View Full Version : Is this game solveable?

05-13-2005, 07:12 PM
Headsup match.

I have 20 red chips, you have 20 blue chips.

We both choose, unseen by the other and simultaneously, how many chips to gamble. These chips are placed into a container, shaken up, and one is selected at random by the referee. The player whose chip was selected wins all the chips in the pot.

His opponent's chips are replaced with chips of his colour, and we repeat the wager.

So, what's the optimal strategy?

My envelope is a horrible mess right now, involving the probability of your opponent playing a certain number of chips.

There's a part of me that says this is a trick question because the marginal return of betting more than 1 chip decreases - you're adding 1 to the prize pool but not increasing your share by 1/x but instead, 1/(x+1) so obviously the best strategy is to keep betting one chip and hope to suckout.

05-13-2005, 07:29 PM
What do you mean by "optimal"?

Under the usual criterion of maximizing your return, it's a non-question. Every possible wager has break-even expectation. You put in X chips, he puts in Y chips; you have X chances to win Y and Y chances to lose X.

The only thing you can choose is how big you want your variance to be. Whooppee.

05-14-2005, 08:04 AM
As Siegmund said, every wager has zero expectation, thus the expectation for any strategy is zero, thus the probability of any player winning is 50% regardless of what strategy he chooses.

05-16-2005, 12:00 PM
Not correct.
If you always put 1 chip, and the opponent always puts all his chips, the probability of win is 0.681
(1/21 + ... + 1/40).

05-16-2005, 12:36 PM
Although the expected value of chips won will always be zero, it doesn't mean that every strategy will produce an equal chance of winning.

Take the following "game" as an example (I realize this game does not involve strategy, but it does involve equal expected value for each trial):

A die is rolled.
If the result is a 1,2,3,4, or 5 player A gets one point
If the result is a 6, player B gets five points

The game is played until one player reaches 5 points.

For each roll, each player has an expected value of 5/6 of a point. However, player B has a 1-(5/6)^5 or ~ 0.598 probability of winning.

05-16-2005, 03:34 PM
Not correct.
If you always put 1 chip, and the opponent always puts all his chips, the probability of win is 0.681
(1/21 + ... + 1/40).

[/ QUOTE ]

That is the wrong sum. (Don't just add those probs.)

Under your strategies, player 2 (always all in) wins if and only if he wins 20 hands in a row. The likelihood of that is
(20/21)*(21/22)*...*(38/39)*(39/40) = 20/40 = 1/2.

This is a symmetric zero-sum game. Therefore, under optimal play, each player will have an equal chance of winning. Proof: Suppose not; then the losing player could simply imitate the winning player's strategy, giving him equal odds.

But this does not prove that all strategies are equally good. That is left as an exercise to the reader. /images/graemlins/smile.gif


05-16-2005, 04:09 PM
In the dice game, it's true that equal expectation doesn't guarantee an equal chance of winning. (Because it the dice game, player A can only win a 5-0 victory, while Player B might achieve anything from a 5-0 to a 5-4 win.)

However, returning to the OP's game:

In his first post it wasn't clear that his goal was to win all 40 chips.

If we make the further assumption that that the two players agree to continue playing until one or the other of them has 40 chips, then the only two possible outcomes of the game are 40-0 and 0-40.

Since the EV of every bet is 0, the EV of the game as a whole is zero; since the EV of the game is zero AND the only two possible outcomes are 'player A wins all the chips' and 'player B wins all the chips', these two outcomes must also be equally likely.

In other words, it is completely irrelevant what strategy either player uses: the probability of winning remains 50-50 no matter what size the bets are.