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The following is a quote from Littlewood's A Mathematician's Miscellany:

(3) The following will probably not stand up to close analysis, but given a little goodwill is entertaining.

There is an indefinite supply of cards marked 1 and 2 on opposite sides, and of cards marked 2 and 3, 3 and 4, and so on. A card is drawn at random by a referee and held between the players A, B so that each sees one side only. Either player may veto the round, but if it is played the player seeing the higher number wins. The point now is that every round is vetoed.

If $A$ sees a 1 the other side is 2 and he must veto. If he sees a 2 the other side is 1 or 3 ; if 1 then B must veto ; if he does not then A must. And so on by induction.

(4) An analogous example (Schrodinger) is as follows.

We have cards similar to those in (3), but this time there are $10^n$ of the ones of type $(n, n+1)$, and the player seeing the lower number wins. A and B may now bet each with a bookie (or for that matter with each other), backing themselves at evens. The position now is that whatever A sees he 'should' bet, and the same is true of $B$, the odds in favour being 9 to 1. Once the monstrous hypothesis has been got across (as it generally has), then, whatever number $n$ $A$ sees, it is 10 times more probable that the other side is $n + 1$ than that it is $n - 1$. (Incidentally, whatever number $N$ is assigned before a card is drawn, it is infinitely probable that the numbers on the card will be greater than $N$.)

While I was able to follow (3), I don't understand (4). Why is it that

whatever number $n$ $A$ sees, it is 10 times more probable that the other side is $n+1$ than that it is $n−1$

The last statement makes no sense to me either. And why is this example analogous to (3)? I would be very grateful for an explicit explanation spelling out what the author means.

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    $\begingroup$ There is 1 0-1 card, 10 1-2 cards, 100 2-3 cards, etc., so if you see (say) a 1, it's 10 times as likely the other side is 2 as that it's zero, etc. $\endgroup$ – Gerry Myerson Dec 22 '17 at 6:40
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    $\begingroup$ Oh, I feel silly now. I interpreted the $10^n$ to mean there were that many total, obviously wrong. Is this a correct interpretation of the point of his statement then- if A sees $n$ and B sees $n+1$, then A should bet on $n+1$ and B on $n+2$? I don't really understand the 9 to 1 business, though. $\endgroup$ – Anu Dec 22 '17 at 6:54
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Perhaps the most important bit is:

The following will probably not stand up to close analysis, but given a little goodwill is entertaining.

The distributions of cards in both cases are not really well-defined. Especially in case (4) if we try to assign a probability to each card (in either scenario), we will find no formulation that satisfies the axioms of probability. In case (3), we can come up with a formulation where the cards are not equally likely, but no formulation where the cards are all equally likely. But, setting that aside...

If we pretend we are A and have seen $n$, then we know that the card is either the card is $(n - 1, n)$ or $(n, n + 1)$. Then we believe that since there are ten times as many cards with $(n, n + 1)$ (there are $10^n$) than there are with $(n - 1, n)$ (there are only $10^{n-1}$), and given the assumption that all cards are equally likely, then it should be that $(n, n + 1)$ is ten times as likely. So, we believe we are ten times as likely to win than to lose.

This is equivalent to $10$ to $1$ odds, by my understanding -- I believe $9$ to $1$ is actually wrong. Odds are not a common way of expressing probability in mathematics, so I suppose the conversion could have been done an additional time as a mistake -- a probability of $\frac{1}{10}$ gives odds of $1$ to $9$, but here the actual probabilities are $\frac{1}{11}$ and $\frac{10}{11}$. The "ten" in "ten times as likely" is already odds.

The scenario is symmetric, so we feel B should believe the same thing. But this seems to be a paradox, because it can't both be true that A is more likely to win and B is more likely to win.

The similarity to (3) is that in both cases we have a sort of contradictory scenario where in (3) both players believe that they are likely to lose (or at least that they will not win) and so they should both veto whereas in (4) both players believe that they are likely to win so they should both bet.

For further reading, I would suggest looking at the two envelopes problem which is (I feel) a similar scenario, but more well-formed. (The usual formulation only has a single participant, but it is easy to imagine one player with each envelope and each deciding whether switching is favorable.)

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  • $\begingroup$ Thank you so much, this makes a lot more sense to me now! $\endgroup$ – Anu Dec 22 '17 at 11:38

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