Toddler poker

In a previous post, I took a look at “baby poker”, a game involving two players rolling a six-sided die. The higher number wins, but players may elect to raise, call, or fold depending on their number (which only they can see). In this post, I’ll take a look at the continuous version of the problem (also appeared in a recent Riddler post!) Here is the full text of the problem:

Toddler poker is played by two players. Each is dealt a “card,” which is actually a number randomly chosen uniformly from the interval [0,1]. (It could be 0.1, or 0.9234781, or 1/π, and so on.) The game starts with each player anteing \$1. Player A can then either “call,” in which case both numbers are shown and the player with the higher number wins the \$2 on the table, or “raise,” betting one more dollar. If A raises, B then has the option to either “call” by matching A’s second dollar, after which the higher number wins the \$4 on the table, or “fold,” in which case A wins but B is out only his original \$1. No other plays are made.

What is the optimal strategy for each player? Under those strategies, how much is a game of toddler poker worth to Player A?

Extra credit: What if the value of the raise is \$k — i.e., players stand to profit \$k instead of \$2 after the raise?

Here is my derivation:
[Show Solution]

If you’d like the tl;dr instead:
[Show Solution]

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]

The war game

This Riddler puzzle is about game theory… War or peace?

Two countries are eyeing each other’s gold. At the beginning of the game, the “strength” of each country’s army is drawn from a continuous uniform distribution and lies somewhere between 0 (very weak) and 1 (very strong). Each country knows its own strength but not that of its opponent. The countries observe their own strength and then simultaneously announce “peace” or “war.”

If both announce “peace,” then they each stay quietly in their own territory, with their own gold, which is worth \$1 trillion (so each “wins” \$1 trillion). If at least one announces “war,” then they go to war, and the country with the stronger army wins the other’s gold. (That is, the stronger country wins \$2 trillion, and the other wins \$0.)

What is the optimal strategy of each country (declaring “peace” or “war”) given its strength?

Extra credit: What if the countries don’t announce at the same time and instead one announces first and the other second? What if the value of winning the war were \$5 trillion rather than \$2 trillion?

Here is my solution for the first part, where both countries declare their intentions simultaneously.
[Show Solution]

Here is my solution for the second part, where the countries declare their intentions sequentially.
[Show Solution]