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# Rolling a 1 or 5

FreeDuck

1
Wed 11 Mar, 2009 08:12 am
Just saw Joe's post -- I am way too slow.
0 Replies

raprap

1
Wed 11 Mar, 2009 08:12 am
probability of expectation is the product of combinations times the probability of rolling only one 1 or 5

P(E)=3!/(1!2!)*(2/6)*(4/6)*(4/6)=3*32/216=96/216=4/9

Rap
0 Replies

joefromchicago

1
Wed 11 Mar, 2009 11:29 am
@oolongteasup,
oolongteasup wrote:

P(none of 1 or5) is 3C0*(2/6)^0* (4/6)^3 = 64/216
P(exactly one of 1 or 5) is 3C1*(2/6)^1*(4/6)^2=96/216
P(exactly two of 1 or 5) is 3C2*(2/6^2*(4/6)^1=48/216
P(exactly three of 1 or 5) is 3C3*(2/6)^3*(4/6)0=8/216

ergo 144/216 , funny how the probabilities add to 1

All that work and you still got it wrong.
0 Replies

oolongteasup

1
Wed 11 Mar, 2009 08:38 pm
my mistake
8 ways to roll all 1's and 5's only 2 of which are all1's or 5's
so 150/216 is correct
0 Replies

maze1on1

1
Thu 7 Jan, 2016 05:15 pm
@joefromchicago,
For 1 dice: 2/6 or 33%
For 2 dice:20/36 or 55.5%
For 3 dice:152/216 or 70.4%

This sounds like a Farkle question 0 Replies

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