Alice and Bob each picks an integer number uniformly at random between 1 and n. Assume that all possible combinations of two numbers are equally likely to be picked. What is the probability that Alice’s number is bigger than Bob’s?

The answer is 1/n

Step-by-step explanation:

First number = Alice's number

Second Number = Bob's Number

There are already 2 good answers if the first number is AT LEAST 1 greater than the second.

However, what if the first number is EXACTLY 1 greater than the second?

I will assume the first number is removed from the array after being selected.

If the first number is 1, then it is impossible, because the second number cannot be less than 1.

If the first number is 2, then the second number could be 1, which has a probability of 1/(n-1)

If the first number is 3, then the second number could be 2, which has a probability of 1/(n-1)

etc…

There is a 1/n probability of each first number selected, so the answer would be:

0*1/n+1/n*1/(n−1)+1/n*1/(n−1)+⋯+1/n*1/(n−1)

There are n terms in that series.

That gives us:

0+1/n(n−1)+1/n(n−1)+⋯+1/n(n−1)

There are n-1 of those identical terms, because 1 of the terms is 0.

(n−1)/n(n−1)=1/n

So the answer is 1/n.

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step-by-step explanation:

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