Nerd in need of help
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April 16th, 2016 at 9:16:20 PM
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So, I play d&d (a tabletop game) and we use 6 20 sided die(6 die each with 20 sides). Someone I play with suggested we use an app to get our initial rolls rather than standard die and his brother apparently got120 (20-20-20-20-20-20), A perfect roll. I'm wondering if anyone here can estimate the probability of that?
I was also wondering if it would be possible to ascertain the probability of getting a total score above 90?
Thanks in advance :)
I was also wondering if it would be possible to ascertain the probability of getting a total score above 90?
Thanks in advance :)
April 16th, 2016 at 9:42:10 PM
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(20-20-20-20-20-20), A perfect roll.
20*20*20*20*20*20 =
20^^6 = 64,000,000
I'll leave the heavy lifting to someone else.
20*20*20*20*20*20 =
20^^6 = 64,000,000
I'll leave the heavy lifting to someone else.
“You don’t bring a bone saw to a negotiation.” - Robert Jordan, former U.S. ambassador to Saudi Arabia
April 17th, 2016 at 1:36:41 AM
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In order to score 90 or more you need to average 15 per dice. 90/15= 6.
There are 6 ways to score 15 or more. 15, 16, 18, 19 and 20.
6/20 * 6/20 * 6/20 * 6/20 * 6/20 * 6/20 = .000729
One time every 1,372 rolls.
This may be well off but to my simple mind it seems pretty logical.
Each dice will average 10.5 per roll. (low of 1 and high of 20 = average of 10.5)
This means scores in the 40 - 80 range should be about 90 - 95 % or above.
There are 6 ways to score 15 or more. 15, 16, 18, 19 and 20.
6/20 * 6/20 * 6/20 * 6/20 * 6/20 * 6/20 = .000729
One time every 1,372 rolls.
This may be well off but to my simple mind it seems pretty logical.
Each dice will average 10.5 per roll. (low of 1 and high of 20 = average of 10.5)
This means scores in the 40 - 80 range should be about 90 - 95 % or above.
April 17th, 2016 at 2:44:11 AM
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Quote: BleedingChipsSlowly(20-20-20-20-20-20), A perfect roll.
20*20*20*20*20*20 =
20^^6 = 64,000,000
I'll leave the heavy lifting to someone else.
I think you did the lifting, 1 in 64 000 000 or 0.000000015625 if you need to see it that way.
Then there are the chances it really happened,
Quote: FattKidd... his brother apparently got ...
Hmmm, I remember when I was a kid and somebody claimed not he, but his brother/friend/dad/whoever did something awesome and how often that got confirmed.
the next time Dame Fortune toys with your heart, your soul and your wallet, raise your glass and praise her thus: “Thanks for nothing, you cold-hearted, evil, damnable, nefarious, low-life, malicious monster from Hell!” She is, after all, stone deaf. ... Arnold Snyder
April 17th, 2016 at 4:40:12 AM
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The answer is for rolls=64000000 gt90=1593130 probability=0.02489265625
The interested reader can verify for themselves using the HTML code provided by the spoiler.
Works out to about 1 roll in 40 will result in a sum greater than 90.
The interested reader can verify for themselves using the HTML code provided by the spoiler.
Works out to about 1 roll in 40 will result in a sum greater than 90.
<!DOCTYPE html>
<html><body><script>
var rolls=0,gt90 = 0;
for (let d1=1;d1<=20;d1++) {
for (let d2=1;d2<=20;d2++) {
for (let d3=1;d3<=20;d3++) {
for (let d4=1;d4<=20;d4++) {
for (let d5=1;d5<=20;d5++) {
for (let d6=1;d6<=20;d6++) {
rolls++;
if (d1+d2+d3+d4+d5+d6>90) {gt90++;}
}}}}}}
document.write(
"The answer is for rolls=" + rolls +
" gt90=" + gt90 +
" probability=" + gt90/rolls
);</script></body></html>
<html><body><script>
var rolls=0,gt90 = 0;
for (let d1=1;d1<=20;d1++) {
for (let d2=1;d2<=20;d2++) {
for (let d3=1;d3<=20;d3++) {
for (let d4=1;d4<=20;d4++) {
for (let d5=1;d5<=20;d5++) {
for (let d6=1;d6<=20;d6++) {
rolls++;
if (d1+d2+d3+d4+d5+d6>90) {gt90++;}
}}}}}}
document.write(
"The answer is for rolls=" + rolls +
" gt90=" + gt90 +
" probability=" + gt90/rolls
);</script></body></html>
Last edited by: BleedingChipsSlowly on Apr 17, 2016
“You don’t bring a bone saw to a negotiation.” - Robert Jordan, former U.S. ambassador to Saudi Arabia
April 17th, 2016 at 7:05:07 AM
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I had this in Excel from long time ago
can't remember why (maybe a test for spreadsheet math)
from here
http://mathforum.org/library/drmath/view/52207.html
(wow! my Excel is really messy. IF I clean it I add it to Google Sheets later)
the distribution
fun dice
never played that game (have some friends that due)
Sally
can't remember why (maybe a test for spreadsheet math)
from here
http://mathforum.org/library/drmath/view/52207.html
(wow! my Excel is really messy. IF I clean it I add it to Google Sheets later)
the distribution
sums from 6 to 120
63 right in the centerso for 91 to 120 =
1,593,130 / 64,000,000 (in agreement with another... Yahoo!!)
(0.024892656 or 1 in 40.17)
<<< >>>
online calculators help too for the distribution
this one looks good to me
http://en.calc-site.com/probabilities/dice_total
and here is what you B after too for percentages
it rounds so B careful
http://anydice.com/
63 right in the center
sum sum ways
6 120 1
7 119 6
8 118 21
9 117 56
10 116 126
11 115 252
12 114 462
13 113 792
14 112 1,287
15 111 2,002
16 110 3,003
17 109 4,368
18 108 6,188
19 107 8,568
20 106 11,628
21 105 15,504
22 104 20,349
23 103 26,334
24 102 33,649
25 101 42,504
26 100 53,124
27 99 65,744
28 98 80,604
29 97 97,944
30 96 117,999
31 95 140,994
32 94 167,139
33 93 196,624
34 92 229,614
35 91 266,244
36 90 306,614
37 89 350,784
38 88 398,769
39 87 450,534
40 86 505,989
41 85 564,984
42 84 627,304
43 83 692,664
44 82 760,704
45 81 830,984
46 80 902,994
47 79 976,164
48 78 1,049,874
49 77 1,123,464
50 76 1,196,244
51 75 1,267,504
52 74 1,336,524
53 73 1,402,584
54 72 1,464,974
55 71 1,523,004
56 70 1,576,014
57 69 1,623,384
58 68 1,664,544
59 67 1,698,984
60 66 1,726,264
61 65 1,746,024
62 64 1,757,994
63 63 1,762,004
1,593,130 / 64,000,000 (in agreement with another... Yahoo!!)
(0.024892656 or 1 in 40.17)
<<< >>>
online calculators help too for the distribution
this one looks good to me
http://en.calc-site.com/probabilities/dice_total
and here is what you B after too for percentages
it rounds so B careful
http://anydice.com/
fun dice
never played that game (have some friends that due)
Sally
Last edited by: mustangsally on Apr 17, 2016
I Heart Vi Hart
April 17th, 2016 at 9:25:39 AM
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this is how math teachers (high school level and above) would like this solved using the normal distribution (as an approximation)Quote: DavidGrantEach dice will average 10.5 per roll. (low of 1 and high of 20 = average of 10.5)
mean = 10.5 * 6
variance = (665/20) * 6
standard deviation = 14.12444689
X=90.5 (so it includes 91)
z = (X - mean) / sd = 1.946978895 (almost 2) and (I never look that up)
math teachers now all smile!
abouts 0.025768632
http://davidmlane.com/hyperstat/z_table.html
https://www.mathsisfun.com/data/standard-normal-distribution.html
Last edited by: mustangsally on Apr 17, 2016
I Heart Vi Hart
