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If you roll two 10-sided dice, what is the prob that the two dice add up to either...
- Thread starter mepoom
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HallsofIvy
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A single 10-sided die can come up 1, 2, 3, 4, 5, 6, 7, 8, 9, or 10. two 10-sided dice can come up
(1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6), (1, 7), (1, 8), (1, 9), (1, 10)
(2, 1), (2, 2), (2, 3), (2, 4), (2, 5), (2, 6), (2, 7), (2, 8), (2, 9), (2, 10)
(3, 1), (3, 2), (3, 3), (3, 4), (3, 5), (3, 6), (3, 7), (3, 8), (3, 9), (3, 10)
etc.
Do you see that there are 10x10= 100 such pairs?
Now, how many of those pairs add up to 3? (1, 2) and (2, 1) certainly but are those the only such pairs?
How many of those pairs add up to 19? (1, 18), (2, 17), (3, 16), ..., (17, 2), (18, 1). How many are there?
(1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6), (1, 7), (1, 8), (1, 9), (1, 10)
(2, 1), (2, 2), (2, 3), (2, 4), (2, 5), (2, 6), (2, 7), (2, 8), (2, 9), (2, 10)
(3, 1), (3, 2), (3, 3), (3, 4), (3, 5), (3, 6), (3, 7), (3, 8), (3, 9), (3, 10)
etc.
Do you see that there are 10x10= 100 such pairs?
Now, how many of those pairs add up to 3? (1, 2) and (2, 1) certainly but are those the only such pairs?
How many of those pairs add up to 19? (1, 18), (2, 17), (3, 16), ..., (17, 2), (18, 1). How many are there?
If you roll two 10-sided dice, what is the prob that the two dice add up to either 19 or 3?
Please help explain me how to solve this question ;(
I figgered i'd go hole hog:
With an n sided die you can roll n different numbers, so with two n sided die you can roll n2 combinations [actually permutations, since order does matter]. To roll a particular number x between 2 (a 1 on each face up side of the die) and 2n (an n on each side), start with y, y = n if x > n or, otherwise, y = x - 1 for the combination of y on one upward face of die 1 and x-y on die 2; next is y-1 on die 1 and x+1-y on die two [indicated as the ordered pair (y-1, x+1-y)], and so forth. The ending numbers will depend again on the size of x. If x > n then the last number will be (x-n, n) and you will have a possible n-(x-n)+1=2n+1-x values. If x \(\displaystyle \le\) n, the ending value is (1, x-1) and there are x-1 values.
Happy figgerin
A single 10-sided die can come up 1, 2, 3, 4, 5, 6, 7, 8, 9, or 10. two 10-sided dice can come up
(1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6), (1, 7), (1, 8), (1, 9), (1, 10)
(2, 1), (2, 2), (2, 3), (2, 4), (2, 5), (2, 6), (2, 7), (2, 8), (2, 9), (2, 10)
(3, 1), (3, 2), (3, 3), (3, 4), (3, 5), (3, 6), (3, 7), (3, 8), (3, 9), (3, 10)
etc.
Do you see that there are 10x10= 100 such pairs?
Now, how many of those pairs add up to 3? (1, 2) and (2, 1) certainly but are those the only such pairs?
How many of those pairs add up to 19? (1, 18), (2, 17), (3, 16), ..., (17, 2), (18, 1). How many are there?
Only (1,2) and (2,1) add up to 3 which is 2/100
Only (9,10) and (10,9) add up to 19 which also is 2/100
So would the answer be 2/100 + 2/100 = 4/100 ?
HallsofIvy
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??? So what happened to (1, 18), (2, 17), (3, 16), etc. that I pointed out in my first response?Only (1,2) and (2,1) add up to 3 which is 2/100
Only (9,10) and (10,9) add up to 19 which also is 2/100
So would the answer be 2/100 + 2/100 = 4/100 ?
??? So what happened to (1, 18), (2, 17), (3, 16), etc. that I pointed out in my first response?
Ten-sided dice have only 1,2,3,4,5,6,7,8,9,and 10. Correct me if I'm wrong.
HallsofIvy
Elite Member
- Joined
- Jan 27, 2012
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Ouch. Now I'm feeling stupid!
(Don't worry, I'm used to it.)
(Don't worry, I'm used to it.)
Ouch. Now I'm feeling stupid!
(Don't worry, I'm used to it.)
Now you know why I said "Of course dum dum" but then I'm used to it too.