probability that the players will exchange their initially drawn number Consider the following two-player game. The players simultaneously draw one sample
each from a continuous random variable X, which follows $Uniform\ [0, 100]$. After observing the value of her own sample, which is private information (that is, opponent
does not observe it), players simultaneously and independently choose one of the following: $SWAP$, $RETAIN$. If both the players choose $SWAP$ then they exchange their initially drawn numbers. Otherwise, if at least one person chooses $RETAIN$, both of
them retain their numbers. A player earns as many Rupees as the number she is holding
at the end of the game.
what is the probability that the players will exchange their initially drawn numbers?
My approach
this is a question from one of master's entrance exams, now i think that the question is incomplete because given the information in the question if the it is up to player to draw any number from the given interval of $ [0,100]$, why would any player choose any number less than $100$. all the players will choose to maximize their payoff and hence they must draw $100$. and hence the required probability should be equal to 0
But if there is a missing point from the question, for instance the players might draw from 101 balls numbered 0 to 100 in that case the question seems like Monty Hall Problem.
If i am wrong then can anyone please help me with this question?

I will solve with $U[0,1]$ as it is slightly nicer.
Let $z$ be the cut-off for Player 1, under which he will want to swap, and similarly $w$ be the cut-off for Player 2.
There are two scenarios: first is $z \leq w$, the second is $z \geq w$.
Let $X$ and $Y$ denote the uniform random variables on $[0,1]$ for Player 1 and player 2 respectively. We will add up the probabilities of the different ways in which Player 1 can win in the first scenario:

*

*$P($P1 wins$ \,| \,X, Y > w) = \frac{(1-w)^2}{2}$

*$P($P1 wins$\,| \,z < X, Y < w)$ = $\frac{(w-z)^2}{2}$

*$P($P1 wins$\,| \,X, Y < z)$ = $\frac{z^2}{2}$

*$P($P1 wins$\,| \,X > w \cap Y < w)$ = $(1-w)w$

*$P($P1 wins$\,| \,X < z \cap z < Y < w)$ = $z(w-z)$

*$P($P1 wins$\,| \,z <X < w \cap Y < z)$ = $(w-z)z$
Adding everything up and simplifying yields
$$ -z^2 + wz + \frac{1}{2}.$$
In a similar way we can find the probability of Player 1 winning when $z \geq w$:
$$-zw + w^2 + \frac{1}{2}.$$
So what do we notice? Firstly, if $z \geq w$, then to maximize the probability of winning, Player 1 must choose $z = w$ since $-zw + w^2 + \frac{1}{2}$ is a decreasing function in $z$ (a straight line with a negative slope), that is his probability of winning will decrease for larger values of $z$.
What if $z \leq w$? In that case the probability of winning for Player 1 is $ -z^2 + wz + \frac{1}{2}$ — a quadratic polynomial with a negative leading coefficient, so it has a maximum value which can be found by completing the square (or using some calculus):
$$-z^2 + wz + \frac{1}{2} = -\left(z-\frac{w}{2}\right)^2 + \frac{w^2}{4} + \frac{1}{2}.$$
So the cut-off value that will maximize the probability of Player 1 to win for a given $w$ is just $w/2$.
At $z = 0$ the probability of Player 1 to win is $-0^2 + w\cdot0 + \frac{1}{2} = \frac{1}{2}.$
At $z = w$ the probability is $-w^2 + w\cdot w + \frac{1}{2} = \frac{1}{2}.$
So for all values of $z \leq w$ Player 1 will have as good or better chances of winning than Player 2.
But we know that both players will try to use the optimal strategy, in which case each player can improve their probability of winning by taking a value less than that of their opponent. Choosing a cut-off greater than $0$ will give your opponent the chance of beating you with a cut-off less than yours. So for optimal play both players will choose a cut-off of $0$. In other words: always retain.
