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gordonm888
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gordonm888
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January 29th, 2021 at 10:15:43 PM permalink
I guess I could pose this as a Math Puzzle, but I'm actually curious if there is any straightforward way to calculate the answer to this question:

What is the probability of getting a sum of 53 on a roll of 15 dice?
So many better men, a few of them friends, are dead. And a thousand thousand slimy things live on, and so do I.
unJon
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January 30th, 2021 at 2:41:59 AM permalink
Quote: gordonm888

I guess I could pose this as a Math Puzzle, but I'm actually curious if there is any straightforward way to calculate the answer to this question:

What is the probability of getting a sum of 53 on a roll of 15 dice?



I didnít know the answer but had an old memory that it was related to a modified Pascal Triangle so googled and got this interesting read.

http://curiouscheetah.com/BlogMath/pascals-triangle-and-dice-rolls/
The race is not always to the swift, nor the battle to the strong; but that is the way to bet.
Wizard
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January 30th, 2021 at 5:36:30 AM permalink
Quote: gordonm888

What is the probability of getting a sum of 53 on a roll of 15 dice?



27,981,391,815 / 6^15 = apx. 0.059511


Exactly the Pascal's Triangle technique described in the post above.
It's not whether you win or lose; it's whether or not you had a good bet.
Ace2
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January 30th, 2021 at 6:28:43 AM permalink
Since 53 is very close to the expected value of 15 * 3.5 = 52.5, the answer can be approximated as :

1 / (2 * pi * 53)^.5 = 5.48 %
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gordonm888
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gordonm888
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January 30th, 2021 at 11:47:38 AM permalink
Thanks for answers, Wizard, Unjon, Ace2. Pascal's triangle looks very useful - I never knew all that.
So many better men, a few of them friends, are dead. And a thousand thousand slimy things live on, and so do I.
ssho88
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January 30th, 2021 at 10:27:32 PM permalink


I used a very basic method(and not a so straightforward way) to solve it, if we arrange the 15 dice(with sum of 53) in ascending order. you can reduce the total no of different combinations to 521 as below :-

comb1 : 1,1,1,1,1,1,1,4,6,6,6,6,6,6,6, total pemutations1 = 15!/7!/1!/7! = 51480
comb2 : 1,1,1,1,1,1,1,5,5,6,6,6,6,6,6, total pemutations2 = 15!/7!/2!/6! = 180180
comb3 : 1,1,1,1,1,1,2,3,6,6,6,6,6,6,6, total pemutations3 = 15!/6!/1!/1!/7! = 360360

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.

comb519 : 3,3,3,3,3,3,3,3,3,4,4,4,4,5,5, total pemutations519 = 15!/9!/4!/2! = 75075
comb520 : 3,3,3,3,3,3,3,3,4,4,4,4,4,4,5, total pemutations520 = 15!/8!/6!/1! = 45045
comb521 : 3,3,3,3,3,3,3,4,4,4,4,4,4,4,4, total pemutations521 = 15!/7!/8! = 6435

All those 521 combinations can be obtained by combinations analysis.

Grand total permutations = pemutations1 +pemutations2 + . . . .pemutations520 + pemutations521 = 27981391815

So Prob = 27981391815/6^15 = 0.059511453



gordonm888
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gordonm888
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January 31st, 2021 at 11:23:55 AM permalink
Quote: ssho88



I used a very basic method(and not a so straightforward way) to solve it, if we arrange the 15 dice(with sum of 53) in ascending order. you can reduce the total no of different combinations to 521 as below :-

comb1 : 1,1,1,1,1,1,1,4,6,6,6,6,6,6,6, total pemutations1 = 15!/7!/1!/7! = 51480
comb2 : 1,1,1,1,1,1,1,5,5,6,6,6,6,6,6, total pemutations2 = 15!/7!/2!/6! = 180180
comb3 : 1,1,1,1,1,1,2,3,6,6,6,6,6,6,6, total pemutations3 = 15!/6!/1!/1!/7! = 360360

.
.
.

comb519 : 3,3,3,3,3,3,3,3,3,4,4,4,4,5,5, total pemutations519 = 15!/9!/4!/2! = 75075
comb520 : 3,3,3,3,3,3,3,3,4,4,4,4,4,4,5, total pemutations520 = 15!/8!/6!/1! = 45045
comb521 : 3,3,3,3,3,3,3,4,4,4,4,4,4,4,4, total pemutations521 = 15!/7!/8! = 6435

All those 521 combinations can be obtained by combinations analysis.

Grand total permutations = pemutations1 +pemutations2 + . . . .pemutations520 + pemutations521 = 27981391815

So Prob = 27981391815/6^15 = 0.059511453






Very nice work. I had thought of this approach. but I had assumed there would be many more combinations - I am surprised there are only 521.

So 15 dice that have 1-6 as possibilities and that add to 53 have 521 distinct combinations. Could we have predicted 521 based on the numbers 15, 6 and 53?
So many better men, a few of them friends, are dead. And a thousand thousand slimy things live on, and so do I.
ssho88
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January 31st, 2021 at 5:40:24 PM permalink
Very good question, but as usual, I can't answer it. Anyone ?
teliot
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January 31st, 2021 at 6:49:56 PM permalink
I thought of a question based on the above and don't know the answer. Roll two dice over and over. Keep track of two sequences. The first sequence is the sum cumulative total of the first dice after each roll. The second sequence is the sum cumulative total of the second dice after each roll. My question is, what is the expected number of roles until these two cumulative totals are equal?

Let me know if this needs more explanation.
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Wizard
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January 31st, 2021 at 7:40:50 PM permalink
I get it. My hunch is the answer is infinity. The paradoxical thing is it must happen eventually.
It's not whether you win or lose; it's whether or not you had a good bet.

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