As usual, Fred is alive at 10 am, and there is an infinite sequence of Grim Reapers, where the nth has an alarm set for 60/n minutes after 10 am, and if the alarm goes off, it checks if Fred is dead, and swings its scythe at Fred if and only if Fred is alive. But here’s the twist. These Grim Reapers are unreliable killers. The probability that the nth Reaper’s swing would succeed in killing Fred is 1/np, where p is some positive real number, the same for each Reaper, and independently of all other relevant events.
Here’s the fun thing. It seems possible for Fred to survive the whole ordeal. All it takes is for every Grim Reaper to fail at killing Fred. Nothing absurd happens then. Moreover, it seems this isn’t the only way for absurdity to be avoided in this case. We could also suppose that the nth Reaper kills Fred, while Reapers n + 1, n + 2, … all fail.
Suppose we adopt what seems the best alternative to Causal Finitism, namely the Inconsistent Pair response to the original Grim Reaper paradox, which says that the reason the original paradox is impossible is simply because it embodies an Inconsistent set of propositions—some Reaper has to kill Fred and none can. If that’s what’s wrong with the original Grim Reaper paradox, then it seems we have to accept my Unreliable Reaper story as possible.
But things are a little bit more complicated. The only way to avoid paradox in the Unreliable Reaper story is if there is some n ≥ 0 such that all the Reapers starting with Reaper n + 1 fail. But now suppose that 0 < p ≤ 1. Then the event that all the Reapers starting with Reaper n + 1 fail is less than or equal to (1−1/(n+1)p)(1−1/(n+2)p)(1−1/(n+3)p)... = 0 (this is because Σk 1/kp = ∞ if p ≤ 1). Thus the probability that we have avoided paradox is 0. Hence, if we have to avoid paradox, a specific zero probability event—namely, the event of paradox-avoidance—has to happen (the probability of a countable disjunction of zero probability events is zero). But if it has to happen, it can’t be probability zero, but must be probability one!
Perhaps here we bring back the Inconsistent Pair response. We say that my Unreliable Reaper story is impossible if p ≤ 1, because if p ≤ 1, then a zero probability event has probability one, which is inconsistent. No such problem occurs if p > 1. Thus, on this version of the Inconsistent Pair response, my Unreliable Reaper story is impossible if the success probability of the nth Reaper is 1/np for p ≤ 1 but possible if p > 1. And that’s pretty counterintuitive.
2 comments:
But is it really "counterintuitive"? For p > 1, the sum converges, so the infinite product lands above zero. That’s it. Fred can survive because probabilities shrink fast enough. For p ≤ 1, the sum diverges, and survival probability gets crushed to zero. This isn’t paradoxical, it’s the textbook nature of p-series and infinite products.
Products of factors between 0 and 1 converging to a positive number? Happens all the time when terms drop off quick.
You seem thrown by infinite attempts not guaranteeing death. Compare it to coin flips or lotteries with static probabilities. Sure, those don’t let you escape forever. But 1/n^p decays, and for p > 1, it decays enough. The p = 1 cutoff is as predictable as gravity: it’s where convergence flips to divergence. No mystery, no intuition-busting twist, just series doing what they do.
Could you explain what is so counterintuitive?
Sure, the math is clear. But it is odd to suppose that there is some kind of a metaphysical restriction on how fast the reliability probabilities of a sequence of beings can go to zero.
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