Wallet Bid: Conditional Expected Total
$n$ people each have an amount of money in their wallets, where each amount is independently drawn from $U(0,1)$. The person with the most money wins everyone's money.
You are person 1. You see $\$x_1$ in your wallet and are told that you won (i.e., your amount was the largest). What is your estimate of the total amount of money across all $n$ wallets?
More precisely, find:
$$E\left[\sum_{i=1}^n X_i \;\middle|\; X_1 = x_1,\; X_1 = \max(X_1, \ldots, X_n)\right]$$
Report the numerical answer when $x_1 = 0.75$ and $n = 15$.
Hints
- Conditioning on $X_1 = \max$, what can you say about the distribution of each other $X_i$? They must be at most $x_1$.
- A $U(0,1)$ variable conditioned on being at most $x_1$ is $U(0, x_1)$, with expected value $x_1/2$. The conditional variables are still independent.
- The total is $x_1 + (n-1) \cdot x_1/2 = (n+1)x_1/2$. Notice this is always less than the unconditional expectation $x_1 + (n-1)/2$ -- knowing you won means others have less.
Worked Solution
How to Think About It: You know two things: your amount is $x_1$, and it is the largest. Conditioning on being the maximum shrinks the support of every other wallet from $U(0,1)$ to $U(0, x_1)$. Each other person's amount is uniformly distributed between 0 and your amount. So the expected total is your amount plus $(n-1)$ copies of the expected value of $U(0, x_1)$, which is $x_1/2$.
Quick Estimate: With $x_1 = 0.75$ and $n = 15$: your contribution is $0.75$, and each of the other 14 people contributes on average $0.75/2 = 0.375$. Total: $0.75 + 14 \times 0.375 = 0.75 + 5.25 = 6.0$.
Approach: Derive the conditional distribution of $X_i \mid X_1 = x_1, X_1 = \max$ and compute the conditional expectation.
Formal Solution:
Let $X_1, X_2, \ldots, X_n \overset{\text{iid}}{\sim} U(0,1)$ and let $T = X_1 + X_2 + \cdots + X_n$.
We want: $$E[T \mid X_1 = x_1, X_1 = \max_i X_i]$$
Step 1: Conditional distribution of each $X_i$ for $i \geq 2$.
Given $X_1 = x_1$ and $X_1 \geq X_i$ for all $i$, each $X_i$ (for $i \geq 2$) must satisfy $0 \leq X_i \leq x_1$. Since $X_i \sim U(0,1)$ conditioned on $X_i \leq x_1$, the conditional distribution is:
$$X_i \mid X_i \leq x_1 \sim U(0, x_1)$$
This follows because for $U(0,1)$, conditioning on $X_i \leq x_1$ gives a uniform distribution on $[0, x_1]$.
Step 2: Conditional independence.
Since the $X_i$'s are originally independent, and we condition each on $X_i \leq x_1$ (with $x_1$ fixed), the conditional variables $X_2, \ldots, X_n \mid X_1 = x_1, X_1 = \max$ are still independent, each $U(0, x_1)$.
Step 3: Compute the expectation.
$$E[X_i \mid X_i \leq x_1] = \frac{x_1}{2} \quad \text{for } i = 2, \ldots, n$$
$$E[T \mid X_1 = x_1, X_1 = \max] = x_1 + (n-1) \cdot \frac{x_1}{2} = x_1\left(1 + \frac{n-1}{2}\right) = \frac{(n+1)x_1}{2}$$
Step 4: Compare with the unconditional case.
If you only knew $X_1 = x_1$ but did NOT know you were the maximum:
$$E[T \mid X_1 = x_1] = x_1 + (n-1) \cdot \frac{1}{2} = x_1 + \frac{n-1}{2}$$
The conditional-on-winning estimate $\frac{(n+1)x_1}{2}$ is always less than the unconditional estimate $x_1 + \frac{n-1}{2}$ (for $x_1 < 1$). This makes sense: knowing you have the maximum tells you everyone else has less than $x_1$, pulling down their expected contributions.
Step 5: Plug in numbers.
With $x_1 = 0.75$ and $n = 15$:
$$E[T] = \frac{(15+1)(0.75)}{2} = \frac{16 \times 0.75}{2} = \frac{12}{2} = 6$$
Answer: The expected total is $\dfrac{(n+1)x_1}{2}$. For $x_1 = 0.75$ and $n = 15$, the answer is $\mathbf{6}$.
Intuition
The central idea is the winner's curse in reverse. Knowing that your value is the highest among $n$ draws tells you that everyone else drew below you. Each competitor's expected amount drops from $1/2$ (the unconditional mean of $U(0,1)$) to $x_1/2$ (the mean of $U(0, x_1)$). This is a pure selection effect -- the same mechanism behind adverse selection in trading. If you are the winning bidder in an auction, the fact that you won is bad news about the item's value (because it means everyone else valued it lower).
The formula $E[T \mid \text{win}] = (n+1)x_1/2$ has a clean structure: it is your amount times $(n+1)/2$, a multiplicative factor that depends only on the number of participants. For a market maker, this is exactly the kind of conditional expectation you need to compute when deciding how to price a winner-take-all contract -- the naive estimate based on your private signal overstates the true expected value because it ignores the selection inherent in winning.