Using the alternation strategy, the value per roll is 10.5/2=5.25. So having rolled once, it's only worth rolling again if you expect to improve the value by 5.25 or more. Since you can only take the current value, not the highest of your rolls, there's no point in replacing the current roll unless it is more than 5.25 below the expected value of your new roll (10.5), so reroll on 5 or lower.
P(r>5)=3/4, so expected number of rolls in a run (before taking the money) is then 4/3. New expected value at the end of a roll is 13 (mid of 6..20). Total cost of a run is (1+4/3) rolls. So expected result is 100*13/(1+4/3)=557 1/4.
The only additional nuance would be that when you are close to the end you might modify your strategy. With 1 turn remaining you must take the value, so with 2 remaining you must take an existing value of 6 or higher or roll-and-take (E=13 for these two rolls, for 13/2=6.5 average). On the 3rd turn from the end the options are roll roll take (E=13) or take roll take (E=r+10.5). So assuming we arrive at this roll with even chance of previously a roll or a take, then we can choose which of these paths.
If we have a previous roll, and it has a value of 3 or more, we take, otherwise we roll. Expected value then is average of 3..20=11.5, so now our last 3 turns expect 22 with probability 9/10 (for rolling 3 or more), or 13 with probability 1/10, or 21.1.
So if we had a take 4 from the end, our expected value is 13, and it was a roll our expected value is 21.1. Truncating at this point gives an expectation of 17.05 for the last 3 rolls, or a grand total of 557.5825.
I'm sure you could continue the analysis backward and maybe generalise an expression.
There is a better strategy that gets you en expected payoff of $560.62.
The basic idea is that suppose on the Nth roll you get an x. Now if you take the money you are left with N-2 moves and $x. If you were to roll again you would be left with N-1 moves. To calculate expected payoff you take the maximum of the two.
We can write a recursive relation for the payoff on the Nth roll.
En = summation of i from 1 to 20 [ max(En-1, i + En-2) ] / 20
with base case E1 = 0 and E2 = 10.5
I couldn't find a simplified formula but simulating it by code I got the expected payoff of $560.62!
I got there an easier way.
Using the alternation strategy, the value per roll is 10.5/2=5.25. So having rolled once, it's only worth rolling again if you expect to improve the value by 5.25 or more. Since you can only take the current value, not the highest of your rolls, there's no point in replacing the current roll unless it is more than 5.25 below the expected value of your new roll (10.5), so reroll on 5 or lower.
P(r>5)=3/4, so expected number of rolls in a run (before taking the money) is then 4/3. New expected value at the end of a roll is 13 (mid of 6..20). Total cost of a run is (1+4/3) rolls. So expected result is 100*13/(1+4/3)=557 1/4.
The only additional nuance would be that when you are close to the end you might modify your strategy. With 1 turn remaining you must take the value, so with 2 remaining you must take an existing value of 6 or higher or roll-and-take (E=13 for these two rolls, for 13/2=6.5 average). On the 3rd turn from the end the options are roll roll take (E=13) or take roll take (E=r+10.5). So assuming we arrive at this roll with even chance of previously a roll or a take, then we can choose which of these paths.
If we have a previous roll, and it has a value of 3 or more, we take, otherwise we roll. Expected value then is average of 3..20=11.5, so now our last 3 turns expect 22 with probability 9/10 (for rolling 3 or more), or 13 with probability 1/10, or 21.1.
So if we had a take 4 from the end, our expected value is 13, and it was a roll our expected value is 21.1. Truncating at this point gives an expectation of 17.05 for the last 3 rolls, or a grand total of 557.5825.
I'm sure you could continue the analysis backward and maybe generalise an expression.
There is a better strategy that gets you en expected payoff of $560.62.
The basic idea is that suppose on the Nth roll you get an x. Now if you take the money you are left with N-2 moves and $x. If you were to roll again you would be left with N-1 moves. To calculate expected payoff you take the maximum of the two.
We can write a recursive relation for the payoff on the Nth roll.
En = summation of i from 1 to 20 [ max(En-1, i + En-2) ] / 20
with base case E1 = 0 and E2 = 10.5
I couldn't find a simplified formula but simulating it by code I got the expected payoff of $560.62!
nice problem!
Very insightful