Mixed strategy equilibrium in a number guessing game

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You and me are guessing numbers between 0 and 100. If I guess higher than you (including ties), you pay me £1. Else I pay you £1.

If the players played according to uniform distribution, then they would win against each other roughly 50% of the time and the payout would be £0 giving a mixed strategy equilibrium. But how can we work out what the mixed strategy equilibrium is explicitly?

The way I would approach this is by constructing a huge 100x100 payoff matrix and solving the corresponding linear program for the mixed strategy. But how can this be done on pen/paper to motivate perhaps both players using a uniform distribution?

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I am unsure whether I am interpreting the posted question correctly.

I am assuming that the posted question is equivalent to the following question. Let $S$ denote the set $\{0,1,2,\cdots,100\}$.

Assume that the variable $a$ is assigned a random value from the set $S$. Assume that the variable $b$ is similarly assigned a random value from the set $S$, such that $a$ and $b$ could be equal (i.e. sampling, with replacement).

What is the probability that $a \geq b$?


There are two ways of approaching the problem.

The first way is to recognize that if there is no tie, then the game is perfectly fair. Further, the probability of a tie is exactly $~\dfrac{1}{101}$.

Therefore, the probability that $a \geq b$ is

$$\left[\frac{1}{2} \times \frac{100}{101}\right] + \frac{1}{101} = \frac{51}{101}.$$


The alternative approach is to recognize that the probability is linear with respect to the value assigned to $b$. Therefore, you can assume, without loss of generality that $b = 50.$

Therefore, $a$ loses if $0 \leq a \leq 49$ and $a$ wins is $50 \leq a \leq 100$.