Estimating XdYbZ

[Extra Anchovies]

How would I determine the expected average value of, for example, 4d6 best 3 or 2d20 best 1? Is there some online calculator that can do this for me? Is there a simple formula I can use to do the calculations on my own (I'm fairly good at mental math)? Can it be done? I'm not looking for specific values for the above examples, but for a general formula or calculator.

[Septimus]

Try http://anydice.com/

Otherwise, try http://dicelog.com/dice with large number of rolls to evaluate the means.

[NichG]

For an off-the-cuff estimate, I generally start from XdYbX, which is easy because its just XdY. Then, every time you double X, you're shifting the mean to roughly about halfway between the previous mean and the maximum.

So e.g. 3d10b3 is about 15, 6d10b3 is about 23, 12d10b3 is about 26.

[T.G. Oskar]

How would I determine the expected average value of, for example, 4d6 best 3 or 2d20 best 2? Is there some online calculator that can do this for me? Is there a simple formula I can use to do the calculations on my own (I'm fairly good at mental math)? Can it be done? I'm not looking for specific values for the above examples, but for a general formula or calculator.

A good rule of thumb (not the exact thing) is that each average, if it ends on ".5", can be expressed as two numbers. For example: the average of a d6 is 3.5, which can be expressed on "half the time lands in 3, half the time in 4". On a 4d6b3 example, assume that you'd end up with four results of 3.5, or one set of 3 and one set of 4. In that case, 4d6b3 would end up on an expected average of 11, because you choose the three best sets - the set of 4s and one of the 3s. Likewise, on a 2d20b1, the result would consider the best of the two results, and if you treat them as 10 and 11, the result will most likely be 11.

Now, this isn't exact, but it works to show at least the tendency that the expected result will be higher than the average result when rolling the total number of "best" results (i.e., rolling 4d6b3 will end with a higher result than 3d6).

[ben-zayb]

3rd post of the thread (http://www.giantitp.com/forums/showthread.php?343968-Needs-moar-math-Reroll-averages-Question-RESOLVED) gives an online calculator.

[GoodbyeSoberDay]

So, anydice is indeed the answer, but you need to know what to type. In this case,

output [highest Z of XdY]

Go to graph if you want to look at it visually.

Here is an example of 4d6b3 (http://anydice.com/program/13e).

[Eldaran]

A good rule of thumb (not the exact thing) is that each average, if it ends on ".5", can be expressed as two numbers. For example: the average of a d6 is 3.5, which can be expressed on "half the time lands in 3, half the time in 4". On a 4d6b3 example, assume that you'd end up with four results of 3.5, or one set of 3 and one set of 4. In that case, 4d6b3 would end up on an expected average of 11, because you choose the three best sets - the set of 4s and one of the 3s. Likewise, on a 2d20b1, the result would consider the best of the two results, and if you treat them as 10 and 11, the result will most likely be 11.

Now, this isn't exact, but it works to show at least the tendency that the expected result will be higher than the average result when rolling the total number of "best" results (i.e., rolling 4d6b3 will end with a higher result than 3d6).

No way this is true. The math makes no sense here.

[Extra Anchovies]

So, anydice is indeed the answer, but you need to know what to type. In this case,


Go to graph if you want to look at it visually.

Here is an example of 4d6b3 (http://anydice.com/program/13e).

Thanks for giving a link with sample syntax! I hadn't figured out how to work anydice before this (they really need an easily findable how-to guide for that site).

[GoodbyeSoberDay]

I had to do a couple of google searches and find an old GitP thread to get that answer, since I don't use anydice with any regularity.

[Jeff the Green]

So, anydice is indeed the answer, but you need to know what to type. In this case,


Go to graph if you want to look at it visually.

Here is an example of 4d6b3 (http://anydice.com/program/13e).

Incidentally, this can't actually find the expected value without some math. The expected value would be the weighted average of the values (i.e. prob_1*1+prob_2*2+prob3_3...). Of course, this isn't the most useful statistic for games; you'll want the mode (which you can find visually), the median (go to "at least" and find the value with the lowest cumulative probability that's greater than 50%; if there's one that has a cumulative probability equal to 50%, it's the average of that one and the one with a cumulative probability immediately greater), or the probability of getting at least or no more than a value.

[Heliomance]

The only thing I've really used anydice for is proving that a roll we experienced in a WoD game I was in had a probability of roughly 1 in 8,000,000.

27 successes on 9 dice. It was insane.

[ThisIsZen]

I got this way wrong yo. Ignore me.

[Heliomance]

This is incorrect, actually. Highest 4 of 2d6 still peaks at 7, like a regular 2d6 roll. I've been trying to figure out how to create a function that would brutal a die roll but thus far anydice's syntax is so limited I can't figure out how to do it.

Of course it does. You're telling it to roll 2d6, then pick the best 4. As it only rolled 2 dice, it picks those two.

[GoodbyeSoberDay]

Incidentally, this can't actually find the expected value without some math. The expected value would be the weighted average of the values (i.e. prob_1*1+prob_2*2+prob3_3...). Of course, this isn't the most useful statistic for games; you'll want the mode (which you can find visually), the median (go to "at least" and find the value with the lowest cumulative probability that's greater than 50%; if there's one that has a cumulative probability equal to 50%, it's the average of that one and the one with a cumulative probability immediately greater), or the probability of getting at least or no more than a value.IMO it's best to just look at the distribution, which is why I cited anydice. The eye can actually interpret visual information like known probability distributions pretty well.
This is incorrect, actually. Highest 4 of 2d6 still peaks at 7, like a regular 2d6 roll. I've been trying to figure out how to create a function that would brutal a die roll but thus far anydice's syntax is so limited I can't figure out how to do it.

EDIT: Based on a google search, the best way to represent XdYbZ would be Xd(Y-Z)+(Z*X). So 1d6b1 would be 1d5+1, 4d6b3 would be 4d3+12, etc.I'm honestly not sure what you mean here. As Heliomance pointed out, 2d6b4 is literally 2d6. Do you mean 2d6 with some re-roll conditions?

[ThisIsZen]

...oh, I think I get what's meant now. I'd thought the 'b' stood for brutal, where you reroll any result equal to or less than the value. It's 'best', isn't it? In which case my posts were based on a fundamental misunderstanding of the desired result, and ignore me. I'll edit my post blank.

EDIT: It even says best in the OP. It's clearly way too late for me.

[A_S]

Incidentally, this can't actually find the expected value without some math. The expected value would be the weighted average of the values (i.e. prob_1*1+prob_2*2+prob3_3...).
Uh, isn't that exactly the number it's displaying right above the graph (12.24 in the linked example)? It's just labelled "average" in the mouseover text, but I assume it means weighted mean.

[Fax Celestis]

I prefer smallroller.

http://www.fnordistan.com/smallroller.html