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Engineer here. This may be a dumb question, but lately I've been wondering about what to make of frequentist probabilities. By definition, the probability of an event is the number of times $k$ it occurs in an infinite number $n$ of tries $$p=\lim_{n\to \infty}\frac{k}{n}$$ Yet in practice, the number of repetitions will always be finite. Now let us assume we can choose between two bets ($1$ and $2$, with winning chances of $p_1=10\%$ and $p_2=50\%$ respectively (e. g. a roulette game with ten numbers where we can bet on a specific number or on even/odd). My first question is: If we play just once, does it actually matter which bet we choose? After all, if I go for bet $1$, I may be the one in ten persons who wins and if I go for bet $2$, I may be one of the five in ten persons who lose and there is no way of knowing what will happen... If we go one step further, one might argue that even if we play many (but obviously still a finite number of) times $n$, the probability of continuously losing $(1-p_{1/2})^n$ remains finite, so no matter which bet we choose, we may still end up as one of the "unlucky guys" (and again, we would have no chance of predicting if that will be the case)... That might lead one to conclude that frequentist probabilities are useless since any individual or group can always end up on the losing side...

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    $\begingroup$ Covid has about 3-5% fatality rate. The usual flu has a fatality rate of about 0.1% (according to my Google search just now). MERS has a fatality rate of 35%. No matter what, you could wind up as one of the unlucky people who dies, yet which disease frightens you most? $\endgroup$
    – Dave
    Oct 10, 2020 at 14:12
  • $\begingroup$ Nice example ;). Well, intuitively I would be most scared of MERS, but do I really have to? As we both pointed out, no matter which bet we make, we never know if we end up on the losing side... $\endgroup$ Oct 10, 2020 at 15:36
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    $\begingroup$ If you state a definition and then steadfastly ignore its premise, then you shouldn't be surprised if the definition doesn't apply. Assuming you you will only once ever do something risky in your life, on that one occasion it may make more sense to consider there are 5 possibilities of winning one bet and there is 1 possibility of winning the other. So you might feel its fair to get back 2 dollars for 1 on the first bet is 'fair' and getting back 10 dollars for 1 on the second is fair. // However, life is full of uncertainty and the frequentist definition is useful over a (finite) lifetime. $\endgroup$
    – BruceET
    Oct 10, 2020 at 16:12

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Now let us assume we can choose between two bets ($1$ and $2$, with winning chances of $p_1=10\%$ and $p_2=50\%$ respectively (e. g. a roulette game with ten numbers where we can bet on a specific number or on even/odd). My first question is: If we play just once, does it actually matter which bet we choose? After all, if I go for bet $1$, I may be the one in ten persons who wins and if I go for bet $2$, I may be one of the five in ten persons who lose and there is no way of knowing what will happen...

Choice matters, definitely. More in general, among disjoint events, less probable event can happen. This is not strange, this is the world of probability. Therefore in gambling, you can win if you bet for less probable event and can lose if you bet for more probable. All gamblers know that. However if the potential gain is the same, all rational gamblers bet on the more probable event. It holds for any sensible definition of rationality. The same story can be easily extended to multiple betting/events.

Moreover this story hold for any definition of probability. It has very few to share with weakness of frequentist approach.

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