if the lottery has one-in-a-million odds, then for every million timelines in which you buy a lottery ticket, in one timeline you’ll win it
I don’t understand this way of thinking about MWI, but in a single universe, you will only win a one-in-a-million lottery one time in a million on average if you play it many, many millions of times. You can easily buy a million lottery tickets and not get a winner at 1-in-a-million odds—in fact the chances of that happening are just short of 37%. Think of how often in a “throw a six to start” game some poor player hasn’t started after six or more turns.
Sums: Chance of no win in a one-off 1-in-N lottery is (N-1)/N. After N tries, the chance of no win is ((N-1)/N)^N—which astonishingly (to me) converges quite rapidly on 1/e, or just short of 37%.
(Thanks to ciphergoth for pointing the convergence out to me elsewhere.)
which astonishingly (to me) converges quite rapidly on 1/e
To make it less surprising that (1-1/n)^n converges, here are two arguments that may help.
First: take logs. You get n log (1-1/n). Now for small x, log(1+x) = x + lower-order terms, so n log (1-1/n) = n (-1/n + lower-order terms) which obviously → −1.
(Is it obvious enough that log(1+x) = x + lower-order terms? The easiest way to prove that might be to say that log x = integral from 1 to x of 1/t, and for x close to 1 this is roughly the integral from 1 to x of 1, or x-1.)
Second: use the binomial theorem. (1-1/n)^n = sum {k from 0 to n} of (-1)^k (n choose k) n^-k. Now (n choose k) = n(n-1)...(n-k+1) / k!, and for small k this is roughly n^k/k!. So for large n, the “early” terms are approximately +- 1/k!. And for large n, the “late” terms are relatively small because of that factor of n^-k. So (handwave handwave) you have roughly sum (-1)^k 1/k! which is the series for exp(-1).
I don’t understand this way of thinking about MWI, but in a single universe, you will only win a one-in-a-million lottery one time in a million on average if you play it many, many millions of times. You can easily buy a million lottery tickets and not get a winner at 1-in-a-million odds—in fact the chances of that happening are just short of 37%. Think of how often in a “throw a six to start” game some poor player hasn’t started after six or more turns.
Sums: Chance of no win in a one-off 1-in-N lottery is (N-1)/N. After N tries, the chance of no win is ((N-1)/N)^N—which astonishingly (to me) converges quite rapidly on 1/e, or just short of 37%.
(Thanks to ciphergoth for pointing the convergence out to me elsewhere.)
To make it less surprising that (1-1/n)^n converges, here are two arguments that may help.
First: take logs. You get n log (1-1/n). Now for small x, log(1+x) = x + lower-order terms, so n log (1-1/n) = n (-1/n + lower-order terms) which obviously → −1.
(Is it obvious enough that log(1+x) = x + lower-order terms? The easiest way to prove that might be to say that log x = integral from 1 to x of 1/t, and for x close to 1 this is roughly the integral from 1 to x of 1, or x-1.)
Second: use the binomial theorem. (1-1/n)^n = sum {k from 0 to n} of (-1)^k (n choose k) n^-k. Now (n choose k) = n(n-1)...(n-k+1) / k!, and for small k this is roughly n^k/k!. So for large n, the “early” terms are approximately +- 1/k!. And for large n, the “late” terms are relatively small because of that factor of n^-k. So (handwave handwave) you have roughly sum (-1)^k 1/k! which is the series for exp(-1).