optimal way to play a game with uniform random variables and opportunity to redraw Suppose you are playing a game where you and an opponent both receive a uniform, random number in interval $[0,1]$. Both of you can choose whether or not to get a redraw after receiving the first number, but neither person knows whether the opponent re-draws, or what the values of the non-final draws are. You can only redraw at most once. In the end, the final numbers are compared, and whoever has the largest number wins. I want to figure out what is the probability I will win?
My intuition is that the strategy should be some number $a$ such that I will redraw if my first number is smaller than $a$. If I assume the opponent has the same strategy but with a number $b$, I can separate into two cases: $a \leq b$ or not. In the former case, there are 3 cases for my final result: my final result is less than a, my final result is in $[a, b]$, or my final result is in $[b, 1]$. The same is true for my opponent giving a total of 9 scenarios between the two of us. If both our answers are in the same interval, I think there is a 50% chance I will win. Otherwise, the intervals are disjoint, and I have either a 0 or 100% chance I will win. Since all 9 of these scenarios are also disjoint, I can add these probabilities together to get an answer for the probability I will win in terms of $a$ and $b$, but when I take a derivative to find a maxima, the answer for $a$ still depends on $b$.
I'm having trouble figuring out what $a$ should be since I don't have a game theory background.
 A: I modeled a discrete approximation of this in Excel as follows, and it looks interesting. The Excel file is here. While this doesn’t answer your question, it suggests the optimal choice for $a$ is over $1/2$, and that there is only one value of $a$ for which your opponent has no winning strategy. (If I had to guess, I’d guess it’s 0.625.)

The game I modeled is the following one:
Aisha and Bob each spin a 26-equal-segment wheel labeled a-z twice. Each player has a secret threshold letter (shown in the example below with Aisha's as m, Bob's as r; these correspond to your $a$ and $b$). If the first spin result is at least the threshold, the player's first spin counts (orange shading); otherwise, the player's second spin counts (pink shading). The winner is the player whose result is highest (closest to z). [I used letters instead of equally-spaced numbers in $[0,1]$ to make it easier to look at and understand.]
The top $26$ by $26$ grid shows Aisha's result for each of the $26^2=676$ equally likely possibilities for the two spins. The bottom grid shows Bob's result.
Columns AO and AP show the distribution of results for each player.
The matrix of numbers at top right tabulates the number of winning games for Aisha among each of the $676^2$ equally likely results of each player spinning twice.
Let me explain the blue square, at the intersection of the m-column and the q-row. Aisha’s result is q $38/676$ of the time, and Bob's is m $17/676$ of the time. So the game outcome is q vs m (Aisha wins) $38x17 = 646$ times (out of all $676^2$ games), so we count $646$ wins for Aisha.
Summing over all $676^2$ games, Aisha wins $0.4734$ of the time with these secret thresholds (m for Aisha, r for Bob). Bob wins $0.4820$ of the time, giving Bob an edge of $0.00853$ wins per game.
What was initially counter-intuitive for me is that given a specific secret threshold for Aisha (say m), the optimal threshold for Bob isn't obvious. (For Aisha's m, Bob's best threshold is p. If Aisha's threshold is e or f, Bob's should be n.)
If Aisha's threshold is q, however, it looks like Bob has no winning strategy, but he can break even by using the same threshold. The lack of a winning strategy isn't surprising, since matching thresholds make it a symmetric game, but the fact that the only threshold that can't be beat is q was unexpected.
Perhaps you or someone else will try to prove that $a=0.625$ is the optimal strategy, guaranteeing at least a break-even game.
