[superprismatic]

Consider the following two-player

game: Each player secretly chooses

a positive integer. The integers

are revealed simultaneously. The

object of the game is to choose an

integer which is either exactly one

larger than your opponent's integer

(in which case, your opponent pays

you $2) or which is at least two

smaller than your opponent's integer

(in which case, your opponent pays

you $1). If you both choose the

same number, the game is a draw and

no money changes hands. So, for

example, if A picks 6 and B picks 5,

B gives A $2; if A picks 6 and B

picks 2, A gives B $1; if A picks 6

and B picks 7, A gives B $2; if A

picks 6 and B picks 8, B gives A $1.

What is the optimal stratagy for

playing this game? Why?

[Guest]

love to see your answer to this one because mine is that you have to learn and know your opponent.

[Guest]

A good strategy would be to pick low numbers

[Guest]

is there a limit to the integers? can they be negative?

[superprismatic]

is there a limit to the integers? can they be negative?

There is no upper limit to the integers. The OP does specify positive integers, so the lower limit is 1.

[Guest]

if you choose the number 2, your opponent can not choose a value 2 lower than you. That means you win if your opponent chooses 1 or a number greater than 3. If you both choose 2, it's a draw. You've essentially limited your opponent's winning numbers to only the number 3.

This is marginally better than choosing 1 and eliminating your chance of winning $2.

[Guest]

if you choose the number 2, your opponent can not choose a value 2 lower than you. That means you win if your opponent chooses 1 or a number greater than 3. If you both choose 2, it's a draw. You've essentially limited your opponent's winning numbers to only the number 3.

This is marginally better than choosing 1 and eliminating your chance of winning $2.

After one round, your opponent would catch on and pick 2 also. Then it would be a draw every round after the intial round (hopefully with your getting $2 as the result). Something about this question reminds me of game theory.

[Guest]

My first thought on this is that the strategy should be the pick one of two or more numbers each time with a certain probability... Its late here so will think about it tomorrow

[Guest]

Hitting a zero each time, win will be dollar each time -Conservative approach

[Guest]

That would work, except zero isn't positive and the rules say 'Each player secretly chooses a positive integer'.

Edited October 8, 2009 by JonTam

[roolstar]

After one round, your opponent would catch on and pick 2 also. Then it would be a draw every round after the intial round (hopefully with your getting $2 as the result). Something about this question reminds me of game theory.

Your idea is on the right track but you missed the right destination.

ROUND 1: You say 2 he says 4 or above (best case scenario) = you get 1$

ROUND 2: You say 2 he says 4 or above again = you get 1$

ROUND 3: You say 2 he says 3 (He picks up your game!!!) = You pay him 2$

Round 4: You say 1 (in case he says 3 again) but he says 2 = you pay him 2$

ROUND 5: you both say 2 = no money exchanged

... And it remains this way until one of you is bored and starts the scenario once again by saying 3 first, and we are back at ROUND 3!!

I fully agree on the game theory analogy!!!

However here, it's not a zero sum game just yet:

By round 5 where the draw starts to repeat itself, YOU are down by 2$. in other words, you a probably more encouraged to say 3 first again, and go back to 2 in the next round, this way you get your 2$ back, and the draw starts repeating itself again.

THERE IS NO POSSIBLE STRATEGY FOR YOU TO MAXIMIZE YOUR WINNING IF YOU ARE PLAYING WITH A PERSON AS SMART AS YOU ARE!!

The Reason is simple: there is nothing you can do to prevent your opponent from following the exact same strategy that you chose and end up in a draw each time!!!!

In other words, when you both are picking your numbers and saying them at the same second, you will surely reach a zero sum game!!!

So to guarantee a win you have to have your choices affects his/hers; and for that you have to decieve him/her. but since he/she is on the SAME INTELLIGENCE LEVEL, you will not be able to do so.

Example of wrong analysis of this problem: We think that if I choose 1 the opponent has to choose 2 to make a win and if I choose 3 the opponent has to...

But this logic is flawed, because your oponent is thinking the same thing (SAME INTELLIGENCE LEVEL) and therefore, his/her strategy will be to choose the same numbers as you do and not react to your choice of numbers!!

This simple game is not so simple after all but it might be a great measure of intelligence!!!

Edited October 9, 2009 by roolstar

[Guest]

What ever strategy A chooses, B can choose the same. So the expected winnings/losses for both are same.

[bushindo]

What ever strategy A chooses, B can choose the same. So the expected winnings/losses for both are same.

It is possible to construct a guaranteed non-losing strategy, regardless of what the other guy does.

Let's say that I only use numbers from 1 to 5. At each turn, I draw a number from the following distribution, where P(n) is the chance of drawing number n

( P(1), P(2), P(3), P(4), P(5) ) = ( 1/16, 5/16, 4/16, 5/16, 1/16 )

A look will show that regardless of what the other guy does or what numbers he pick, I never have less than a 1/2 chance of winning. I suspect that there are slightly more optimal probability distribution. I'll update with more info later.

Edited October 9, 2009 by bushindo

[bushindo]

It is possible to construct a guaranteed non-losing strategy, regardless of what the other guy does.

Let's say that I only use numbers from 1 to 5. At each turn, I draw a number from the following distribution, where P(n) is the chance of drawing number n

( P(1), P(2), P(3), P(4), P(5) ) = ( 1/16, 5/16, 4/16, 5/16, 1/16 )

A look will show that regardless of what the other guy does or what numbers he pick, I never have less than a 1/2 chance of winning. I suspect that there are slightly more optimal probability distribution. I'll update with more info later.

Sorry for not updating earlier as promised. I was away at camp for three days without access to internet.

As far as memoryless strategy goes, the strategy posted in the previous post guarantees a expected loss of 0 if the opponents plays numbers between 1-5. If the opponent plays numbers > 5, then the previous strategy has positive expecting winnings. This strategy works regardless of whatever the strategy the opponent does. I was thinking that slightly more optimal strategy might exist that has worst case expected loss of 0, and positive expected winnings against some numbers inside the set of 1,2,...5. However, no such strategy exist, since that would contradict with the strategy in the previous post. The strategy posted is the only one that has non-losing property. Any other strategy, i.e. any probability distribution in 5-dimensional vector, can be beaten by some other strategy vector in the same space.

That said, an optimal strategy for this game is probably to start playing with the guaranteed non-losing strategy above, and observe the opponent's moves. Once you have a good idea of his strategy (i.e. the probability distribution of his numbers, or whether he uses information from previous rounds ), you can take advantage of it, assuming that he doesn't use the same non-losing strategy.

Edited October 12, 2009 by bushindo