# The Resistor Game Show

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.

• Isn't this some form of prisoner's dilemma for which one can only talk about Nash equilibria? Commented Jan 15, 2022 at 8:37
• @WhatsUp It's not quite the prisoner's dilemma. In the prisoner's dilemma, you're always better off defecting no matter what the other person does. In this, there's not a single resistance that's always the best choice. Commented Jan 15, 2022 at 17:09
• I see it more as an example of the Tragedy of the commons. If you knew what the others would pick, your optimum choice would be rot13(gur fhz bs gurve erfvfgnaprf) so if this were played for several rounds, there would be a race to the bottom with the worst outcome for everyone, unless the players all cooperate to make a pact and enforce it through means separate from the game itself. Commented Jan 16, 2022 at 9:14
• Does the "contestants can talk to each other..." part have anything to do with the solution? Talk is cheap; since words have no affect on payout, they do not affect any strategy. Why do you mention talking in the puzzle description? Just to confuse us? Commented Mar 28, 2022 at 18:00
• @MikeEarnest I don't think talking is necessary. I just mentioned it in case someone else comes up with an answer different than the one I'm thinking of that does need it. Commented Mar 28, 2022 at 20:35

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.

• I think I spotted one flaw in this: rot13(Pbafvqre gur pnfr jurer a rdhnyf guerr, naq nyy guerr cynlref pubbfr n erfvfgnapr bs bar zvyyvba gjb uhaqerq rvtugl gubhfnaq buzf. Vfa'g guvf n Anfu rdhvyvoevhz gung tvirf gurz rnpu bar prag?) Commented Mar 29, 2022 at 4:38