Suppose you can play on either of 2 slot machines:
 Slot machine A pays out with probability one in a million.
 Slot machine B pays out with probability one in 10.
Are you more likely to get a payout with one million attempts with slot machine A or with 10 attempts on slot machine B?
Have a think about this before scrolling down.

















I was prompted to think about this question by the following tweet, which includes both the answer and the relevant calculations.
I find it mildly interesting that if you have a million chances at a oneinamillion shot, it's less likely to happen than if you have 10 chances at a onein10 shot, or four chances at a oneinfour shot, etc.
1((3/4)^4)=0.68
1((9/10)^10)=0.65
(999999/1000000)^1000000=0.63— Tom Chivers (@TomChivers) November 4, 2019
So, there’s a bigger probability (0.65) that you’ll get a payout from 10 spins of slot machine B than from a million spins of slot machine A (probability 0.63).
Hopefully, the calculations above are selfexplanatory. But just in case, here’s the detail. Suppose you have N attempts to win with a slot machine that pays out with probability 1/N.
 First we’ll calculate the probability of zero payouts in the N spins.
2. This means we get a zero payout on every spin.
3. The probability of a zero payout on one spin is one minus the probability of a win: 1 – 1/N.
4. So the probability of no payout on all the spins is
5. And the probability of at least one payout is
As explained in the tweet, with N=10 this gives 0.65 and with N=1,000,000 it gives 0.63. The tweet’s author explains in a followup tweet that he was expecting the same answer both ways.
But I do find it interesting! intuitively I assumed that three chances at a oneinthree shot would be equivalent to two chances at a oneintwo shot, but they're not. If you're offered two chances to flip heads or six chances to roll a six, you should definitely choose the coin
— Tom Chivers (@TomChivers) November 4, 2019
But as someone in the discussion pointed out, that logic can’t be right. Suppose you had one attempt with slot machine C which paid out with probability 1. In other words, N=1 in my example above. Then, of course, you’d be bound to get a payout, so the probability of at least one payout is 1. So, although it’s initially perhaps surprising that you’re more likely to get a payout with 10 shots at slot machine B than with a million shots at slot machine A, the dependence on N becomes obvious when you look at the extreme case of slot machine C.
Footnote: What does stay the same in each case however is the average number of times you will win. With N shots at a slot machine with win probability 1/N, you will win on average once for any choice of N. Sometimes you’ll win more often, and sometimes you may not win at all (except when N=1). But the average number of wins if you play many times will always be 1.