Game of Mediocrity

Logic; Difficulty: easy

Are you Mediocre Enough To Win?

The Riddler Classic from this week is an easy but interesting game theory/strategy question that illustrates the idea of Nash Equilibrium. The question could have been made even more interesting with a variation that I discuss.

In the three-player Game of Mediocrity, you win by not winning too much.

Each round, every player secretly picks a number from 0 to 10. The numbers are simultaneously revealed, and the median number wins that number of points. (If two or more players pick the same number, then the winner is randomly selected from among them.)

After five rounds, the winner is whoever has the median number of points. (Again, if two or more players have the same score, then the winner is randomly selected from among them.)

With one round remaining, players A, B and C have 6, 8 and 10 points, respectively. Player A sighs and writes down “3,” but fails to do so in secret. Players B and C both see player A’s number (and both see that the other saw A’s number), and will take care to write their own numbers in secret. Assuming everyone plays to win, what numbers should B and C choose?

A more Interesting question: How would the game play-out had B and C not seen A’s number? Could A have played the last round better?

*spoiler alert* solution and answer follow in the expandable sections below.

Solution to the problem as posed

Player B wins the game outright (no ties needing resolution) if C wins the last round round or B themselves win the round with 0 or 1 points.

Player C wins the game directly only if A wins the last round by 3 or more points.

So clearly neither player B not player C wants A to win the last round since that would mean they both lose the game!

For player C to force A to win the game by more than 2 points, C has to pick 10. Any other choice leaves open the possibility of B picking 10 and forcing either a win for C or B, which would mean C loses the game! 10 is thus a dominant strategy for C over all choices except 0,1,2 when B picks 2. But if C acts based on B picking 2 and chooses 0 or 1, B would then as well choose 0 or 1 ensuring B wins the game! So C is better off sticking to 10.

Similarly, B picking any number other than 10 with C having picked 10 results in either B winning the round with more than 2 points or A winning the round, both of which imply B loses the overall game!

Solution: B and C each picking 10 is the only rational choice for both of them. This of course means they head to the coin toss playoff with each then having a final 50% shot at winning the game!

This is the only Nash Equilibrium (Remember the movie: A Beautiful Mind?): Either B or C deviating from a choice of 10 given the other has picked 10, results in lower payoff for them: in this case certain loss vs. retaining a 50% chance of winning the overall game.

What about any other choices?: We have seen that 10 is a dominant strategy (almost) for C with no other dominant choices. While B does not have a single dominant strategy (e.g. B could pick 0,1,or 2 if C picked 0), B’s hand is forced by A’s choice of 10. This is seen in the IEDS solution below.

Here’s the full table of payoffs (win probabilities) for different choices of B and C and iterated solution showing (10,10) as the Nash equilibrium.

Cells in gray represent dominated strategies and the remaining are equilibrium candidates.

Player B ↓ Player C→	0	1	2	3	4	5	6	7	8	9	10
0	(1,0)	(1,0)	(1,0)	(0.5, 0)	(0,0)	(0,0)	(0,0)	(0,0)	(0,0)	(0,0)	(0,0)
1	(1,0)	(1,0)	(1,0)	(0.5, 0)	(0,0)	(0,0)	(0,0)	(0,0)	(0,0)	(0,0)	(0,0)
2	(0.5,0.5)	(0.5,0.5)	(0.75,0.25)	(0.5, 0)	(0,0)	(0,0)	(0,0)	(0,0)	(0,0)	(0,0)	(0,0)
3	(0, 0.5)	(0, 0.5)	(0, 0.5)	(0.33, 0.33)	(0,0.5)	(0,0.5)	(0,0.5)	(0,0.5)	(0,0.5)	(0,0.5)	(0,0.5)
4	(0,0)	(0,0)	(0,0)	(0.5,0)	(0.5, 0.5)	(0,1)	(0,1)	(0,1)	(0,1)	(0,1)	(0,1)
5	(0,0)	(0,0)	(0,0)	(0.5,0)	(1,0)	(0.5, 0.5)	(0,1)	(0,1)	(0,1)	(0,1)	(0,1)
6	(0,0)	(0,0)	(0,0)	(0.5,0)	(1,0)	(1,0)	(0.5, 0.5)	(0,1)	(0,1)	(0,1)	(0,1)
7	(0,0)	(0,0)	(0,0)	(0.5,0)	(1,0)	(1,0)	(1,0)	(0.5, 0.5)	(0,1)	(0,1)	(0,1)
8	(0,0)	(0,0)	(0,0)	(0.5,0)	(1,0)	(1,0)	(1,0)	(1,0)	(0.5, 0.5)	(0,1)	(0,1)
9	(0,0)	(0,0)	(0,0)	(0.5,0)	(1,0)	(1,0)	(1,0)	(1,0)	(1,0)	(0.5, 0.5)	(0,1)
10	(0,0)	(0,0)	(0,0)	(0.5,0)	(1,0)	(1,0)	(1,0)	(1,0)	(1,0)	(1,0)	(0.5, 0.5)*

Table below shows result after removing all dominated strategies; next all non equilibrium strategies are struck through. The remaining cells highlighted in green are Nash equilibrium strategies. Since all potential strategies for B other than picking 10 result in zero payoff, B will pick 10, forcing equilibrium at (10,10) highlighted in yellow instead of the others.

Player B ↓ Player C→	0	1	2	3	10
0	(1,0)	(1,0)	(1,0)	(0.5, 0)	(0,0)
1	(1,0)	(1,0)	(1,0)	(0.5, 0)	(0,0)
2	(0.5,0.5)	(0.5,0.5)	(0.75,0.25)	(0.5, 0)	(0,0)
10	(0,0)	(0,0)	(0,0)	(0.5,0)	(0.5, 0.5)*

Solution: What should player A do?

How would this game play out if B and C had not seen A’s choice

For A to have a chance of winning the game, A would have to win the round by 2,3, or 4 points. This obviously limits A’s choices to those one of those three numbers!

This does not meaningfully change anything for B and C who would then battle it out for victory with the same 50% chance by each of them choosing 10.

Is there any other possibility?:

Since the question allows for the possibility of A revealing their choice (deliberately or inadvertently), the smart thing for A to do would be to pick 2 and slyly reveal the choice to B and C.

What this does is now give B a choice of either picking 2 or 10 with either choice resulting in the same 50% chance of winning the game for B while C continues to be best served by picking 10. Specifically:

If A picks 2 and B picks 2, the final outcomes considering all the tie breakers, are:

A wins with probability 25%

B wins with probability 50%

C wins with probability 25%

If A picks 2 and B picks 10, with C at 10, the final outcomes considering all the tie breakers are:

A wins with probability 0%

B wins with probability 50%

C wins with probability 50%

B is agnostic to the two choices and assuming B decides by a coin flip, the final equally likely choices would be (2,2,10) OR (2,10,10) and final outcomes would be: