Intuitively, this seems wrong.
If I know my opponent had 5 tries to roll, surely their expected value is better than 51?
So even if I was down to one roll only, I think stopping at 51 on roll #4 is a mistake.
I know how to calculate the highest value expectation if you use all 5 rolls, but the game theory part of it is not something I’ve ever learned. This does sound like something where there is some equilibrium (both players only have their own info) but I don’t know how to calculate those.
My guess is it’s a concave down increasing curve between 50 and 100 as you increase rolls. (Stepwise since only integers are valid)
This only gets you the expected value if you use all rolls. I suspect the difficulty in this one is actually determining when is optimal to stop
Reroll 1-62, hold 63+.
Basic level zero strategy is to reroll 1-50, hold 51+. You can easily show that this gives you an expected roll value of 63 and all other strategies give less.
Level 1 strategy would then be to reroll 1-62. If you hold 51-62 against a level zero player they’ll beat you with 63+ at least half the time. If you hold 63 you’ll only be 50/50 but your expectation only goes down from there, and you’ll be a favorite vs anything other than level zero strategy, so hold. The cost of this strategy is you’re deviating from max roll EV. So can it be exploited by a level 2 strategy?
If you look for an equilibrium strategy you’ll find it’s also at 62. The roll EV of dumping 1-62 is still 62.28. If you were to model each successive strategy level as a continuous function and dump below the opponent’s roll EV you’ll find it converges pretty quickly to 62.2497. At least it looks like it converges. I won’t bother with the math to prove it because the real game is discrete. Level 1 strategy is to dump up to 62 to try to beat 63, level 2 strategy needs to beat 62.28 so… you’re still dumping up to 62.
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Now do it for arbitrary N dice ;)
Should already be done, N*0.622497
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I don’t think that’s true.
Let’s assume opponent is keeping 51 and higher on first roll.
We just do 10K (or whatever) trials where we keep everything 51 and higher on the first roll, then 52 and higher on the first roll , and so on. I guess you could do lower numbers to, but seems pointless. See which strategy wins the most. That should be the answer.
You can also alter opponent’s strategy if you like.
Weirdly, the equilibrium is to reroll 61s and keep 62s. When you reroll at N, you lose (N - 50) win rate against his rerolls, which happen N% of the time (assuming mirror strategy at equilibrium). You gain (100 - N) / 2 against his stays, which happen (100 - N)% of the time. Setting those equal gives you a messy quadratic, but if you just guess and check, you get the above strategy as optimal. Against an opponent who uses the naive strategy, you’d want to reroll more often, because rerolling wins against stays, and he stays too often and with weaker numbers.
To solve this from a game theory perspective you wa t to write down each player’s best response function to the other’s reroll threshold and find where they intersect to find the Nash equilibrium. This assumes that a pure strategy with a simple threshold dominates other strategies. Will look into this later if I have time.
Grunching, put me down for 200 on the over.
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I think this is the same result I got, although it appears your math is for a dice that rolls 0-99 and I assumed 1-100.
This sounds like the correct approach, but I think everyone is just assuming that opponent is re-rolling 50 or below. If we just assume that, it is a much easier and different problem.
With a finite number of outcomes you’re better off mapping every possible outcome and tallying the results. Feel free to verify but comparing a reroll at 50 vs reroll at 62 strategy, every 10,000 games the results should be:
50 strat: 4898.5-4982.5-119
62 strat: 4982.5-4898.5-119
I don’t think the Nash Eq applies because this has nothing to do with information at least the way problem was stated. Both people are just maximizing expectation. Now if you added something to the set up, like “other player goes first and you know whether they stopped on one roll or took both rolls” that would change your strategy obviously. As the problem is stated, there’s no better strategy than simply maximizing your expectation.
The equilibrium concept still applies when players move simultaneously, as in rock paper scissors or this game. But you may be right that choosing the reroll rule that maximize the expected value of your own roll is the dominant strategy no matter what your opponent does. If that’s right then it is the Nash equilibrium. I haven’t thought about it long enough to be sure.
Back in the day if you had a tough dice rolling problem you had to take it to your buddy Pierre de Fermat. After he died you were stuck that jerk, Newton.
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Expectation of winning or expectation of final roll value? Because I believe these are not the same. See the 10,000 trial results I posted above. Rerolling at 50 has the highest roll expectation, but that strategy loses to rerolling at 62. The explanation for this is that some of the higher roll EV of the 50 strat comes by holding 51-62, but with either strat expecting a roll value of >62, 51-62 will more often be a losing roll.
With certain parameters it could take some work to find the right balance between throwing away a roll you’re unlikely to beat vs hoping for a roll that’s more likely to win, but in this case both considerations drive the answer into a pretty narrow range where the hold point is certainly above 62 and less than 63.
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I didn’t understand this notation. I need the Ben Garrison version.
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Win-Loss-Tie expected record per 10,000 games between a player who rerolls 50 and below vs a player who rerolls 62 and below.
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How did you calculate/do this?
I gotta start working, but, I will write a program for this tonight that allows setting the number of rerolls allowed and picking a number. Get your python installed.