Considering the Probability of the Series

Question

As we know the probability of a series of events (independent) is p1 times p2 times Pn. Let's say that there are two gamblers and that they are both playing a game that requires guessing a number from 1 to 100 accurately. One gambler has lost several times in a row. The other gambler has only lost once. The probability of gambler one continuing a series of losses is low. The series favors a win for him on the next try. For gambler two, the probability of the series indicates less of a chance of a win. Disregarding the personal series of each gambler, on the next turn each gambler has the same probability of guessing correctly. If I were copying either one gambler or the other, I would choose to go with gambler one. Probability seems relative to the participant. I think I have an idea about how to resolve this paradox but I thought I would throw it out there. What is your opinion?

Answer 1

So I take it that Gambler 1 (G1) has lost several times in a row. Gamber 2 (G2) has so far lost only once—in the absence of other information, I'm also assuming that G2 has also only gone once.

Given what you've said, given their histories, on the next turn, the probability of G1 guessing correctly is > 1/2 and the probability of G2 guessing correctly is < p(G1 guesses correctly). So (p(G1 guesses correctly) | G1's history) > 1/2 and (p(G1 guesses correctly) | G1's history) > (p(G2 guesses correctly) | G2's history).

But you also say that on the next turn, independent of their histories, p(G1) = p(G2).

Since the probability of G1 and G2 guessing correctly is independent both of each other and of history, there's no reason to prefer either one. In fact, you say "One gambler has lost several times in a row. The other gambler has only lost once. The probability of gambler one continuing a series of losses is low. The series favors a win for him on the next try." But given the other information, this is to commit the (aptly-named) gambler's fallacy.