# Baby poker

Another game theory problem from the Riddler. This game is a simplified version of poker, but captures some interesting behaviors!

Baby poker is played by two players, each holding a single die in a cup. The game starts with each player anteing \$1. Then both shake their die, roll it, and look at their own die only. Player A can then either “call,” in which case both dice are shown and the player with the higher number wins the \$2 on the table, or Player A can “raise,” betting one more dollar. If A raises, then B has the option to either “call” by matching A’s second dollar, after which the higher number wins the \$4 on the table, or B can “fold,” in which case A wins but B is out only his original \$1. No other plays are made, and if the dice match, a called pot is split equally.

What is the optimal strategy for each player? Under those strategies, how much is a game of baby poker worth to Player A? In other words, how much should A pay B beforehand to make it a fair game?

If you’re interested in the derivation (and maybe learning about some game theory along the way), you can read my full solution here:
[Show Solution]

This alternate solution was proposed by a commenter named Chris. Same answer, but a simpler argument!
[Show Solution]

If you’d just like to know the answer along with a brief explanation, here is the tl;dr version:
[Show Solution]

## 11 thoughts on “Baby poker”

1. Hector Pefo says:

Wonderful, thank you! I’d figured out that there was no pure-strategy solution (there’s a quick argument that an optimal pure B strategy against any pure or mixed A strategy has to be a threshold for calling–there are only 7 such strategies–and that the best threshold strategy against A’s best strategy against a threshold strategy is never the same threshold strategy). But I was out of my depth looking for mixed strategies non-numerically. Would you mind giving a brief description of how you found the solutions to the dual LPs?

BTW, the solution for A given in your tl;dr version differs from that in the extended version. I am sure the extended one (always call with a 4) is correct.

Also, I have trouble wrapping my head around the exact significance for NE to rational choice in a one-shot game like this. If B adopts any of the equilibrium strategies you identify, then A’s expectation for a strategy of the form [p 0 0 0 1 1] is 5/54 independently of p. It’s true that B’s best strategies against A include those equilibrium ones only for p = 2/3, which is what makes that A’s NE strategy. But that makes me doubt that it follows from a strategy being part of a NE that a player has more reason to follow it than non-NE strategies. If A knows for sure that B will choose a NE strategy, maximizing expectation only motivates choosing some [p 0 0 0 1 1] strategy or other. And if A doesn’t know for sure that B will go NE, it’s not clear in what sense it is optimal for A to do so. Sure, he won’t wish he’d chosen a different one if B happens to choose the best counter-strategy to his, as might happen if he picks a non-NE strategy, but why should an expectation-maximizer be especially worried about that?

1. Thanks for the comment; I fixed the typo.

I square it as follows: as you pointed out, A has several possible best responses to B’s Nash strategies, i.e. by choosing a different p. But if A made such a choice, B would have an incentive to deviate from its Nash strategy and improve their expected winnings. The only way A can be assured that they will win 5/54 on average is if they play p=2/3. It’s true that A doesn’t know for sure if B will play a Nash strategy, but if A plays p=2/3, then it’s in B’s best interest to play a Nash strategy as well. If B does anything else, then A will win even more. Likewise, if B plays a Nash strategy, they are guaranteed to lose no more than 5/54 on average. There is no guarantee that A will follow suit and play Nash, but if A plays anything else, B will win even more. I hope that makes sense; I apologize if I misunderstood your question!

1. Hector Pefo says:

Thanks. What you say is correct, but my point is that it’s a minimax justification rather than a maximize-expectation one, and I would think that what we mean by the “optimal” strategy for a player is the one that maximizes expectation, not the one that minimizes possible losses. Maybe there’s an argument for the principle “where expectation is not defined, minimize possible losses”, but it obviously can’t be an argument based on the expected benefit of adopting it!

2. Mark Rickert says:

I second this: “Would you mind giving a brief description of how you found the solutions to the dual LPs?”

I’d be interested in knowing what tool(s) you used (open source solver?).

Maybe that’s what’s in Baby poker revisited?

Thanks!

1. Hi Mark,

Baby Poker Revisited is a different problem altogether — I’m hiding it for now and will probably post at a later date.

I used the JuMP modeling language in Julia together with the CLP open source solver to solve the LPs. I used a random perturbation method to determine whether solutions were unique and to find the full set of solutions in cases where it was not.

I didn’t give more details in my post because it was already getting a little long!

If there is interest I could post code/etc.

1. Mark Rickert says:

Thanks. After posting that request and doing some web searching, I downloaded JuMP and got a single result that fell into the range you specified.

3. Chris says:

Great post! I followed a similar approach but instead formed the LP in the space of the strategy vectors. Basically, I used the function W as the objective and then maximized with respect to p and minimized with respect to q, with p and q each constrained to [0,1]^6. (The order of min and max still doesn’t matter because of duality.) I rewrote the inner problem using LP duality and then was left with a single LP. I think this can generalize a little better because it’s a smaller LP than when you enumerate all the possible pure strategies, so you can solve it quickly if you use n-sided dice. I thought it might be instructive in case anyone’s interested. You get the same answer either way, obviously. Thanks again for your post!

1. That’s an excellent point! I somehow convinced myself that I had to work in the augmented space because the p’s and q’s don’t sum to 1. In fact, all I had to do was remove the constraint that they sum to 1! That’s quite nice… and you’re right, much easier to generalize. Much easier to interpret as well, since I had to do all this extra work to tease out the multiple solutions in the higher dimensional space. Thanks for the comment.

2. I updated my solution to include your approach with the more efficient LPs. Thanks again! I learned something today.

1. Chris says:

You’re welcome! Thank you for your blog posts! You give great explanations.