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What is the probability of rolling three eight-sided dice one by one and getting three strictly increasing numbers? A) 1/512 B) 1/64 C) 1/128 D) 1/256

User Jian Huang
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1 Answer

4 votes

Answer: 7/64

Step-by-step explanation:

Because the numbers must strictly increase, we must have something like {1,2,3} or {4,5,6}. The three numbers need to be different.

If we temporarily ignore the "strictly increasing" requirement, then we have 8*7*6 = 336 different permutations of three different values when selecting from a pool of 8. Alternatively, you can use the nPr formula with n = 8 and r = 3.

This would be out of 8^3 = 512 different ways to roll 3 eight-sided dice (where repeats can be allowed now).

The probability of getting three different values is 336/512 = 21/32.

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The fraction 21/32 represents the probability of getting something like {1,2,3} but it also considers {3,1,2} and {3,2,1} as well.

There are 3*2*1 = 6 ways to arrange the set {a,b,c}. Of those 6 permutations, there's only one that has them in strictly increasing order.

It means we have 1/6 chance of picking the strictly increasing sequence of the permutations of {a,b,c}.

Therefore, we must multiply the previous result 21/32 by 1/6 to get the final answer.

We get the final answer of (21/32)*(1/6) = 7/64

Unfortunately, 7/64 isn't listed as one of the answers. I suppose 1/64 is close, but not really. I would ask your teacher for more clarification. It's possible that your teacher made a typo somewhere.

User Patie
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