Two players take turns beside a growing pot. The player whose turn it is can take the bigger share and end the game, or pass and let the pot double for the other player. Reasoning backward from the last move says take at once. Real people usually pass for a while, and a small doubt about the other player is enough to make that sensible.
You are P1 and move first. At every node the pot is split 80/20: whoever takes gets 80% and the other player gets 20%. Passing doubles the pot and hands the move to the other side. If nobody takes, the final pot is split 80/20 in favor of the player who would have moved next. Pick an opponent and see how much you walk away with.
How to read this: play runs left to right. At each circle the mover can Take (T, down, ends the game) or Pass (P, right). The two numbers under each T are what you (green, top) and the opponent (amber, below) get if the game stops there.
The whole game is one list. take[k] is what both players get if the mover stops at node k, and the last entry is what happens if everyone passes. Its mover index n % 2 belongs to the player who would move next, which is exactly the flipped split.
Rational plays the subgame-perfect strategy from Lab 2: take at every node. Trusting passes until its own final move, then takes, because no later move can reward a pass there. Random flips a fair coin at each node.
Start at the last node, where the choice is easy because nothing comes after it. Record what the mover picks and what both players end up with. Then step one node back: that mover now knows what passing leads to. Repeat until you reach node 1. This is backward induction, the same idea as 004 · Dynamic programming filling a table from the base case.
How to read this: solving starts at the right. The box above each circle shows what both players end up with from there (you in green, opponent in amber).
Try this: press Step and follow the gold edges as they spread back to node 1.
One pass from right to left, so the cost is O(n). Uses centipede() from Lab 1.
Taking gives the mover 80% of the pot Pk. If the next player takes, the mover gets 20% of a pot that is g times bigger. For every growth on this slider, 0.8 beats 0.2·g, so the last mover takes, so the one before takes, and so on back to node 1.
A subgame-perfect equilibrium is a strategy pair that is a Nash equilibrium in every subgame, which here means at every node. With no ties, backward induction finds exactly one: both players take everywhere, and the game ends at node 1 with $4 and $1.
A simplified model. You are P1 and plan the node where you will take. Your opponent is an altruist who always passes with probability q, and otherwise a selfish player who takes at a fixed node. Each plan's expected payoff is a weighted average of how it does against the two types. Bars use a log scale. The pot doubles each pass.
How to read this: each bar is a plan: "@3" means take at node 3. Taller means more money on average. The strip below shows which plan is best for every chance q that your opponent is an altruist; the ▲ marks the current q.
A stop of None means never take. Uses centipede() from Lab 1. Node indices in the code start at 0, so your first move is k = 0.
Robert Rosenthal introduced the game in 1981. In 1992 Richard McKelvey and Thomas Palfrey ran it in the lab with an 80/20 split and a doubling pot, the same rules used on this page. Most players passed at least once, and games that ended at the first node were rare.
Their explanation used the idea in this lab: if a few players might be altruists, a selfish player can do better by passing early. The pot grows so fast that a tiny chance of a big payoff outweighs a likely small loss. Their full model also lets selfish players imitate altruists, which this simplified version leaves out.