Posted May 29, 2012 By Presh Talwalkar. Read about me , or email me .

Alice, Bob, and Charlie are chosen to participate in an auction for an economics class.

Each submits a secret bid of 1, 2, or 3. The person with the lowest and unique* bid wins the auction and gets a prize of $1.

*Example: If Alice and Bob bid 1, and Charlie bids 2, then Charlie wins. Although Alice and Bob bid lower numbers, their bids were not unique so they do not win.

(Also: if they all bid the same number, no one wins)

If each plays the same mixed strategy, what is the optimal strategy? (what is the symmetric mixed Nash equilibrium?)

Fair warning: this game is harder to analyze than it sounds!

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"All will be well if you use your mind for your decisions, and mind only your decisions." Since 2007, I have devoted my life to sharing the joy of game theory and mathematics. MindYourDecisions now has over 1,000 free articles with no ads thanks to community support! Help out and get early access to posts with a pledge on Patreon. .

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Setting up the problem

I read about this game’s solution in Comments on “Reverse auction: the lowest positive integer game” by Adrian P. Flitney. I will use similar notation in this post.

We are seeking an optimal mixed strategy for each player. We will write the triple (a, b, c) to denote the probabilities that a player picks the choices 1, 2, and 3, respectively.

Since the three probabilities must sum to 1, we have that c = 1 – a – b. So our probability triple is (a, b, 1 – a – b).

The way we will analyze the problem is as follows. We will write out Alice’s expectation for playing the game, conditional on Bob and Charlie playing with probabilities (a, b, 1 – a – b).

Then we will maximize that expression by using the first order conditions and solving for the best mixed strategy.

The optimization problem

The payoff to the game for Alice is the following:

Payoff = Pr(playing 1)Pr(winning at 1) + Pr(playing 2)Pr(winning at 2) + Pr(playing 3)Pr(winning at 3) Payoff = Pr(playing 1)Pr(others play > 1) + Pr(playing 2)Pr(others both play 1 or both play 3) + Pr(playing 3)Pr(others both play 1 or both play 2)

Now we substitute for the other player’s probability as each player plays (a, b, 1 – a – b). We get:

Payoff = Pr(playing 1)(1 – a)2 + Pr(playing 2)(a2 + (1 – a – b)2) + Pr(playing 3)(a2 + b2)

Now at this point we are almost there.

We could substitute in the probabilities Alice plays her numbers as (a, b, 1 – a – b) into the above expression. But I have done this and it gets quite complicated.

The paper I cited above uses a neat trick. Instead just write out Alice’s probabilities as (x, y, 1 – x – y) and then later set x = a, y = b.

Why would you do this? The reason is we are going to take partial derivatives with respect to Alice’s probabilities. It is notationally more convenient to distinguish Alice’s probabilities, take the derivative, and then set the probabilities equal later than to substitute the probabilities immediately.

You can trust me on that or do the algebra yourself.

So finally, we have Alice’s expectation from playing the game is:

Payoff = x(1 – a)2 + y(a2 + (1 – a – b)2) + (1 – x – y)(a2 + b2)

We will optimize this expression with respect to its partial derivatives and solve.

The solution

The first order conditions are:

d(payoff)/dx = (1 – a)2 – (a2 + b2) = 0 d(payoff)/dy = a2 + (1 – a – b)2 – a2 – b2

d(payoff)/dy = a2 + 1 – 2a – 2b + 2ab = 0

Solving these equations takes a bit of work. I was lazy and just threw it into WolframAlpha:

The only suitable solution, since these are probabilities, is the final one listed.

The answer is therefore:

Play “1” with probability 46.4%

Play “2” with probability 26.8%

Play “3” with probability 26.8%

The expected payoff to each player is 0.287.

Whew, that was a lot of work for a seemingly simple auction.

Conclusions

There are a few lessons that I learned from this example, including:

–Unique lowest bid auctions are tremendously complicated to analyze. Remember we only solved for the symmetric equilibrium. There are in fact many asymmetric equilibria that were not even analyzed.

–In a normal lowest bid auction, it is never sensible to bid the highest value. In a lowest unique bid auction, it can be sensible because you win when others bid the same lower numbers.

–The players slightly increase their payout by bidding “3” on occasion. If all three bid the numbers “1” and “2” with equal chance, their expected payout would be 0.25. This is slightly lower than the 0.287 expected payoff each player could get in the symmetric Nash equilibrium.

And if you want to take this example a bit further, you might conclude the following:

–Stay away from unique lowest bid sites (they are too much like a lottery)