Mathematical types - stat problem!

[Morrus]

I roll initiative once, on average I'll roll 10.5.

If I get to roll twice and pick the highest, what's my average result?

If I get to roll twice and pick the highest, and then I get to reroll one of those, does it make a difference if I simply roll three times and pick the highest or if I roll twice and then reroll the lower of the two? What's my average result using each of those two strategies (assuming it's different)?

 

[Abisashi]

If you roll two dice and take the highest, the average result will be 13.825.

It doesn't matter if you reroll one of those dice or roll a new one; the dice don't remember that they were rolled for initiative before, so that won't affect the third result. (unless you have to take the third die roll even if it isn't higher).

The average for rolling three dice is 15.4875.

 

[Harlekin]

I roll initiative once, on average I'll roll 10.5.

If I get to roll twice and pick the highest, what's my average result?

If I get to roll twice and pick the highest, and then I get to reroll one of those, does it make a difference if I simply roll three times and pick the highest or if I roll twice and then reroll the lower of the two? What's my average result using each of those two strategies (assuming it's different)?

I assume the extra reroll replaces the dice it rolled, even if it's worse than the original? Otherwise, this is the same as just rolling three dice and picking the highest result.

If my understanding of your question is correct, then it clearly matters which die you reroll, and you always want to reroll the lower die.

 

[Morrus]

I assume the extra reroll replaces the dice it rolled, even if it's worse than the original? Otherwise, this is the same as just rolling three dice and picking the highest result.

If my understanding of your question is correct, then it clearly matters which die you reroll, and you always want to reroll the lower die.

Yep.

Initially I roll two dice and take the highest,

Then I have the option of rerolling one of them, and must accept that result - for that die only. The one I didn't reroll stands, and I get to pick between that one and the replacement reroll of the other die.

I hope I exlained that right! I'm looking for the average results of the various stages (1 die, 2 dice, then rerolling one and accepting it and choosing between it and the non-rerolled second die).

 

[Abisashi]

Yep.

Initially I roll two dice and take the highest,

Then I have the option of rerolling one of them, and must accept that result - for that die only. The one I didn't reroll stands, and I get to pick between that one and the replacement reroll of the other die.

You roll two dice, and get the results X and Y, where X >= Y. You re-roll Y and get Z. If Z is greater than X, you use Z, otherwise you use X.

You roll three dice, and get X, Y, and Z, where X >= Y. If Z is greater than X, you use Z, otherwise you use X.

Assuming that's what you are saying (I think it is), then the two cases have the same result.

Edit: I can't read, this is exactly what you said. I'll leave this here in case anyone else can't read.

 

[Abisashi]

In case you missed my edit earlier, the result from rolling three dice is 15.4875.

 

[Roman]

If you roll two dice and take the highest, the average result will be 13.825.
...
The average for rolling three dice is 15.4875.

Would it be possible to show your method of calculation? Don't worry, I trust you on the result, but it has been a long time since I have done probability, so I want to re-learn how to do this kind of thing on my own by following your working. Thanks!

Also, if you have the time, would it be possible to do the same for 4d6 drop the lowest? Thanks!

 

[Elric]

More generally, if you roll two X-sided dice and take the higher number, your average is 2/3*X + 0.5 - 1/(6*X).

If you roll three X-sided dice and take the highest number, your average is 3/4*X + 0.5 - 1/(4*X).

You can plug in X=20 to these formulas and see that you get the exact answers that Abisashi did.

Would it be possible to show your method of calculation? Don't worry, I trust you on the result, but it has been a long time since I have done probability, so I want to re-learn how to do this kind of thing on my own by following your working. Thanks!

For the case of rolling two dice, imagine drawing out a grid of all the possible results, with one dice as the up/down (1 to 20) and the other dice as left/right (1 to 20). To get a result of 1, both dice need to come up as 1, so that's the top left corner- 1 outcome out of the total number of outcomes. To get a result of 2, you can either have (1,2), (2,1), or (2,2)- that's 3 outcomes. To get a result of 3, you can have (1,3), (2,3), (3,3), (3,2), or (3,1); 5 outcomes. Indeed, the number of possibilities that get you a result of i is 2i-1.

The expected result is just the sum, from 1 to 20, of (Result of taking the higher of the two die rolls) * (the number of combination of die rolls that get you this result), divided by the number of outcomes, since all combinations of die rolls are equally likely. So you're going to add up 1*1 + 2*3 + 3*5 + ... + 20*39 and then divide by (20*20) possibilities, which gives you 5530/400= 13.825.

The general formula above is what you get if you replace 20 with X, replace 39 with (2X-1), and use a few formulas for sums to get the answer.

 

[Abisashi]

Would it be possible to show your method of calculation? Don't worry, I trust you on the result, but it has been a long time since I have done probability, so I want to re-learn how to do this kind of thing on my own by following your working. Thanks!

I cheated; for two dice, I made a table in excel with 1-20 along the top and the same along the side, then in the 400 squares between them I had it find the larger of the corresponding column and row numbers (the MAX function in excel). I added up all of the middle numbers and divided by 400.

For three dice, I did the same as for two, but with 20 tables, and each middle box was the max of the column, row, and table number. I added up all of the middle boxes and divided by 8,000.

For me, the easiest way to do these kinds of tests is to use excel or perl. I did all of this in less than 5 minutes. I didn't save the excel spreadsheet, but Elric has kindly provided a more straightforward way of finding the answer anyway.

Also, if you have the time, would it be possible to do the same for 4d6 drop the lowest? Thanks!

I'll do this later tonight.

 

[Abisashi]

For 4d6 drop the lowest:

I did 100,000 rolls and got an average of 12.26443

I ran through all possible combinations and got: 12.2446

So, the brute force method was off by 0.162%, and it would have taken another 60 seconds to get it even closer. Thinking is bad. Excel is good.