What is the probability of an individual in a game where six of thirty-six are to be chosen, losing 100 times?

Imagine that a probability of certain event happening is P.
Probability that such event will NOT happen is then 1 - P.
Probability that such event will happen 3 times in a row is P^3 (P to the power of 3).

Now if your chance to win a game (once) is P, then your chance to lose a game (once) is 1 - P, and that means that your chance to lose 100 times in a row in that game is (1 - P)^100.

Now all that's left to do is calculate the probability of winning a game once. So if you choose 6 numbers out of 36, then all 6 of the numbers have to be "chosen" by "the game" in order for you to win. That means that you have 6/36 chance of that happening, which is equal to 1/6. Now that means that you have 5/6 chance of losing in a game, and that means that your chance of losing 100 times in a row is (5/6)^100 = 1.2*10^(-8) = 0.000000012 = 0.0000012% which is quite marginal.

Also, notice that there are no games (in real world) that would give you more than 50% chance of getting out with more cash than you brought. If that was the case, casinos and other stuff wouldn't survive. There have been enough smart mathematicians that would discover it, abuse it, and it would get closed pretty quick.

Also, I must say I was a bit wrong up there, you can play some games that would give you more than 50% chance of winning, but it would be something like: coming in with 1000$, and you have a "safe" way to get away with 50$, which I wouldn't say is worth the risk.

Kresho