It seems like your best strategy is to bet when you have a 3 or 4. In random dealing that means you’ll be betting in 50% of the deals. I’m assuming both players follow the same strategy.
25% of the time, you’ll have a four and it’s an automatic win for you. You win one token against a one or two and two tokens against a three.
25% of the time, you have a three. You have a 2/3 chance of winning and a 1/3 chance of losing to a four. That means you’ll win more times than you’ll lose in the long run. But you’ll break even compared to your opponent because you only win one token when you win but he wins two tokens when he wins.
You don’t get to control the strategy of your opponent though, they are free to employ any strategy they want.
So eg, if your strategy is to “always ask for the token”, then I could combat that by always asking for the token if my card is a 2, 3 or 4, but never if I’m holding a 1. That would give me a >50% chance of beating you.
But if you both always ask, then why would you ask for the token if you have a one? You’re guaranteed to lose. The only way betting on a one makes any sense is if you know your opponent doesn’t bet on twos - in that case, you have a 1/3 chance of beating him with a bluff.
I don’t think that matters. It is equivalent to getting +2 from the pot. The question basically boils down to if your expectation for each play is greater than or equal to the expectation your opponent has given the best strategy to counter your strategy.
For example, if you bet only when you have a 4, your opponents best strategy is to bet all of the time your expectation is (1/4)x2 - (3/4)x1 or -1/4*. Your opponent’s expectation is (3/4)(1/3)x-2 + (3/4)(2/3)x1 + (1/4)x1 = +1/4
Let’s turn it around. If you bet all the time and your opponent only bets 3s and 4s then their expectation is [(1/4)x(2/3)x2 + (1/4)x(1/3)x-2]*** + (1/4)x2**** - (1/2)x-1 = 1/6. Therefore your expectation is -1/6
If you bet 3 and 4, your opponent bets on 4. 1/12 of the time he wins 2, 2/12 of the time he wins 1 and 5/12 of the time you win 1 making your expectation 1/12.
If instead he extends out to betting 2, 3 and 4 then
1/12 of the time he wins 2 and 2/12 of the time he wins 1 (He has a 4)
1/12 of the time you win 2 and 2/12 of the time he wins 1 (He has a 3)
2/12 of the time you win 2 and 1/12 of the time he wins 1 (He has a 2)
2/12 of the time you win 1 (he has a 1)
Your expectation is 1/12
If he plays the same strategy as you the expectation is 0 and if he bets all the numbers your expectation is 1/6.
So betting every 3 and 4 and not 1 and 2 is a winning strategy.
And clearly betting 2, 3, 4 is not as a defense of betting 3 and 4 only gives you an expectation of -1/12.
*1/4 of the time you have a 4 and win 2 chips. 3/4 of the time you don’t have a 4 and your opponent wins 1 chip simply by betting.
**3/4 of the time your opponent doesn’t have a 4 and bets. 1/3 of THOSE times you have a 4 and win 2 chips but 2/3 of the time you don’t have a 4 and your opponent wins 1 chip. 1/4 of the time your opponent has the 4 and wins 1 chip.
*** Betting on a 3
**** Betting on a 4
***** Not betting on a 1 or 2
God I hope I did the math right!
If by “fixed strategy” you mean non-mixed strategy, then one must solve the game to find out. If, however, mixed strategies are allowed then the thread-title must have a Yes answer, since the game is symmetric and will terminate.
(In many games of this type, the optimal mixed strategy will involve bluffing, but it can be easily seen that bluffing has no benefit in this game.)
What makes this game difficult is that each round depends on the score. The optimal strategy when each player has 199 tokens need not be the same as when each player has 100 tokens. However, in the early game, when the 200-token goal is far away, the chance of ultimate success will depend almost linearly on token difference so that, at least for the early game, we need only maximize the expected token difference.
When you win, your gain relative to opponent is +1 if he passes, +3 if he plays.
I think your analysis is correct when the called-win is only worth +2 folded-wins. But, a score of 102-to-99 is three times as good as a score of 101-to-100 (relative to 100-to-100).
Using that +3 value for a called-win; your expectation (betting 3 and 4) is +1/2 against opponent who plays all; +1/4 against opponent who plays 2,3,4; 0 against opponent who plays 3,4; 0 against opponent who plays 4 only.
Note that, although bet 3 and 4 is the best strategy, betting only 4 yields as good a result! (Assume opponent tries to counter by betting 2,3,4; he wins 1 in cases (2,3) and (3,3), but loses 3 instead of 1 in case (2,4).)
It’s easy to show that an optimum strategy exists: In any given situation, you could in principle calculate your odds of winning given each strategy for you and each strategy for your opponent, and choose the strategy which maximized your odds of winning. And given that the game is perfectly symmetric, the optimum strategy must of course result in a 50% chance of victory.
It’s worth noting, by the way, that the game described here is a simplified version of Goofspiel, which I believe has been solved by computer.
However, if a perfect strategy exists, guaranteeing a 50% win rate, it must be assumed that your opponent also knows and follows that strategy. So, in the long run, you are guaranteed a 50% win rate, but can never exceed that.
Same as any other game of chance, and by locking into a formula that is dictated by the random fall of a card (always bet 3-4, never bet 1-2), this then simply becomes a game of chance. Your opponent always bets 3-4 and never bets 1-2, and is also guaranteed 50% win rate.
Suppose the opposition tries the naive stragety and always cut their losses on receipt on 1 and 2, and do not play cards on 1 or 2.
So whats the outcomes
Oponent: You result
If you “know” that your opponent always cuts their losses when they get a 1 or 2 but always plays cards on a 3 or better,
THEN you know to always play cards - never cut your losses. This is because you may win the one token every time they get a 1 or 2.
However, should they deduce that you always play cards, then they may also defeat your strategy by always playing cards. At which point , you can know that you can defeat their strategy by not playing cards when you get a 1.
For any one round, you may have a good chance to win,
but perfect logicians who also count the cards (“you mostly always cut your losses on a 1 !” ) facing off may deduce that they should more vary between playing and cutting their losses on receipt of 1 and 2, and the randomness there means the result is 50/50…
But the game is equal to both players. I don’t see how either player can have a better strategy. Obviously your opponent might just play poorly but taking advantage of a weak opponent isn’t really a strategy. And you ruled out any strategy based on “reading” your opponent.
OP never clarified whether by “fixed” strategy he meant “pure” (i.e., non-mixed) strategy, but something of possible interest occurred to me. As usual, let’s assume players’ token counts are equal and far from their goal, so we needn’t worry about asymmetry or endgames.
This game is one of a broad class of games that always have a pure strategy solution! This will apply even if the payoffs or number of cards are changed.
I won’t try to prove this fact, or even articulate it rigorously. It follows from the fact that strategies are non-antagonistic: if our opponent plays looser, we should also loosen (rather than tighten).
(The same loosen–>loosen strategy applies to all poker variations, but any more complicated than OP’s will admit bluffing, which will lead to strategy mixing.)
That simple games with “monotony” properties like OP’s always have pure-strategy solutions may be “well known.” But only well known to those who know such things well, which evidently does not include me.