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I've got a question about a game taken from a book called Rachunek prawdopodobieństwa dla (prawie) każdego by Jacek Jakubowski and Rafał Sztencel.

Adam and Bolek have a machine that generates a pair of random natural numbers that are next to each other (like $1$ and $2$ or $55$ and $54$) and each boy gets a different number from the pair. They know the number of their opponent but they don't know their own numbers.

Here are the rules:

  • The winner is the one with greatest number.
  • The loser must give the winner as many coins as great is the loser's number (so if the loser has number $40$, he has to give 40 coins to the winner).
  • If a player doesn't like his number he can decide to get a new pair of random numbers.

Both players will never want to generate a new pair, because they think like this: There is a $50\%$ chance that I am the winner. And if the number of my opponent is $k$ then I'll win $k$ coins or lose $k-1$ coins. So my overall score would be $\frac{k - (k-1)}{2} = 0.5$ of a coin.

This strategy is of course based on wrong assumptions - the set of numbers that can be generated by the machine cannot be infinite. Additionally, it can't be that both players are winners of the game.

My questions are:

  • What would be their strategy if they knew that the set is finite?

  • Is their strategy truly wrong even if the set is infinite? If their strategy turns out to be a winning one, how could it be that both of them are going to be the winners of the game?
    Apparently, this is not true because:

    "One can define a uniform distribution on some infinite sets (uniform on $[0,1][0,1]$ is totally legit). But one cannot define a uniform distribution on an infinite yet countable set, such as the natural numbers."
    ~ Clement C.

    Here is an additional reference with a good explaination.

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Mateusz Piotrowski
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    When can a player decide to get new numbers? Is it possible that player A says 'I am happy with thr numbers' then player B says 'I am also happy with the numbers' and then A says 'oh, but then I want new numbers' before any numbers are revealed? – Elsa Jan 10 '16 at 16:12
  • @Elsa. Well, a player may want to change the numbers when he gets his number. It isn't specified in the book if they can make a decision based on the opponent's decision. – Mateusz Piotrowski Jan 10 '16 at 16:17
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    You could read up on the two envelopes problem I don't see why they would or would not want to get new numbers-the same calculation after switching applies so switching (in the reasoning presented) is a nonevent. That is why the usual problem is posed with one player able to switch envelopes – Ross Millikan Jan 10 '16 at 16:22

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In the original question, the lack of a uniform distribution on the naturals saves you. The probability of all the numbers has to sum to $1$, so the probability has to decrease with $n$ eventually.

If you limit the number to the range $[1,N]$, a player who sees $N$ will certainly want to change numbers as he knows he is losing. If either player can force a change, a player who sees $N-1$ should request a change, because either his opponent sees $N$ and will also request a change or his opponent sees $N-2$ and is winning. Then a player who sees $N-2 \dots$

Ross Millikan
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  • So the game wouldn't end if the set is finite and both players can request a change? It makes a lot of sense. Even if they don't know what is the upper bound of the set they can still request a change until the opponent has a number $0$. But then the player with $0$ would request a change ... One way or another, this game is quite broken. – Mateusz Piotrowski Jan 10 '16 at 16:36
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    You have to decide if only one change is allowed or if they can keep requesting one. If they only get one chance to switch and you want to do the calculation the way you did in the question, you have to think about how the end effects. Clearly you can just rely on the symmetry to say neither one is winning before they look at the numbers. – Ross Millikan Jan 10 '16 at 16:42