counting and probability
posted by zamanofdoom .
A fair, sixsided die is rolled eight times, to form an eightdigit number. What is the probability that the resulting number is a multiple of 8? Express your answer as a common fraction.

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I need help 2
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Total number of numbers using 16
= 6^8 = 1679616
For any number to be divisible by 8, its last 3 digits must be divisible by 8
but these last two digits must contain only the digits from 1 to 6, with no zeros.
possible cases are:
112 136 144 152
216 224 232 256 264
312 336 344 352
416 424 432 456 464
512 536 544 552
616 624 632 656 664
Hoping that I didn't miss any, I count 27
so the front 5 numbers could be anything
there are 6^5 or 7776 different front numbers, each of those could have 27 different last three numbers
so there are 7776x27 or 209952 which are divisible by 8
prob that a number is a multiple of 8
= 209952/1679616
26244/209952
= 6561/52488
= 1/8 < Very suspicious 
BOB is right.
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