Choose Room for Maximum Reward in Game of 4 Rooms

[Happiness]

Suppose you are in a game with 4 rooms, A, B, C and D, with the rewards $30, $50, $70 and $100, respectively. There are 8 people in the game, and the reward in each room will be equally shared among those who choose the room. You will not know the choice of other players until you have made yours. Which room should you choose to maximise your reward?

For example
If all 8 people choose room D, then each person will only get $\frac{100}{8}=12.5$, which is even lower than the reward of room A.

Spoiler: A conjecture

If the number of players in the game is very large, would it be true that the population of players will "naturally" (or automatically) adjust itself such that everyone gets roughly the same reward, and so it doesn't matter which room you choose?

My attempt Please vote before looking at my answer, as it may affect your decision!

Spoiler: Please vote first!

Assuming the above conjecture is true, the distribution of players is 1A 2B 2C 3D, where the succeeding letter is the letter of a room and the preceding number is the number of players in that room. Then the rewards (per person) are $30, $25, $35, $33.33, respectively. So the best choice is room C.

I think {$30, $25, $35, $33.33} has the least variance (the conjecture demands the variance to be minimum). I checked it against the other distribution 1A 1B 2C 4D (it gives rewards $30, $50, $35, $25), which has a larger variance.

But if everyone thinks like me, then everyone would choose C? And everyone will only get $\frac{70}{8}=8.75$.

 

[StoneTemplePython]

A conjecture
If the number of players in the game is very large, would it be true that the population of players will "naturally" (or automatically) adjust itself such that everyone gets roughly the same reward, and so it doesn't matter which room you choose?

If you are looking at stable solutions -- i.e. where given your expectation of the other player's strategies, you cannot improve by switching (and this holds for each player involved). Then yes, and this is a Nash Equilibrium. It's been a while but I think you can model this using percentages and a 4x4 matrix, where you have $p_i$ being you with your strategy and $p_j$ being the others in the population. (One technical nit: if there are multiple Nash Equilibria, the interpretation as points of stability gets to be a bit challenged but I think there's only one NE so not to worry here.)

If your question is does this "naturally" happen in the real world? Yes and no. Frequently there are very large gaps between what is predicted in game theory and what actually happens in the real world, but its still a useful way to start thinking about things. A fun game is to say there's a number between [0,100] that everyone involved will write down. There could be multiple winners but... Whoever guesses closest to half of the average of everyone else's guess wins. Empirically the Nash Equilibrium is not the winner here.

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edit:

@Happiness

apparently edited the original post and turned it into a voting problem with spoiler tags right about the time I submitted my post. I don't think the new shoe fits but that's life.

 

[willem2]

It may be interesting to consider a simpler problem. Suppose you have just 2 rooms and 2 players and the rewards are 2 and 3.
Suppose player 1 chooses 3 with probability p₁ and player 2 chooses 3 with probability p₂.
The payoff for player 1: P₁ = 1+ 2 p[/SUB]1 + 2 p₁ + p₂ -5/2 p₁p₂
The playoff for plater 2: P₂ = 1 + 2 p₂ + p₁ -5/2 p₁p₂
To get a nash equilibrium you need $\frac {dP_1} {dp_1} = 0$ and $\frac {dP_2} {dp_2} = 0$
You get this when p = 4/5 and the reward for both players is 9/5.
Because it is a symmetric game, you should find p=q.
You should be able to find an equilibrium distribution for 10 players with the same strategy of choosing the rooms with probabilities $p_1, p_2, p_3$ and $1 - p_1 - p_2 - p_3$ using some sort of hill climbing algorithm. THe rewards as a function of the probabilities are very large degree 10 polynomials, so solving them directly seems hopeless.
This problem is known as minority game or El Farol Bar problem https://en.wikipedia.org/wiki/El_Farol_Bar_problem and seems to have a rather large literature because it's useful for resource allocation.

 

[mfb]

If your question is does this "naturally" happen in the real world? Yes and no. Frequently there are very large gaps between what is predicted in game theory and what actually happens in the real world, but its still a useful way to start thinking about things. A fun game is to say there's a number between [0,100] that everyone involved will write down. There could be multiple winners but... Whoever guesses closest to half of the average of everyone else's guess wins. Empirically the Nash Equilibrium is not the winner here.

The Nash equilibrium is "everyone picks 0", the price is shared, and deviating from that strategy is bad.To solve the original problem;

For a large number N of players, choose a room with a reward R_i with a probability proportional to R_i. Define $R=\sum_{i=1}^n R_i$. Assume the others pick the rooms i with probability p_i each, where $\sum_{i=1}^n p_i = 1$. Then your expected reward is $$\frac{1}{R} \sum_{i=1}^n \frac{R_i^2}{N p_i}$$
To see where this is minimized, consider two rooms only, without loss of generality we can choose the first two. Let $c=p_1+p_2$ be fixed. Your expected reward from these two is $$\frac{1}{RN} \left(\frac{R_1^2}{p_1}+\frac{R_2^2}{c-p_1}\right)$$
This is minimized at $p_1=\frac{cR_1}{R_1+R_2}$, i. e. if the rest of the players follow the same strategy. As the total payout is always constant (for many players, every room will be paid out), and a deviation from the strategy of others means an increased gain for you, deviating from this strategy must reduce the expected gain. Therefore this strategy is the Nash equilibrium. As expectation value, you are guaranteed to get at least the total reward divided by the total number of players, and if the other players don't follow this strategy you get even more.

With 8 players, there is a small risk that some rooms are not picked and you picking the room changes the money every player in the room gets notably, which makes this derivation only an approximation, but I would expect it to be good enough for practical purposes.

 

[StoneTemplePython]

The Nash equilibrium is "everyone picks 0", the price is shared, and deviating from that strategy is bad.

I agree about the Nash Equilibrium. And in this game that Nash Equilibrium is the only point of stability. However, I don't think your normative statement is per se true though. Or put differently, strategy is formulated at the individual player level, and the question as always is: 'what is your best response given what you know/expect from your opponents.'
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I did this game when I took game theory and something in the teens was a winner. (I vaguely recall a discussion that this is common, but emphasis on vague.)

More interestingly, I used to subscribe to the FT and they ran this contest a couple years ago. Apparently they actually ran it first in 1997 and gave the contest winner 2 international business airline tickets. The more recent one had a nice travel bag and an autographed copy of "Misbehaving" by Richard Thaler (winner of this year's nobel in econ). The winning guesses were 13 and 12.

(for the more recent content, I seem to recall commentary on Tim Hartford's podcast that there were a lot of guesses of 100, and that was attributed to people enjoying causing havoc, and something like one guy commandeered an army of bot computers that all submitted 100, hoping to flood the results in his favor, or something like that.)

There's a nice discussion by Thaler here:
https://www.ft.com/content/6149527a-25b8-11e5-bd83-71cb60e8f08c

(ctrl + f: "try out this puzzle that Tim Harford recently" to get to this puzzle about 1/3 the way down. I no longer subscribe but was able to access this i.e. not behind the pay wall.)

 

[mfb]

Some people make irrational choices, and sometimes you can abuse this, sure.
If, for some reason, people prefer the 100 room more than they should, you can use this by preferring the other rooms. That only works if you know more than the other players, however.