cooperation game

I'm currently doing some research calculating strategies of two-player games with imperfect information distribution (for example poker).

Reading about Nash equilbria (in short summary: no player can improve his strategy knowing all other players strategies), I remember a coorporation game from my childhood (and could not solve by means of logic).

The game is as follows: Two players each pick a natural number from 1 to 10, and reveal it at the same time. If both numbers are off by at most 1, both players win.
What is the best strategy for the one player, under the conditions that the other player will play the best strategy (and vice versa) ?

In my naive days my logic was like this:
Obviously you should never pick 1, since 2 is a dominant strategy over 1 (2 will win on others pick 1-3, while 1 will win only on 1-2).
So you can assume that you (and likewise the other player) should never pick 1.
But then, if the other player will never pick 1, 3 is a dominant strategy over 2, since 3 will win on 2-4, while 2 will win on 1-3 (but 1 is never played).
With the same logic going further, n+1 will be a dominant strategy over n, since n-1 is never played.
By this conclusion the best strategy should be 10, which is a contradiction since 9 is always a dominant strategy over 10 - so the initial statement "Obviously you should never pick 1" must be wrong.

Now, what is the best strategy ? I.e. what is the Nash equilibrium for this game, and is it a unique equilibrium ? Should you pick 1 and 10 with non-zero probability although you know that 2 and 9 are a dominant choice ? Maybe the best strategy would be to pick either 5 or 6, but by which logic ? - obvisously you win each time if played by both player, but this is also true for the strategy "always pick 1".

This reminds me a little of a this puzzle:

A drill seargent says to his platoon that one day of the week (Mon-Fri) he will do a surprise inspection. He will choose the day so that the platoon will never know for sure when it happens. What is the probability of the inspection for each day?

To get back to the question at hand, my inclination would be to pick 5 or 6 for the reasons stated above, and the logic of the inspection puzzle at least seems to push away from the extreme ends.

Quote: Wizard

This reminds me a little of a this puzzle:

A drill seargent says to his platoon that one day of the week (Mon-Fri) he will do a surprise inspection. He will choose the day so that the platoon will never know for sure when it happens. What is the probability of the inspection for each day?

At the beigining of the week it should be 1/7, after sunday 1/6, and so on. Or so I think, surely I am missing some smarter-than-me-math.

Quote: AZDuffman

At the beigining of the week it should be 1/7, after sunday 1/6, and so on. Or so I think, surely I am missing some smarter-than-me-math.

First, it was supposed to be a five-day week, for simplicity. You might check this hint, but you won't get full credit for the right answer if you do.

I suppose the boss could add in a factor to the equation: his awareness of the employees/platoon's general sloth or maticulousness could weigh the desicision to accelerate or delay the hardship of the look-see. Lazy underlings get a full-treatment Monday white-glove inspection, and goody-two-shoes workers get a Thursday pat-on-the-head, and a pizza & beer party.

Quote: AZDuffman

Or so I think, surely I am missing some smarter-than-me-math.

( I heard the puzzle in a version of a prisoner awaiting his execution day somewhere within this week, but the judge promised him he will never know which day it is).

Mango got it right.

For those who still don't give up, the perspective is from early Monday. Yes, I think I've heard the prisoner on death row version as well.

There is no unique Nash equilibrium to the original puzzle. Many mixed & pure strategies can't be improved upon:

example: Player A chooses 3. Player B's best response is any mix of 2,3,4. Now Player A cannot improve by switching. Neither can B. Nash equilibrium found.

It's an interesting paradox, but ultimately, not that interesting a math puzzle.

Quote: MangoJ

I'm currently doing some research calculating strategies of two-player games with imperfect information distribution (for example poker).

Reading about Nash equilbria (in short summary: no player can improve his strategy knowing all other players strategies), I remember a coorporation game from my childhood (and could not solve by means of logic).

The game is as follows: Two players each pick a natural number from 1 to 10, and reveal it at the same time. If both numbers are off by at most 1, both players win.
What is the best strategy for the one player, under the conditions that the other player will play the best strategy (and vice versa) ?

Here's what I get, although my logic is probably suspect:
Let P(1), P(2), ..., P(10) be the probability of you choosing 1, 2, ..., 10 respectively; the sum of the P values = 1.
The probability of winning is the same regardless of what the other player chooses if all ten of these sums are equal:
1: P(1) + P(2)
2: P(1) + P(2) + P(3)
3: P(2) + P(3) + P(4)
4: P(3) + P(4) + P(5)
5: P(4) + P(5) + P(6)
6: P(5) + P(6) + P(7)
7: P(6) + P(7) + P(8)
8: P(7) + P(8) + P(9)
9: P(8) + P(9) + P(10)
10: P(9) + P(10)
#1 = #2 only if P(3) = 0; #9 = #10 only if P(8) = 0
#6 = #7 only if P(5) = P(8), so P(5) = 0
#3 = #4 only if P(2) = P(5), so P(2) = 0
#4 = #5 only if P(6) = P(3), so P(6) = 0
#7 = #8 only if P(9) = P(6), so P(9) = 0
This reduces the sums to #s 1 and 2 = P(1), #s 3-5 = P(4), #s 6-8 = P(7), and #s 9-10 = P(10).
#2 = #3 only if P(1) = P(4)
#5 = #6 only if P(4) = P(7)
#8 = #9 only if P(7) = P(10)
This is possible only if P(1) = P(4) = P(7) = P(10) = 1/4.
The strategy appears to be choose 1, 4, 7, or 10 equally.
Assuming the other player does the same thing, both players must choose the same number to win; this happens 1/4 of the time.

Well, all those puzzles are all fun to talk about, but what is the solution ?

I'm not a mathematician, but I heard a Nash equilibrium are guaranteed under very general premises.

Using the approach that a expected result of a good strategy should be identical to all of the other players action,
one should pick the numbers 1, 4, 7 and 10 each with same probability 1/4. With this strategy, regardless of the other players strategy, the expected win is always 1/4.

If this is indeed the Nash equilibrium, it suggests some "funny" moves (picking 1 and 10 instead of 2 and 9).
This reminds me of the Cold War, where risking the global atomic war suddenly became an "option".