Suppose Alice, Bob and Carol are playing a number guessing game, in which Alice, Bob and Carol pick a number from $[0, 1]$ sequentially i.e. first Alice then Bob and then Carol. After that, a random number $X$ is drawn from the uniform distribution $U[0, 1]$, and the person whose number is closest to $X$ wins (in terms of absolute differences). Suppose all the three players are smart and want to win, then which number should Alice pick?
An obvious choice seems to be $0.5$, and we can indeed prove that in this case Bob and Carol will have to both pick $0.5$ also - let's say Bob pick any $a<0.5$, then Carol would just pick any $b>0.5$ such that $|b-0.5|<|a-0.5|$ to have a strictly greater probability to win than Bob, which implies Bob wouldn't pick any number other than $0.5$, and thus Carol also has to pick $0.5$ too. So, Alice picking $0.5$ will definitely result in a draw.
My question is, does there exist other choices that can give Alice a strict advantage over Bob and Carol?
Let's assume that draws are prohibited, but infinitesimals (suggestively named $\epsilon$ here) can be used as tie-breakers.
Carol can always enforce a winning probability $\ge \frac14$: If Alice and Bob pick $0<x<y<1$, then Carol can pick $x-\epsilon$ or $y+\epsilon$ or any number in $(x,y)$ to win with probabilities $\approx x$, $\approx 1-y$, or $\frac{y-x}2$, respectively. At least one of these is $\ge\frac14$.
Wlog Alice picks $a\le \frac12$. If Bob picks $b<a$, then Carol can maximize her winning probability by picking $a+\epsilon$. This results in Alice winning with probability $\approx \frac{a-b}2$, Bob with probability $\approx\frac{a+b}2$, and Carol with probability $\approx 1-a$. This also implies that Bob will take $b=a-\epsilon$ to win with probability $\approx a$, i.e., Alice will win with probability $\approx 0$. Bad idea.
Alice must try to convince Bob to pick $b>a$.
Assume Alice picks $\frac 14<a<\frac12$. Then if Bob picks $b=1-a+\epsilon$, Carol will choose between $c=a-\epsilon'$ to win with $\approx a$, or $c=b+\epsilon'$ to win with $\approx a-\epsilon<a$, or $a<c<b$ to win with $\frac{b-a}2=\frac{1-\epsilon}2-a<a$. So she will choose the first option, leading to Alice winning with $\approx\frac{b-a}2\approx \frac12-a$, Bob with $\approx\frac12$, Carol with $\approx a$. Any other choice by Bob will only decrease his chances. Apparently Alice wants to minimize $a$ and should pick $a=\frac14+\epsilon$ in this scenario. As Bob wins $\approx \frac12$, he will indeed happily comply and not pick $b<a$.
If Alice picks $a<\frac14$, Bob can pick $b=\frac{a+2}3$, which will guarantee that Bob gets at least $\frac{1-a}3$. However, by allowing Carol to find her optimum in $(a,b)$, the winning probabilities of Bob (and Alice) are not guaranteed to be a specific value. Rather, they depend on a choice by Carol that is indifferent to her (Carol's) result. Alice is only guaranteed to win with at least $a$, but it could be higher. However, Bob might enter a coalition and promise to pick $b$ in a way that makes Alice's chance $>\frac14$ if Carol plays optimally. To this end, Bob must make $1-b>\frac{b-a}2$ (to lure Carol into picking $b+\epsilon$) and $\frac{a+b}2>\frac14$(to keep his promise). So $\frac{2+a}3>b>\frac12-a $. Bob's chances will then amount to $\approx \frac{b-a}2<\frac{1-a}3$. Even with small $a$, this is worse than the at least $\frac{a1-a}3$ Bob achieves without such a coalition.
In the end, Alice can guarantee only $\approx\frac14$.