Scenario: there is a machine at a casino that hits jackpot 1/100 times it is used. The probability that one does NOT hit jackpot on their first spin is .99^1, the second .99^2, and on their nth .99^n (hoping my math is right). As the number of non-winning spins increases, many people would say the machine is “due” because the probability of the losing streak continuing gets lower and lower, but AFAIK that is not valid. Why is that?
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Because the trials are independent.
What’s the probability of flipping a coin tails twice in a row (assuming you only flip twice). Well, it’s 0.25. There’s a 0.5 chance of the first flip coming up tails, and a 0.5 chance of the second coin coming up tails.
But suppose you’ve *already* flipped tails once. What’s the probability the second flip comes to tails? Well it’s a coin, so there’s a 0.5 chance. The first flip is irrelevant. The coin doesn’t *remember* the result of the prior flip.
Thus we say that the flips are independent. The result of one flip has no bearing on the result of the next. The odds of getting two tails are 0.25, but only before the first flip. *After* the first flip comes up tails, we already have one guaranteed tail. All that’s left is the second flip, and like all coin flips, there’s going to be an equal chance of getting either possible result.
So if you’re in a casino right when it opens, you can look at a slot machine and say *the odds of it not paying out at least once in the next hour are a billion to one”, and be absolutely right. But, if you’ve been playing for 58 minutes with no wins, that has no bearing on what happens in the next two minutes. The incredibly unlikely event of not winning anything in 58 minutes has come to pass, but that doesn’t change how the machine works. It’s the past. And the next time you pull the lever, the odds of that *individual pull* winning is exactly the same as it was for all the other pulls.
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