What is the Average Result of 2d20, Drop the Lowest.
- Thread starter DM_Matt
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pawsplay
Hero
Henry said:
Somwhat complicated.
For each result of die A, determine the odds that the other die is lower. If so, add the result of die A. For instance, for 1, you don't add anything. For a 5, you add 5 four times, for 1, 2, 3, and 4.
Repeat for die B.
Add all the results where the two rolls are the same.
Divide by the number of permutations (number of sides times number of sides)
(as best as I can figure):
Expected value is the sum of the probabilities (a number less than one) times the value matched with that probability. For example, if I have coin with a 5 on one side and a 10 on the other, assuming 50/50, then the expected value is 5*.5 + 10*.5 = 7.5. In this case "expected value" is a bad term since you can't actually get this value. But neither can you roll a 13.825 (or whatever) on a d20. I don't remember the right term, maybe I'll just call it "average value".
So in the case of the 2d20 throw away the lowest, you could do the same thing. The chance of getting a 1 is .0025 (1 in 400).
The chance of getting a 2 is the chance of rolling 2-2, 2-1, or 1-2. That's .0075.
The chance of getting a 3 is the chance of rolling 3-1, 3-2, 3-3, 2-3, 1-3. That's .0125.
Maybe there's a pattern here, I don't know right now. When you're done, you could add them up:
1*.0025 + 2*.0075 + 3*.0125 + ... (and so on)
The sum would be the average/expected value.
For any given result, X, the chance of producing that result using the 2d20 scheme described seems to me to be a combinatorics question - but I'd have to think more about it.
It probably would be faster just to write a random number generator program, run it 10,000 times, add up the results, and divide by 10,000 to get a good approximation.
Expected value is the sum of the probabilities (a number less than one) times the value matched with that probability. For example, if I have coin with a 5 on one side and a 10 on the other, assuming 50/50, then the expected value is 5*.5 + 10*.5 = 7.5. In this case "expected value" is a bad term since you can't actually get this value. But neither can you roll a 13.825 (or whatever) on a d20. I don't remember the right term, maybe I'll just call it "average value".
So in the case of the 2d20 throw away the lowest, you could do the same thing. The chance of getting a 1 is .0025 (1 in 400).
The chance of getting a 2 is the chance of rolling 2-2, 2-1, or 1-2. That's .0075.
The chance of getting a 3 is the chance of rolling 3-1, 3-2, 3-3, 2-3, 1-3. That's .0125.
Maybe there's a pattern here, I don't know right now. When you're done, you could add them up:
1*.0025 + 2*.0075 + 3*.0125 + ... (and so on)
The sum would be the average/expected value.
For any given result, X, the chance of producing that result using the 2d20 scheme described seems to me to be a combinatorics question - but I'd have to think more about it.
It probably would be faster just to write a random number generator program, run it 10,000 times, add up the results, and divide by 10,000 to get a good approximation.
AntiStateQuixote
Enemy of the State
OK, it's actually a reasonably simple formula, but I can't express it here because I don't know how to do mathematical symbols on the PC. What I need is a capital sigma with series expressions.
So, here it is in words:
where x is the number of sides on your die:
fn(x): sum over the series 1 . . . x mulitplying each number by 2 and subtract one then divide the summation result by the square of x.
So, you will have [(1 * 1) + (2 * 3) + (3 * 5) + (4 * 7) + . . . + (x * (2x-2))] / (x^2).
That will give you the average of the rolls of two dice with x sides when you drop the lowest result.
So, here it is in words:
where x is the number of sides on your die:
fn(x): sum over the series 1 . . . x mulitplying each number by 2 and subtract one then divide the summation result by the square of x.
So, you will have [(1 * 1) + (2 * 3) + (3 * 5) + (4 * 7) + . . . + (x * (2x-2))] / (x^2).
That will give you the average of the rolls of two dice with x sides when you drop the lowest result.
a formula
for "higher of two dn":
The possible results are 1,2,3,...,n
The probability of getting 1 is 1/n^2
The probability of getting 2 is 3/n^2
The probability of getting 3 is 5/n^2
The probability of getting k is (2k-1)/n^2
The probability of getting n is (2n-1)/n^2
Now the expected value (mathematical jargon for mean average in this context) is
the sum from k=1 to k=n of k(2k-1)/n^2
Distribute the 1/n^2 and focus on the sum of k(2k-1). This is the sum of 2k^2-k.
The sum of k, k=1 to k=n, is n(n+1)/2
The sum of k^2, k=1 to k=n, is n(n+1)(2n+1)/6
So the sum of 2k^2-k is
2n(n+1)(2n+1)/6 - n(n+1)/2
= [n(n+1)/2][(4n+2)/3 - 1]
= [n(n+1)/2][(8n-2)/6]
= n(n+1)(8n-2)/12
Now remember that 1/n^2 we put aside a little earlier. Then we have
(n+1)(8n-2)/12n
This is the formula. Checking it for n=20, to see if it agrees with the people who did that case by itself, we get
(21)(158)/240=13.825
which agrees with what two people got from calculating this specific case. So I believe my algebra is correct.
for "higher of two dn":
The possible results are 1,2,3,...,n
The probability of getting 1 is 1/n^2
The probability of getting 2 is 3/n^2
The probability of getting 3 is 5/n^2
The probability of getting k is (2k-1)/n^2
The probability of getting n is (2n-1)/n^2
Now the expected value (mathematical jargon for mean average in this context) is
the sum from k=1 to k=n of k(2k-1)/n^2
Distribute the 1/n^2 and focus on the sum of k(2k-1). This is the sum of 2k^2-k.
The sum of k, k=1 to k=n, is n(n+1)/2
The sum of k^2, k=1 to k=n, is n(n+1)(2n+1)/6
So the sum of 2k^2-k is
2n(n+1)(2n+1)/6 - n(n+1)/2
= [n(n+1)/2][(4n+2)/3 - 1]
= [n(n+1)/2][(8n-2)/6]
= n(n+1)(8n-2)/12
Now remember that 1/n^2 we put aside a little earlier. Then we have
(n+1)(8n-2)/12n
This is the formula. Checking it for n=20, to see if it agrees with the people who did that case by itself, we get
(21)(158)/240=13.825
which agrees with what two people got from calculating this specific case. So I believe my algebra is correct.
gizmo33 said:
If you're going to take that approach, why use random numbers? It's no sweat to have the program (or spreadsheet -- it's not hard to implement) systematically generate all 400 possibilities. That's both more efficient than 10,000 random cases, and guaranteed to give you the correct answer instead of merely a good approximation.
the Jester
Legend
This thread is both fascinating and silly. 
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