Consider a game with two players.

Each player starts with a certain amount of tokens (say 100 each).

Each turn, they gain a set number of tokens (say 10 each).

They then bid a whole number of tokens, between 0 and the amount they currently have. Each player bids without knowing what the other player has bid (a 'blind simultaneous bid').

Each player loses the number of tokens they bid.

If one player bid more than the other, that player gains a point.

IMPORTANT NOTE: points and tokens are completely separate.

The game ends when one player's point total is ahead of the other player's point total by a certain amount (say 8).

Players don't lose by running out of tokens.

My question is this: is there a strategy such that no other strategy has a more than 50% chance of beating it?

I think the answer is yes, and I think it would involve bidding a number of tokens randomly chosen from a range of numbers. But I have no idea how to start working out what that range would be.


1 Answer 1


This isn't an answer really, but its also too long for a comment.

This is an interesting little game; did you define it yourself or did you find it somewhere? As presented, it is really a 3 parameter family of games defined by the starting number of tokens $s$, the tokens gained per turn $g$, and the difference in points required to win $d$.

I am going to assume in what follows that players do not gain tokens on the very first turn of play; without loss of generality we can simply absorb such earnings into the starting amount.

An immediate thought is that depending on the parameters you play with, there might not be an end to your game! If we have that $d>s>g$, the game will never end so long as one player simply bids $g$ each turn. Thus, in all of these instances there is in fact a non-losing strategy, which is stronger than what you asked for.

This makes me believe that there is probably going to be some kind of case analysis for the various relations between your parameters. We have already addressed one of them, and there are two more that we can answer pretty quickly; if $d>g>s$ or $g>d>s$ then the game also never ends, and again we have non-losing strategies where you bid $s$ on the first turn and $g$ every turn thereafter. Note that these two cases cannot happen if you gain tokens on the first turn.

Unfortunately these cases are not the ones we are interested in; requiring that the winning difference be larger than the starting number of tokens makes for dull play. But there are other easy cases to tackle, like $d=1$. This too has a non-losing strategy, and results in a never ending game.

What about a small game, like $(s,g,d)=(3,1,2)$? This game tree is already quite large, and I do not feel like exploring it right now.

Nothing I have said so far is useful for the actual interesting cases of the game, but hopefully I have highlighted that any kind of analysis you do on the strategies for playing this game, randomized or not, is going to be very sensitive to certain parameter relations. An exact analysis is likely to be very time consuming even for small games, and so going forward you may want to try and explore these games algorithmically or collect data "experimentally" by making computers play against each other.

  • $\begingroup$ Thanks. I thought of this myself, but there are very similar ideas out there. $\endgroup$ Commented Apr 21, 2019 at 22:26
  • $\begingroup$ I'm not sure that bidding g every turn would really win. What if the opponent bid g+1 every time they had that many counters, and 0 otherwise? $\endgroup$ Commented Apr 22, 2019 at 15:27

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