game with a not too obvious solution

well · #1 10-09-2003, 01:46 PM

When someone asked me to solve this game, I first thought there was no solution. Later on I realized there was, and I was surprised...

The ante for this game is $1 for both you (I) and your opponent (II). II takes two pieces of paper and writes secretly on both of them a whole number between minus and plus infinity. II puts the papers face down on the table. I take one of them (my dicision), and have a look at it. I may now either keep it and give the other one to II, or give it back to II and take the other one.
The one with the highest number wins the pot.

Is there a strategy that makes the EV for I strictly positive? If so, give the strategy or an example of such a strategy.

As you suspect, the opposite of what you expected is true: there is...

Next Time.

pudley4 · #2 10-09-2003, 03:40 PM

Always switch if you pick a negative number. Never switch if you pick a positive number.

well · #3 10-09-2003, 06:31 PM

That's not a right observation!

Since your opponent may choose the numbers himself, he only has to pick two numbers with the same sign and there it is: EV=0, wich means no strictly positive EV.

Next Time.

1800GAMBLER · #4 10-09-2003, 09:59 PM

If he filled the size of the page with numbers and it's a minus choose the other one. If he did it with a positive keep it.

If he filled 3/4 of the page and then looked tired or bored of this game (unsurprising) follow the above pattern.

1800GAMBLER · #5 10-09-2003, 10:05 PM

Seriously though. Since you have no idea if is going to be tricky with the signs you know he isn't going to be tricky with the length of the numbers.

Hence if you see it and think he writen something which is longer than his average and the number and it is positive then go less. If it's above average length and minus go higher.

Meanwhile you have no EV if you think the number is smaller than his average.

Copernicus · #6 10-10-2003, 07:28 AM

If you use words like "If you think", then the answer cant result in a strictly positive EV, which I cant find, and somehow doubt exists!

1800GAMBLER · #7 10-10-2003, 07:48 AM

After playing this game once you can get an average of the length of his numbers. First game go by half the length of the card.

Othernote:

If you think it's 70% a player has a worse hand and,
If you think it's 30% a player has a better hand.

Then it's EV when you bet the river. That can be 'strictly' EV by grouping together all the possible hands.

well · #8 10-10-2003, 08:24 AM

Hadn't time to overthink all of your suggestions, but most of them depend on analysis of the opponent. I am looking for a solution that on EVERY imaginable opponent results in a strictly positive EV.
Hint: you may have to use a calculator and/or a random number generator before you make your action.

Maybe this will help you all.

Next Time.

irchans · #9 10-10-2003, 10:12 AM

switch with probability f(x) where f(x) is strictly decreasing, 1>f(x)>0, and x is the amount written. For example f(x) = 1/(1+exp(x)).

Copernicus · #10 10-10-2003, 10:38 AM

If II generates his numbers randomly, and given that the tails are infinite, I cant think of any probability function that will have an EV other than 0. No matter how large or small the first number he picks is, there is still a "50/50" chance that the next number will be greater or smaller. If II can have a strategy where even he cant predict the relationship between the two numbers, how can I do better than 0 EV?

If the tails were finite but very large then a f(x) like you are proposing might work.