Seems like an easier problem. Opponent’s EV is 50.5 after reroll, hold 51+ and dump 50-. Also massive disadvantage for player 1.
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I guess @econophile is also asking what P2 should do when finding out that P1 didn’t reroll. Intuition says P2 will need to be greedier in that case. To figure out how much will require math again.
Sure, deciding what to do if they reroll is straightforward. But if they don’t reroll, it’s trickier. And it seems like player 1 should change their reroll strategy with these rules.
I thought I might be missing something. Yeah that makes things more interesting.
Yeah, I think both players reroll a lot. Maybe P1 rerolls up to 66 or so and P2 rerolls up to 75? Need to do some math to get some real numbers…
EDIT: Nah, I’m thinking about this all wrong.
Something tells me that “d10” might be biased here.
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JFC, 10 sided die. Joke, get it?
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I’ll just go home.
sir, this is a Wendy’s
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I mean if you are player 1 you could choose to stand on something in the 40s to try and get player 2 to re-roll after hitting 60 or something like that.
Also to make the game better, make it best of 10 cycles, alternate first to roll each cycle.
I think usually problems like this can be reduced to a trivial case then solved by induction. Solve it for D3 then the solution for D100 kinda drops out.
Staying with 40 is definitely wrong. You win 37.5% of the times you choose to reroll, and you will win less than that with a 40 against any sane strategy.
Alright, my buddy Donkey is threatening to solve this problem. We can’t have this, I must solve it first.
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Yup I was telling him I think it’s just a tweak to my previous solution. But actually I gtg, so now the drunkard will get a head start.
We all enjoy a bit of a “head” start.
Ok, guys, recognising this 37% number you keep throwing around as 1/e…
I think we can solve this through backward induction. Call player 1 A and player 2 B.
If A rerolls, then the conditional probability of each of A’s outcomes is 1/100, and B can just choose the highest value response to this distribution. As our work on the previous problem shows, this response is B rerolling everything 50 and below. On this branch, A’s equity is 37.5% and B’s equity is 62.5%.
If A does not reroll, the conditional distribution of A’s outcomes depends on their strategy for rerolling. If their strategy takes the form reroll if and only if the initial roll is less than or equal to T, then the conditional probability of A’s outcome is 1/(100 - T) for values {T+1, …, 100} and 0 for all other outcomes. If B knows T, then B can set it own reroll strategy to maximize its equity on this branch. For each threshold T, we can determine B’s best response, and the equity that A gets with that best response. Call this value E(T)
Then A’s overall equity looks like (T/100)*0.375 + ((100-T)/100)*E(T). Then A’s problem is basically to choose the value of T that maximizes this expression. If we find that, I think we will be at a point where A and B can do no better against each other than their current strategies, although I would want to double check.
You should be able to express E(T) in terms of B’s reroll percentage. If B rerolls on X and below in response to T and A has N, then (multiplying by 100 for fewer fractions) 100 * E(T) = X* (N - 50) / 100 + (100 - X) * E(N | X). E(N | X) is the probability that B’s die between X and 100, which is (N - X) / (100 - X), so 100 * E(T) = X * (N - 50) / 100 + (N - X).
At equilibrium, A should be indifferent between keeping his worst die and rerolling it, so 100 * E(T) = 37.5 = X * T / 100 - X / 2 + T - X = X * T / 100 - 3 * X / 2 + T.
Now we need to express X in terms of T. If A stays on >T, B should stay when his chances of beating A’s distribution with his current die are greater than the chances of a random die winning. So, for a die of N, (N - T) / (100 - T) > (100 - T) / 100. The cutoff is X = (100 - T) ^ 2 / 100 + T.
Soooo, we have 37.5 = ( (100 - T) ^ 2 / 100 + T) * T / 100 - 3 * ( (100 - T) ^ 2 / 100 + T) / 2 + T.
Wolfram Alpha says T = 91.46 and X = 92.19, which is obviously wrong though, so I must have made some math errors somewhere. 
You just outed yourself as not an engineer.
I have no idea how to quantity, but there is some margin where if we stand, we can get the other player to re-roll after a first roll above 50 and improve your odds (in the game version where you are player 1 and player 2 knows if you stayed or re-rolled).
So basically if we stand, then player 2 will assume we have a “good” number and re-roll even if they get some number that stands.
So yes we are a dog on the re-roll, but we gain significant equity by getting player 2 to abandon their “winning” first roll.
I think the math depends on the margin that P2 will abandon. If P2 gives up 10% of their “winners”
And re-rolls, as long as we don’t give up 10% on the re-roll we are ahead??