"Probability help please" Topic
7 Posts
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| Schogun | 24 Nov 2024 10:28 a.m. PST |
A model has three rolls to hit with 1d20. The first roll for success is 9 or less (45%), second is 7 or less (35%), third is 5 or less (25%). What are the odds of getting 1/2/3 hits? Thanks |
| BillyNM | 24 Nov 2024 11:51 a.m. PST |
For one success it is: 45% x (100-35)% x (100-25)% = 21.9375%
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(100-45)% x 35% X (100-25)% = 14.4375%
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(100-45)% x (100-35)% x 25% = 8.9375% = 45.3125% For two successes it is: 45% x 35% x (100-25)% = 11.8125%
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45% x (100-35)% x 25% = 7.3125%
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(100-45)% x 35% x 25% = 4.8125% = 23.9375% For three successes it is: 45% x 35% x 25% = 3.9375% For zero successes it is: (100-45)% x (100x35)% x (100-25%) = 26.8125% You calculate the probability for any particular set of results on the three dice by multiplying the probablity of each result. To find the probability for all the results giving the same score you sum them. |
| Schogun | 24 Nov 2024 12:49 p.m. PST |
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| Dagwood | 25 Nov 2024 12:02 a.m. PST |
Beware of the zero. For 9 or less to be 45%,the zero must count as HIGH, i.e., 20
Otherwise 9 or less is 50%. |
GildasFacit  | 25 Nov 2024 1:23 a.m. PST |
My D20 are numbered 1-20. |
| Dagwood | 25 Nov 2024 12:07 p.m. PST |
I was thinking of D10s numbered 0 – 9. Not sure I have any D20s ! |
Wolfhag  | 26 Nov 2024 12:58 p.m. PST |
Wouldn't you use a binomial table with one die roll telling you the # of hits? Wolfhag |
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