Probability of winning a game: frequentist vs Bayesian approach

Question

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?

Answer 1

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?

Answer 2

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!

Answer 3

the rules of the game are not important,

As noted, the rules certainly are important. But given your example, I'm going to make the safe assumption that the game favors Alice to the point that she is beating Bob 5 to 3. So the game is either pure luck (say pick a number from 1 to 1000, if it's 1-625, Alice gets a point, otherwise Bob does), Alice is better than Bob by this degree, or some combination thereof.

Anyway, there is a PDF and a YouTube video that cover this exact example. They both arrive at the answer of 1/11. The specific game is one of luck - a ball rolling onto a table, and coming to rest at one spot. We don't know where the line demarcating Alice's zone is vs Bob's, but that's the whole point.

The primary formula is:

$$ P(\text{Bob wins|a=5,b=3}) = \frac{P(\text{a=5,b=3|Bob wins)P(Bob wins)})}{P(\text{a=5,b=3})} $$

The denominator:

$$ P(\text{a=5,b=3}) = \int_{0}^{1} \binom{8}{3}x^3(1-x)^5 \,dx $$

$$ P(\text{a=5,b=3}) = \frac{1}{9} $$

And the numerator:

$$ P(\text{a=5,b=3|Bob wins)}P(\text{Bob wins}) = \int_{0}^{1} \binom{8}{3}x^6(1-x)^5 \,dx $$

$$ P(\text{a=5,b=3|Bob wins)}P(\text{Bob wins}) = \frac{1}{99} $$

So the final calculation is:

$$ \frac{\frac{1}{99}}{\frac{1}{9}} = \frac{1}{11} $$