My crazy die has one side with one pip on it, two with two pips , and three on each of three remaining sides.
I tossed it 7 times in a row and wrote down the result of those tosses as a 7 digit number.
What is the probability that this number
a. Is bigger than 1231111?
b. Contains at least one triplet of identical digits?
c. Is divisible by 3?
d. Exhibits all the 3 features mentioned above?
For part a. I found the probability the number is less than or equal to 1231111 then subtracted from 1:
If it starts 11 it is guaranteed less. Probability 1/6*1/6 = 1/36
If it starts 121 or 122 it is guaranteed less. Probability 1/6*1/3*(1/6+1/3) = 1/36
If it starts 123 it must continue 1231111 to be equal. Probability 1/6*1/3*1/2*(1/6)^4 = 1/46656
These sum to 2593/46656
The complement of this is 44063/46656
For part c. I considered how many 3's there are and then what combinations of 1's and 2's keep the divisibility by 3:
7 3's probability 1/128
this guarantees divisibility so 1/128*1 =
1/1286 3's probability 7/128
cannot be divisible so 7/128*0 =
05 3's probability 21/128
the other digits must be a 1 and a 2 (since they are not 3 they are in a 1:2 ratio) 1/3*2/3*(2C1) = 4/9
4/9*21/128 =
7/964 3's probability 35/128
the other digits must be 111 or 222 (since they are not 3 they are in a 1:2 ratio) (1/3)^3*(2/3)^3=1/3
1/3*35/128 =
35/3843 3's probability 35/128
the other digits must be 1122 (since they are not 3 they are in a 1:2 ratio) (1/3)^2*(2/3)^2*(4C2) = 8/27
8/27*35/128 =
35/4322 3's probability 21/128
the other digits must be 11112 or 12222
21/128*10/27 =
35/5761 3 probability 21/128
the other digits must be a 111111, 111222, or 222222
25/81*7/128 =
175/103680 3's probability 1/128
the other digits must be a 1111122 or 1122222
28/81*1/128 =
7/2592
The sum of the bold numbers is the total probability for part c:
3455/10368
I won't be doing parts b or d by hand. You'd have to be careful about overcounting. For example: 1111222 contains 3 triplets but counts once.

Posted by Jer
on 20140204 13:09:46 