2 Player All-Pay Auction

Question

Consider a 2 player all-pay auction, bidding for $1.

Each submits a bid that is a real number, thus $S_i=[0,\infty)$. The player with a higher bid wins $1, but both players must pay the submitted bid.

Player $i$'s payoff function is:

$v_i(s_i,s_{j})=$

$-s_i$ if $s_i<s_{j}$

($\frac{1}{2}-s_i$) if $s_i=s_{j}$

($1-s_i$) if $s_i>s_j$.

Suppose player $j$ plays a mixed strategy in which she is uniformly choosing a bid between 0 and 1.

I am only interested in the expected payoff from player $i$'s bidding less than 1, because if $s_i$>1, he would win, but also incur a negative payoff.

My question:

Why is Pr$(s_i=s_j)$=0 for any $s_i\in[0,1]$ if player $j$'s mixed strategy is U[0,1]?

My confusion is: the PDF of player $j$ is 1 for any $s_j\in(0,1)$. So if I only consider $s_i<1$ and Pr$(s_i=s_j)$, how do I justify it becomes $0$?

Answer 1

This seems like a basic issue about probability calculus/theory.

The intuition behind the uniform distribution over $[a,b]$ is that all outcomes between and $a$ and $b$ are equally likely. Because of this, one can get an intuitive grasp about the probability that the outcome $x$ falls in a subinterval $[a_1,b_1] \subset [a,b]$. As the ratio of their length is $$ \frac{b_1 - a_1}{b - a}, $$ the subinterval $[a_1,b_1]$ contains $(b_1 - a_1)/(b-a)$ share of all the points in $[a,b]$. Each point is an equally likely outcome, therefore $$ P\left(x \in [a_1,b_1]\right) = \frac{b_1 - a_1}{b - a}. $$ A problem arises if the subinterval is in fact a single point, $a_1$. What is $P\left(x \in \left\{a_1\right\}\right)$?
By the above intuitive reasoning $$ P\left(x \in [a_1,a_1]\right) = \frac{a_1 - a_1}{b - a} = 0. $$ A more rigorous approach:
Let us denote this probability $P\left(x \in \left\{a_1\right\}\right)$ by $\alpha$. As the distribution is uniform, all points are equally likely outcomes and hence for any $a_i \in [a,b]$ we have $$ P\left(x \in \left\{a_i\right\}\right) = \alpha. $$ There are infinitely many (disjoint) points in $[a,b]$. Let $A$ be an arbitrary set of infinitely many such points, that is $$ A = \bigcup_i^{\infty} a_i. $$ Then as probability is sigma additive over disjoint sets $$ P\left(x \in A \right) = P\left(x \in \bigcup_i^{\infty} a_i \right) = \sum_i^{\infty} P\left(x \in \left\{a_i\right\}\right) = \sum_i^{\infty} \alpha. $$ However as $A \subseteq [a,b]$, $P\left(x \in A \right) \leq 1$. But for any $\alpha > 0$ the sum $\sum_i^{\infty} \alpha$ would be infinite. Therefore $\alpha$ cannot be positive, it is zero.