The Resistor Game Show

Question

You were one of $n$ contestants chosen to participate in a game show. Each of you will secretly select a resistor of your choice, with a resistance between 0.01 and 10,000,000 ohms. Once everyone has made their choice, the resistors will be revealed, connected in series, and attached to an ideal 240 volt power supply. Each contestant will then be paid \$1 per watt of power that their resistor dissipated, rounded to the nearest cent.

For example, assume that $n = 3$, you chose a 2.2 ohm resistor, and the other contestants chose a 4.7 and 6.8 ohm resistor. You'd win \$675.16, and they'd win \$1,442.38 and \$2,086.85, respectively.

Contestants may talk to each other at any time during the game, but are not required to be truthful. If you want to win as much money as possible for yourself, and are indifferent to what anyone else wins, what should you do? Note that the answer depends on the value of $n$.

Hint:

The rounding of the prize money is important. The best strategy would be qualitatively different if it weren't rounded.

Answer 1

Here's my stab at it:

First, note that the power dissipated in resistor $i$ in a series circuit is

$$ P_i = \frac{V_0^2 R_i}{\left(\sum_j R_j\right)^2}, $$ which implies that for any player, $$\frac{\partial P_i}{\partial R_i} = V_0^2 \frac{ \left( \sum_{j\neq i} R_j \right) - R_i}{\left(\sum_j R_j\right)^3}.$$ This equation implies that as any player increases their resistance, the power dissipated by their resistor increases up until their resistance $R_i$ is equal to the sum of the other players' resistances. Increasing $R_i$ beyond this point leads to a decrease in $P_i$.

This means that for $n > 2$, and assuming completely rational players,

The players will all pick a 10 MΩ resistor. We want to have our resistor being the sum of all the other resistors. So no matter what first-order strategy we assume about the players' proclivities in picking resistances, we want to pick a resistance greater than all of those resistances. But all of the other players are rational, and they want to choose a resistance greater than ours; which means that we need to pick a resistance greater than that resistance; and so on, until we all hit the maximum possible resistance. At that point, any one of us decreasing our resistance just means that we get less money, so nobody will have any incentive to do so. This state is an equilibrium.

The expected payoff for all players is then

$ (0.576¢)/ n^2$, which rounds off to zero for $n > 2$.

For $n = 2$, however,

Both players will pick a 0.01–Ω resistor. In this case, both are aware that this will give them the maximum possible payout, and neither one can increase their payout by changing their resistance.

Under this strategy, both players can then expect to win

$1.44 million.