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"Probability help please" Topic


7 Posts

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Schogun24 Nov 2024 11:28 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

BillyNM24 Nov 2024 12:51 p.m. PST

For one success it is:

45% x (100-35)% x (100-25)% = 21.9375%
+
(100-45)% x 35% X (100-25)% = 14.4375%
+
(100-45)% x (100-35)% x 25% = 8.9375%

= 45.3125%

For two successes it is:

45% x 35% x (100-25)% = 11.8125%
+
45% x (100-35)% x 25% = 7.3125%
+
(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.

Schogun24 Nov 2024 1:49 p.m. PST

@BillyNM -- Thanks!

Dagwood25 Nov 2024 1:02 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 Sponsoring Member of TMP25 Nov 2024 2:23 a.m. PST

My D20 are numbered 1-20.

Dagwood25 Nov 2024 1:07 p.m. PST

I was thinking of D10s numbered 0 – 9.

Not sure I have any D20s !

Wolfhag Supporting Member of TMP26 Nov 2024 1:58 p.m. PST

Wouldn't you use a binomial table with one die roll telling you the # of hits?

Wolfhag

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