I ran across this question:
Alice and Bob have fair 30-sided and 20-sided dice, respectively. The goal is to obtain the largest possible value. Alice rolls her die one time. However, Bob can roll his die twice and keep the maximum of the two values. In the event of a tie, Bob is the winner. Find the probability Alice is the winner.
The answer to this is $\frac{647}{1200}$, you basically use LOTP and find the pmf.
Then I ran into this question:
Alice and Bob have fair 30-sided and 20-sided dice, respectively. The goal for each player is to have the largest value on their die. Alice and Bob both roll their dice. However, Bob has the option to re-roll his die in the event that he is unhappy with the outcome. He can see Alice's die beforehand. Bob then keeps the value of the new die roll. In the event of a tie, Bob is the winner. Assuming optimal play by Bob, find the probability Alice is the winner.
The answer to this is $\frac{17}{30}$.
Why are these not the exact same question? Both just check to see if the max of Bob exceeds Alice, right? In the second one, it isn't phrased like that specifically, but I feel like it should be the same.
EDIT: HERE IS THE PROOF FOR $\frac{17}{30}$. To be honest, I don't understand the second half of it [EDIT 2: Seems like the provided answer does not answer the question first asked.
Let $W$ be the event Alice wins. First, we want to find $P(W)$. Let $A$ and $B$ be value of Alice's roll and Bob's first roll, respectively. The key here is to condition on whether or not $A < 20$. Namely, this means $P(W)= P(W| A \leq 20) P (A \leq 20) + P(W | A > 20) P (A > 20)$.
We can quickly see that $P(A \leq 20) = \frac{2}{3}$, as this accounts for 20 of 30 equally-likely values. If $A > 20$, then clearly Alice wins regardless of what Bob obtains, so $P(W | A > 20) = 1$.
If $A < 20$, then we are really rolling 2 independent 20-sided fair dice. Namely, to compute $P(W | A < 20)$, we see that the probability of a tie is $\frac{1}{20}$, as there are 20 values the can agree upon out of $20^2 = 400$ outcomes. Therefore, the probability the two rolls are different is $\frac{19}{20}$.Because of the fact that the two dice are symmetric, $P(A > B) = \frac{19}{20}\frac{1}{2} = \frac{19}{40}$ as when the two rolls are not the same, it is equally-likely whether $A > B$ or $A < B$.
Combining these all together, we see that $P(W) = \frac{1}{3} 1 + \frac{2}{3}\frac{19}{40}$. Now, Bob is going to roll again whenever $P(W | B = b) = 1- \frac{b}{30} > \frac{39}{60}$. This is because he wants to minimize Alice's chances of winning. It is easy to see that $P(W | B = b) = 1 - \frac{b}{30}$, as Aice just needs to roll a value of at least $b + 1$.
We just need to find the maximum value of $b$, say $b^*$ such that $P(W | B = b) = 1 - \frac{b}{30} > \frac{39}{60}$. Solving this inequality, we get that $b \leq 10.5$, so $b^* = 10$ is the maximum value that Bob rolls again. Let $W_1$ be the event that Alice wins in this new game where Bob can roll again. We have that
$P(W_1) = P(W) + \frac{10}{20}(P(W) - P(W | B \leq 10))$.
We get this because of the fact that Bob rolls again with probability $\frac{1}{2}$, which is when he rolls at most 10. Then, the probability Alice wins is going to change by $P(W) - P(W | B \leq 10)$.
To calculate $P(W | B < 10)$, this is simply just average all the cases where $B = b$ for $1 \leq b \leq 10$, as when $B \leq 10$, the value is uniformly distributed on the first 10 positive integers. Therefore,
$P(W | B < 10) = \frac{1}{10}\sum_{b = 1}^10 1 - \frac{b}{30} = \frac{49}{60}$
Therefore, we get that
$P(W) = \frac{39}{60} + \frac{1}{2}(\frac{39}{60} - \frac{49}{60}) = \frac{17}{30}$
Proof of equivalence I:
Of course, in the "max" version, Bob wins iff the max is at least as great as Alice's roll.
Now, consider the "option" version, and let Bob's rolls be $b_1,b_2$, and Alice's be $a$. Of course, if $b_1≥a$, Bob wins. But in that case, $\max(b_1, b_2)≥b_1≥a$ so the max wins. If $b_1<a$, Bob will take $b_2$ instead and then wins iff $b_2≥a$. But in that case $\max(b_1, b_2)=b_2≥a$. Thus, Bob's win is the same as saying that $\max(b_1, b_2)≥a$.
Proof II (rephrasing I): Bob wins iff at least one of $b_1,b_2$ is $≥a$. That's the same as requiring that $\max(b_1, b_2)≥a$.
Unless I am missing something, the two versions of the game are equivalent. Of course, it is very different if, in the option version, Bob doesn't see Alice's toss. In that case, if Bob's $b_1=6$ or whatever, he might feel obliged to go with the second roll. That could cost him the game, in the scenario where $b_1≥a$ but $b_2<a$. Of course, if Bob sees that $b_1≥a$, the game is over.
For completeness, here is the calculation for the "max" version, yielding $\frac {647}{1200}$ as expected.
Post edit: it appears that the confusion does indeed concern the rule regarding what Bob knows. If Bob sees Alice's toss, then the games are equivalent. If Bob does not see Alice's toss, then that gives Alice a clear advantage (since Bob might well discard a winner). For that problem we consider the probability that Bob ends up with a particular value (assuming that Bob plays optimally and re-rolls if he gets $≤10$ on the first try). For that one $p_i=\frac 12\times \frac 1{20}$ if $i≤10$ since the only way to get such a bad roll is to discard the first (probability $\frac 12$) and then roll $i$ (probability $\frac 1{20}$). If $i>10$ then $p_i=\frac 1{20}+\frac 12\times \frac 1{20}$ since Bob might get $i$ initially or on the reroll. Thus the probability that Bob wins this version is $$\sum_{i=1}^{20}p_i\times \frac i{30}=\frac {13}{30}$$ as desired.