Say you (person A) and person B put down 10 dollars each. There's a bag with numbers uniformly distributed from [0, 100]. Whoever has the higher number gets to keep the $20. How much would you pay to have a chance at drawing again (you don't get to see the first draw)?
I'm super confused how to answer this question. What I was thinking is that I'm trying to solve for the price Y, in this equation, where X is the value of my first draw (I'm confused because I don't get to see the first draw):
(X/100) * (20) + ((100-X)/100)(-20+Y) = 0
Having a chance at drawing again is equivalent to getting to draw twice and keep the higher number. Therefore, if you have a chance to draw again, you will have a $2/3$ chance of winning the game (since three draws are done without replacement, and there is a $2/3$ chance that the highest draw is yours). Therefore, your probability of winning increases by $1/6$, which is worth $20/6 = 10/3$ dollars.