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In roullette (say European with 37 slots) when the wheel is fair and the success is predefined as hitting a specific number (say $13$) the number of successful outcomes in $n$ spins/trials should follow the binomial distribution with $p=1/37,$ I figured.

But what happens if instead of selecting $13$ as a success a priori, one performs $n$ trials first, finds the slot with maximum hits (say the maximum is unique) and defines this slot as a success. Would the expected number of hits in this slot still be binomially distributed with the same $n$ and $p$ and can this still be used to estimate the probability of the observed number of hits in this slot?

Michael Hardy
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Maciej Tomczak
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  • When the wheel is fair, as you suppose, then the results of previous trials are irrelevant, because future trials will be independent of them and follow the same distribution. – whuber Oct 04 '23 at 19:16
  • Sure, but the issue I can't figure out is slot selection. If the slot of interest is predetermined, I can use use the binomial distribution to estimate the probability of observing as high (or as low) of a value as it was observed. But by letting the outcome of n trials decide on the slot (the one with most hits) I intuitively think I may be skewing such analysis because the proces always selects the unusual slot which will never experience a number of hits lower than n/37, I think. – Maciej Tomczak Oct 05 '23 at 02:36

1 Answers1

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To begin with, you should first note that if the Roulette wheel actually is fair then (by assumption) the outcomes are independent uniform random variables. In this situation the results of previous trials are irrelevant to prediction of future results. If you are looking for the distribution of the maximum-count statistic in a fair process, then this is the distribution of the following random variable based on a multinomial distribution with uniform probabilities:

$$\text{Maxcount}(n,m) \equiv \max (N_1,...,N_m) \quad \quad \quad (N_1,...,N_m) \sim \text{Mu}(n, (\tfrac{1}{m}, ..., \tfrac{1}{m})).$$

You can find an algorithm for this distribution in Bonetti et al (2019). The probability functions for this distribution are available as the "max count" distribution (see e.g., the function dmaxcount) in the occupancy package in R.

In practice, we might think that a mechanism like this is fair, but it is possible that it might actually have some bias towards particular outcomes. This belief was examined in a series of papers on the "gambler's fallacy" and broader methods of binomial prediction under the Bayesian paradigm (see O'Neill and Puza 2005; O'Neill 2012; O'Neill 2015). These papers consider the case where a mechanism is designed to give "fair" uniformly distributed outcomes but there is a possibility of bias in the mechanism. If you hold the a priori belief that the direction of bias is symmetric then this leads to an optimal prediction method where you predict the outcome that has occurred the most in the observed data (which these papers call the "frequent outcome approach"). If you have a look at the linked papers you will see that this leads to some particular forms for the prediction accuracy, and the accuracy will converge towards "perfect prediction" (assuming a priori knowledge of the bias in the mechanism) over a large number of trials.

Ben
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  • Many thanks for this! But say that the wheel is not biased at all. And I'm interested in the probability of observing x hits, but in the slot that, as I found after the trials, happened to receive the maximum number of hits. Unlike a predetermined slot, the max slot would never experience lower value than the expected n/37. The binomial model that should work for any predetermined slot doesn't seem to be applicable for determining how likely is the maximum frequency observed should the wheel be perfectly fair. But I don't know what could be more appropriate... – Maciej Tomczak Oct 05 '23 at 03:01
  • I have updated the answer to add this information. – Ben Oct 05 '23 at 03:10