Winners and losers

I have tried three times before to formulate a paradox that comes up when countably infinitely many people roll dice, looking for the most compelling version. Let me try one more time. You observe Jones participating in the following game. She rolls a die, but doesn't get to see the result. Over the next six days, on day n she is said to be a winner if n is what she rolled. Thus, Jones will be a winner on exactly one of the next six days. Only after the end of the six days will Jones find out on which day she was a winner. Let's say it's now day n. What's the probability that Jones is a winner now? Surely it's 1/6.

But now you find out that infinitely many people independently participated in this game. And that, completely unsurprisingly, each of the possible six die outcomes were had by infinitely many players.

Furthermore, you find out that the organizers set up the following thing. On each of the next six days, they paired off the players, with the following properties:

1. On any given day, every player is paired with some other player.
2. No player is ever paired twice with the same player.
3. A winner is always paired with a loser.

(Assuming that each die outcome was rolled by infinitely many people, it is possible to have such a pairing.)

So, it's day n, and you observe Jones and Smith paired together. What probability should you assign to Jones being a winner?

Suggestion 1: Jones and Smith are exactly on par, so you should assign equal probability to each being the winner. So you assign 1/2 to each. But now repeat this every day. By the same reasoning, on each of the six days, you have probability 1/2 that Jones is the winner on that day. But you can't assign probability 1/2 to Jones being the winner on each of the six days, since Jones is certain to be the winner on exactly one of the six days, and so your probabilities will be inconsistent.

Suggestion 2: Stick to your guns. Assign probability 1/6 to Jones being the winner. But what goes for Jones goes for Smith. So you should assign 1/6 to Smith being the winner. But this probability assignment is inconsistent as well, since it's certain that one of the two is the winner.

Well, that didn't work out too well, did it. Maybe it's time to get a bit more elaborate in our suggestons.

Suggestion 3: Constantly shifting probabilities. Let's say on day 1 you observe Jones and Smith paired together. You assign 1/2 to each being the winner on day 1. Then on day 2, you observe Jones and Kowalska paired together. You likewise assign 1/2 to Jones being the winner on day 2. But since you're not certain that Jones didn't win on days 3,4,5 or 6, you downgrade your probability that Jones won on day 1. I don't know what you do about days 3,4,5 or 6. But what do you do about Smith? On day 2, you observe Smith and Ahsan paired together. If you downgraded your probability that Jones was the winner on day 1, you should also downgrade your probability that Smith was. But that results in inconsistent probabilities, too. Moreover, this suggestion crucially depends on the days being sequential. But we can come up with a variant story on which all the pairings are simultaneous. For instance, let's suppose that the organizers put strings, with attached tags printed 1,2,3,4,5 or 6, between pairs of participants—a nasty tangle of them—with the property that each participant has six strings attached to her, with the property that the string tagged n joins a participant who rolled n with one who didn't, and that each participant has coming out of her strings numbered 1,2,3,4,5 or 6. Jones and Smith have string number 1 joining them. What's the probability that Jones rolled 1? The same issues come up, but simultaneously.

Suggestion 4: No numerical probabilities. Once you see the pairings, you no longer assign numerical probabilities to who is the winner when. Perhaps you assign interval-valued probabilities, presumably assigning the interval [1/6,1/2] to each ``x is a winner on day n'' hypothesis. Or perhaps you have no probability assignments, simply sticking to non-probabilistic claims like that Jones is a winner on exactly one of the six days, and ditto Smith, and Jones is a winner on day 1 if and only if Smith is not a winner on day 1.

This may be what we have to do. It tracks the fact that the most obvious ways to arrange a pairing satisfying (1)—(3) all require a numerical ordering on the participants, and we are given no information about such numerical orderings, but there is no meaningful uniform probability distribution over infinite orderings.

But Suggestion 4 doesn't come cheap, either. We still have to answer questions about what bets are prudent under the circumstances.

Moreover, we have to ask at which stage of the experiment you drop your numerical probabilities. The radical suggestion is that you do so when you find out that there are infinitely many die rollers. This is radical: it means that we can't do epistemic probabilities if we know we live in a universe with infinitely many die rollers. That may be true (and if true, it means that scientific arguments for an infinite multiverse are self-defeating), but it's really radical.

The less radical options are: (a) drop your numerical probability assignments, say for Jones, when you find out that a pairing satisfying (1)—(3) will actually be announced and (b) drop your numerical probability only when you actually see the pairings.

I think (a) might actually be better. For as soon as you know that such a pairing will be announced, you can stipulate a name, say "JonesPair1", to refer to the first person that Jones will be paired with. And it would be odd if you then changed your probabilities for Jones and JonesPair1 on day 1 just because you found out that JonesPair1 had the legal name "Smith", that she was female, etc.