Timeline for Probability of one horse finishing ahead of another
Current License: CC BY-SA 4.0
6 events
| when toggle format | what | by | license | comment | |
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| Jan 30, 2023 at 16:21 | comment | added | Henry | @Silverfish: I think it is the memorylessness of exponential distributions which leads to "the second horse sampled with probabilities proportional to the weights of all the horses except that already selected as the first horse" sort of assumption I made. But you could also get sort of assumption that without requiring independent exponential distributions | |
| Jan 30, 2023 at 16:09 | comment | added | Silverfish | "This works" = "by the following math, this explicit probability distribution produces the same result as the vaguer 'weighting' argument" ≠ "this is a realistic model of horse races" (as I hope my comment on it being an unreasonable model makes very clear!) | |
| Jan 30, 2023 at 15:35 | comment | added | Sextus Empiricus | @Silverfish is the assumption of exponential distributed finishing times really an argument that it works? | |
| Jan 30, 2023 at 15:07 | comment | added | Silverfish | This works because if finishing time $T_A\sim\operatorname{Exp}(\lambda_A)$ etc, then the fastest finishing time of the other horses is another exponential random variable $T_{others}=\min(T_B,T_C,T_D,T_E)\sim\operatorname{Exp}(\lambda_B+\lambda_C+\lambda_D+\lambda_E)$ and the probability A wins is given by the comparison $\Pr(T_A < T_{others})=\frac{\lambda_A}{\lambda_A+\lambda_{others}}=\frac{\lambda_A}{\lambda_A+\lambda_B+\lambda_C+\lambda_D+\lambda_E}$, equivalent to the weighting answer | |
| Jan 30, 2023 at 14:57 | comment | added | Silverfish | "If you make the strong assumption that the probabilities are in effect weights, and the first horse is sampled with probabilities proportional to the weights of all the horses" - an equivalent (but also unreasonable for horse races!) assumption is that the horses' finishing times are independent exponentially distributed variables, with rate parameter $\lambda$ equal (or proportional, doesn't matter provided you scale them all in up the same way) to the desired probability of winning. In other words, the mean finishing time for each horse is the reciprocal of their probability of winning. | |
| Jan 29, 2023 at 18:27 | history | answered | Henry | CC BY-SA 4.0 |