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Alice and Bob play the game - the rules of the game are not important, and after 8 rounds Alice has 5 points and Bob has 3 points. Every round one of 2 players gets 1 point and the winner of the game is the one who gets 6 points first.

What is expected probability of Alice winning the game considering:

  1. Frequenstist approach
  2. Bayes approach

I calculated probability by summing up probabilities of Alice wins at least once in 3 rounds:

  1. W = 5/8
  2. LW = 3/8*5/8
  3. LLW = 3/8 * 3/8 * 5/8

The total sum is 94.7%

Would it be correct considering this variant as frequentist approach? How to calculate probability using Bayes approach?

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    $\begingroup$ The rules of the game appear to matter a lot. Your answer seems to make the strong assumption that the success probability is the same in every round, which (I assume) Is why you decided that the maximum likelihood estimator of the probability in the past rounds (assumed to be the same) is how you'd want to predict success in future rounds. This would arguably be one frequentist approach, if you add an assumption of both players having the same (fixed true) probability to win in each round (incl. future rounds). $\endgroup$
    – Björn
    Sep 25, 2023 at 21:27
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    $\begingroup$ What matters most is not whether your approach is frequentist or Bayesian, but how much information you interpret out of the sentence "the rules of the game are not important, and after 8 rounds Alice has 5 points and Bob has 3 points". You used this sentence to approximate Alice's win probability in each round as 5/8, and it is not at all obvious that this follows from the sibylline hypothesis "the rules of the game are not important". $\endgroup$
    – Stef
    Sep 26, 2023 at 9:49
  • $\begingroup$ It could be a game of pure luck where each player has 1/2 chance of winning each round, and Alice just got lucky in the first rounds. Or it could be a game of both skill and luck, in which case Alice's performance in the first rounds leads us to believe that she is likely to be more skilled than Bob. You can incorporate this interpretation in your reasoning, regardless of whether you use a frequentist or a Bayesian approach. $\endgroup$
    – Stef
    Sep 26, 2023 at 9:53
  • $\begingroup$ The "Rule of Succession" by Pierre-Simon Laplace (from 1814) states that the probability of Alice winning the next game is (5+1)/(8+2) = 6/10. Laplace was a Bayesian and used a uniform prior probability. $\endgroup$ Sep 26, 2023 at 10:05
  • $\begingroup$ The Pascal-Fermat problem of points would have changed your calculation to $\frac12+\frac12 \times \frac12+ \frac12 \times \frac12 \times \frac12 = \frac{7}{8}=0.875$, so there is more than one frequentist approach. $\endgroup$
    – Henry
    Sep 26, 2023 at 10:16

2 Answers 2

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As @Björn mentioned in his comment, the rules of the game are not at all negligible. The most plausible scenario seems to be a simple coin toss, where heads gives Alice a point and tails a point for Bob.

The data in this setup clearly follows a binomial distribution, i.e. $$ P(\left\{ \text{Alice has 5 points} \right\} \Leftrightarrow \left\{\#Heads = 5 \right\}|\theta) = \frac{8!}{5!(8-5)!} \, \theta^{5} (1 - \theta)^{8-5} $$ (It might seem like overkill for your problem, but the expression will play a role below.) If your prior is simply "the coin is fair", then $\theta=\frac{1}{2}$. Since the game will end within the next three tosses (either Bob gets three points by tossing tails three times, or else Alice wins), we have $$ P(\text{Bob wins}) = (1 - \theta)^3 = \frac{1}{8} \\ P(\text{Alice wins}) = \theta^{1} + \theta^{1}(1-\theta)^{1} + \theta^{1}(1-\theta)^{2} = \frac{7}{8} = 1 - P(\text{Bob wins}) $$ So far there isn't really anything Bayesian or frequentist about this, just simple probabilistic reasoning.

A Bayesian "extension" of this would be to reconsider the assumed value of $\theta$ in light of the observed data, i.e. finding $P(\theta|\text{Alice has 5 points})$ via Bayes' rule. Of course, if your prior $P(\theta)$ in this calculation is essentially a point mass at $\frac{1}{2}$, it won't be swayed by the data. Hence, a Bayesian would probably pick a more "open-minded" prior, such as Beta distribution, which when taken together with the binomial likelihood function above neatly completes the Beta-Binomial model that will conveniently lead to a Beta posterior. (You can find a worked-out example of this model (with Python code) here.)

With that posterior you could recalculate $P(\text{Alice wins})$ and $P(\text{Bob wins})$ if necessary. The MLE solution would be to find the arg max of the binomial likelihood function, which as you've hinted at is $\theta_{MLE} = \frac{5}{8}$. With these estimate of $\theta$ you would find the $P(\text{Alice wins})$ as in your question.

But this brings me back to the opening comment, on why the context of the game matters. In a real life scenario, would 5 heads in 8 tosses really lead you to question the fairness of a coin?

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  • $\begingroup$ Regarding your last sentence: if I already know that the game really is a coin toss, then I probably have a prior with a very strong belief that the coin is fair; but if I only know that they "played a game", it makes sense to model the game as Bernoulli experiment, ie a coin toss with a possibly-unfair coin, where the unfairness of the coin reflects the skill difference between the two players. $\endgroup$
    – Stef
    Sep 26, 2023 at 15:00
  • $\begingroup$ I agree, it does make sense to assign non-zero probability to the unfairness of the coin. But based on your common sense (i.e., understanding of basic physics) you should be biased towards $\theta=.5$ (i.e., a prior). MLE, on the other hand, is effectively the MAP of a flat prior that considers every $\theta$ between 0 and 1 equally plausible. I guess that's ultimately the main difference between the Bayesian and non-Bayesian approach to your question. $\endgroup$
    – Durden
    Sep 26, 2023 at 15:16
  • $\begingroup$ @Durden That is the curiosity of MAP: it uses Bayesian words like a posteriori, but is not the choice a Bayesian decision rule would make (at least not for a continuous posterior distribution) even if a Bayesian reluctantly had to make a point estimate rather than keeping the whole posterior distribution. $\endgroup$
    – Henry
    Sep 26, 2023 at 15:27
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Durden's answer is totally correct from a mathematical standpoint. I will reframe the question from another point of view.

From a Frequentist approach, you know the results up to round eight. You know that Alice won 5 rounds and Bob 3. The best guess you can get (only considering this) is that the probability of Alice winning a round is 5/8. This might be a good guess or not, for example Alice may have got lucky and Bob unlucky or some other factors may have played in.

The Bayesian approach tries to account for new information that you may have. I will include an example to avoid being too general and not specific enough:

Imagine that Alice and Bob are playing online at some videogame (some about fighting so we can get the win/loss result). Imagine you are Bob's sibling and know that, if Bob is losing, he will ask a friend to play for him and this friend is the best player of the world. In this case, this new information is very useful as it dramatically changes the result you expect in the next rounds (you would expect for "Bob" to win each of the three remaining games).

Frequentist rely on the data they have observed, while Bayesian statistics consider your beliefs/knowledge about it. In Bayesian statistics, in some models, the final result comes from a weighted average between the results from the frequentist approach and the "weight" you give to the information that you know. If you want to know more, I recommend you to watch the first few videos of "Bayesian Statistics: From Concept to Data Analysis" from Coursera.

Hope this helps!

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  • $\begingroup$ Perhaps. Though $5$ wins out of $8$ for Alice would not reject the null hypothesis that each has probability $\frac12$ of winning a round, so there is more than one frequentist approach $\endgroup$
    – Henry
    Sep 26, 2023 at 10:18
  • $\begingroup$ That approach would be classical, assigning equal probabilities to all the different possibilities (assigning 1/6 to all values from 1 to 6 in a dice throw). Frequentist approach is: given an experiment and some data from the results, which probabilities can be assigned according to it. If you rolled a dice (not necessary that it is a fair dice) 1000 times and you got 300 "1s", the probability of "1s" would be 0.3. It could not be enough to reject the null hypothesis but I think that entering now in nullhypothesis might be a bit too much as the questions is about basics of these approaches $\endgroup$ Sep 26, 2023 at 11:07
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    $\begingroup$ Supposed I flipped a coin just once and it came up heads. It would be peculiar to then assume it was a double-headed coin for future predictions even if this was the maximum likelihood estimate. $\endgroup$
    – Henry
    Sep 26, 2023 at 11:14

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