You have three dice with different numbers of sides. One is a six sided die, one is an 8 sided die and the other is 20 sided. If you roll all three, what is the probability of rolling a collective value of less than 20?
Answers (1)
I broke this down into the probabilities of rolling over a 20 by the different probabilities of every combonation of rolling the 6 sided and 8 sided dice then combaring probabilty of final dice.
* __ 1 2 3 4 5 6
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1__ 2 3 4 5 6 7
2__ 3 4 5 6 7 8
3__ 4 5 6 7 8 9
4__ 5 6 7 8 9 10
5__ 6 7 8 9 10 11
6__ 7 8 9 10 11 12
7__ 8 9 10 11 12 13
8__ 9 10 11 12 13 14
p6, p8:
1[2] = requires 19,20 = 2/20 = .1 1*.1/48 = .002083
2[3] = requires 18,19,20 = 3/20 = .15 2*.15/48 = .00625
3[4] = requires 17,18,19,20 = 4/20 = .2 3*.2/48 = .0125
4[5] = requires 16,17,18...20 = 5/20 = .25 4*.25/48 = .020833
5[6] = requires 15,16,17...20 = 6/20 = .3 5*.3/48 = .03125
6[7] = requires 14,15,16...20 = 7/20 = .35 6*.35/48 = .04375
6[8] = requires 13,14,15...20 = 8/20 = .4 6*.4/48 = .05
6[9] = requires 12,13,14...20 = 9/20 = .45 6*.45/48 = .05625
5[10] = requires 11,12,13...20 = 10/20 = .5 5*.5/48 = .052083
4[11] = requires 10,11,12...20 = 11/20 = .55 4*.55/48 = .045833
3[12] = requires 9,10,11...20 = 12/20 = .6 3*.6/48 = .0375
2[13] = requires 8,9,10...20 = 13/20 = .65 2*.65/48 = .027083
1[14] = requires 7,8,9...20 = 14/20 = .7 1*.7/48 = .014583
+_______
total probability = .4 = 40%
That is the exact answer I know there are faster methods but I was bored
and I dont remember them.