Just Some Game Theory

A famous artist is auctioning off his masterpiece, which everyone values at $2:

There are two people who will bid, Alice and Bob. They have to put a whole dollar amount ($0, $1, $2, $3, ...) into an envelope (in secret).

Both the winner and loser will have to pay whatever they put in the envelope; the highest bidder will get the painting, while the loser won't receive anything. If there is a tie, the painting is given to one of them at random.

Assuming both people are trying to maximize their expected wealth, what would you if you were Alice?

#Logic

Easy Math Editor

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Comments

Suppose Alice and Bob put $A$ and $B$ dollars in the envelope respectively. If Alice wins, then her wealth would increase by $2 - A$ ( She gets a painting worth $2, but spends $ $A$ ). If she loses, then her wealth would increase by $- A$ . Note that if $A \geq 3$ , then Alice's wealth would be decrease irrespective of whether she gets the painting or not since $2 - A < 0$ . Therefore Alice should not put more than 2 dollars in the envelope. By symmetry, Bob should also not put more than 2 dollars.

Putting $2 in the envelope is not a good idea either. If she wins, then her wealth remains the same. If she loses, then her wealth decreases. There is no positive outcome and there is risk of losing money, therefore Alice should not put $2 in the envelope. Again by symmetry, Bob should also not put $2.

We will now look at the remaining cases, that is $A, B \in \{ 0, 1 \}$ , in the form of a matrix

$\begin{array}{|c|c|c|} \hline & 0 & 1 \\ \hline 0 & (2, 0) / (0, 2) = (1,1) & (0, 1) \\ \hline 1 & (1, 0) & (1, {-1}) / ({-1}, 1) = (0,0) \\ \hline \end{array}$

We see that putting $0 in the envelope is the dominant strategy for both players. It is also the only Nash equilibrium in this game. I feel this strategy is the best for Alice since she has a chance of getting $2, which is not possible in any other case. Also, there is no negative payoff; Alice does not risk losing any wealth even if she loses the game.

Pranshu Gaba - 5 years, 3 months ago

If the players are risk-averse, then having 50/50 chance of getting $2 (if they both bid $0) is worse than having a sure $1 (if one player bids $1). Thus it becomes a coordination game, both of them trying to cooperate to bid different amounts.

Ivan Koswara - 5 years, 3 months ago

How will the players decide who pays $1 and who pays $0? Note that they are putting the amount in the envelope secretly. Will the players be willing to pay a dollar when they do not know what the other player is bidding?

Pranshu Gaba - 5 years, 3 months ago

@Pranshu Gaba – That's exactly the main discussion of cooperation game; look it up in Wikipedia.

Ivan Koswara - 5 years, 3 months ago

Since the game setting is symmetric, Alice and Bob will bid the same amount and this is a common knowledge to them.
Since they are trying to maximize their expected wealth, they will both bid $0.

展豪張 - 5 years, 3 months ago

Keep in mind they don't have to bid a fixed amount; for example, they could each choose to bid $\$X$ with probability $P(X).$

Eli Ross Staff - 5 years, 3 months ago

So they will each bid at random according to their strategy with probability?

展豪張 - 5 years, 3 months ago

@展豪張 – In general, that's how all strategies work. You can think of a "fixed" strategy as a special case of this more general form, where $P(Y) = 1$ for some $Y.$

Eli Ross Staff - 5 years, 3 months ago

@Eli Ross – Oh I see...... I have only considered the "fixed" strategy...

展豪張 - 5 years, 3 months ago

@Eli Ross – I liked this problem. Do we a wiki on game theory on Brilliant?

Swapnil Das - 5 years, 3 months ago

Math	Appears as
Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$