Go to a casino. Bet $1 on something with a 50% chance of winning. If you win, you have won $1; try again. If you lose, double your bet size (which means that winning will leave you having won $1 total over the sequence of doubled bets) and repeat.
One argument says that in the long run, you will come out a winner, because every bet you make is part of a sequence and at the end of that sequence, you are $1 richer. Another argument says that in the long run, you will only break even, because each bet has a 50% chance of winning and a 50% chance of losing the same amount of money.
Of course, the answer is that you can’t increase your bet infinitely, and when you stop increasing your bet, the statistical loss at the point where you stop increasing your bet exactly makes up for the statistical win all the other times you finished the sequence and won $1.
Furthermore, if you could increase your bet infinitely, this problem wouldn’t happen, but if you could increase your bet infinitely, the expectation isn’t well defined, because you are trying to compute it for a non-converging infinite series.
All this problem is is the same idea applied to probability of death instead of expectation of win. If the madman ever runs out of people, the overall probability depends exactly on what the madman does when he runs out of people (since it’s not as well defined as it is for bets). If the madman never runs out of people, the probability involves a non-converging infinite series and so is not well defined.
If this is a metaphor for extinction, then when the madman runs out of people, he keeps rolling the dice on the remaining people until it eventually comes up snake eyes, in which case the chance of extinction is 100%. On the other hand, they can last arbitrarily long given an arbitrarily small probability of extinction.
An earlier version of my analysis (the previous blog post) looked at the case of finite n and found, as you suggest, that the possibility of running out of people to kidnap is an important consideration. You can choose the number of batches n to be so large that it is virtually certain a priori that the madman will eventually murder:
P(eventually murders) = 1 - epsilon for some small epsilon
However, it turns out that conditioning on the fact that you are kidnapped changes the probability dramatically:
P(eventually murders | you are kidnapped) = about 10⁄9 * 1⁄36
The reason for this is that there are about 9 times as many people in the final batch as in all other batches combined, so the fact that you are kidnapped is strong evidence that the madman is on his last batch of potential victims.
Go to a casino. Bet $1 on something with a 50% chance of winning. If you win, you have won $1; try again. If you lose, double your bet size (which means that winning will leave you having won $1 total over the sequence of doubled bets) and repeat.
One argument says that in the long run, you will come out a winner, because every bet you make is part of a sequence and at the end of that sequence, you are $1 richer. Another argument says that in the long run, you will only break even, because each bet has a 50% chance of winning and a 50% chance of losing the same amount of money.
Of course, the answer is that you can’t increase your bet infinitely, and when you stop increasing your bet, the statistical loss at the point where you stop increasing your bet exactly makes up for the statistical win all the other times you finished the sequence and won $1.
Furthermore, if you could increase your bet infinitely, this problem wouldn’t happen, but if you could increase your bet infinitely, the expectation isn’t well defined, because you are trying to compute it for a non-converging infinite series.
All this problem is is the same idea applied to probability of death instead of expectation of win. If the madman ever runs out of people, the overall probability depends exactly on what the madman does when he runs out of people (since it’s not as well defined as it is for bets). If the madman never runs out of people, the probability involves a non-converging infinite series and so is not well defined.
If this is a metaphor for extinction, then when the madman runs out of people, he keeps rolling the dice on the remaining people until it eventually comes up snake eyes, in which case the chance of extinction is 100%. On the other hand, they can last arbitrarily long given an arbitrarily small probability of extinction.
An earlier version of my analysis (the previous blog post) looked at the case of finite n and found, as you suggest, that the possibility of running out of people to kidnap is an important consideration. You can choose the number of batches n to be so large that it is virtually certain a priori that the madman will eventually murder:
P(eventually murders) = 1 - epsilon for some small epsilon
However, it turns out that conditioning on the fact that you are kidnapped changes the probability dramatically:
P(eventually murders | you are kidnapped) = about 10⁄9 * 1⁄36
The reason for this is that there are about 9 times as many people in the final batch as in all other batches combined, so the fact that you are kidnapped is strong evidence that the madman is on his last batch of potential victims.