about the win probabilities. For instance, if you have $q(5)-q(3)\gt q(4)-q(2)$, you're assuming an $A_5$ has a better chance of beating an $A_3$ than an $A_4$ has of beating an $A_2$. Even if you think you know such things, you should also fit a model in which the $q(i)$ are estimated, just to see whether your assumption is even approximately true. | |
| Nov 18, 2011 at 17:20 | comment | added | alan | So now, post battle, we have the knowledge "Ai beats Aj", based on probability p(i, j). We also have the value q(j) - q(i) to inform the size difference. | |
| Nov 18, 2011 at 17:11 | comment | added | alan | Yes, that can be assumed. | |
| Nov 18, 2011 at 17:04 | comment | added | whuber♦ | OK, that's good. To be precise, you appear to assume all battle outcomes are independent and that the chance of an $A_i$ beating an $A_j$ never changes. Call these chances $p(i,j)$. Without any loss of generality we can assume $A_i$ is the smaller of the two--$1\le i\lt j\le 5$--and you are assuming $p(i,j)\lt 1/2$ in all such cases. Do you also assume there is some numerical attribute $q(i)$ and some function $f$ such that $p(i,j)=f(q(j)-q(i))$; in other words, that the battle probabilities can be determined by a single number associated with each character size? | |
| Nov 18, 2011 at 16:14 | comment | added | alan | Can we assume that if the character size is bigger, there is a higher probability of winning the battle. A size difference is extracted as an output of the battle (call it DH) | |
| Nov 18, 2011 at 16:08 | comment | added | whuber♦ | Unless you can say more about the process that determines the winner in a battle, this question has no right answer. In fact, there might not exist a ranking at all. For instance, the players could be rolling intransitive dice to determine the battle outcomes. | |
| Nov 18, 2011 at 14:33 | history | asked | alan | CC BY-SA 3.0 |