Problem Statement
Suppose Jimmy and Simon are selecting uniformly random numbers.
Jimmy selects from the set \(\{1, 2, \ldots, 1000\}\),
and Simon selects from \(\{1, 2, \ldots, 3000\}\).
Compute the probability that Simon chooses a strictly larger number than Jimmy.
Original Problem Link: Click here
Solution
Let’s break this down based on Simon’s value. Clearly, if Simon picks anything greater than 1000, he automatically wins, since Jimmy can’t go beyond 1000.
So let’s first consider two cases:
Case 1: Simon picks a number from 1001 to 3000
These are all guaranteed wins, since Jimmy picks from only 1 to 1000.
Total such numbers = \(3000 - 1000 = 2000\)
So the probability Simon picks such a number is:
\[P(\text{Simon > 1000}) = \frac{2000}{3000} = \frac{2}{3}\]And for each such pick, probability that Simon wins = 1.
So this contributes \(\frac{2}{3} \times 1 = \frac{2}{3}\) to the final answer.
Case 2: Simon picks a number from 1 to 1000
This is the more interesting case.
Let’s say Simon picks a number \(x \in \{1, 2, \ldots, 1000\}\).
This happens with probability \(\frac{1}{3000}\) for each value of \(x\).
Given this, Jimmy can pick any number from 1 to 1000, uniformly.
The probability that Jimmy picks a strictly smaller number than \(x\) is:
So for a fixed \(x\), the probability that Simon wins is \(\frac{x - 1}{1000}\).
Now sum this over all values of \(x\) from 1 to 1000:
But:
\[\sum_{x=1}^{1000} (x - 1) = \sum_{i=0}^{999} i = \frac{999 \cdot 1000}{2}\]So the contribution from this case is:
\[\frac{1}{3000 \cdot 1000} \cdot \frac{999 \cdot 1000}{2} = \frac{999}{6000}\]Final Step: Total Probability
\[P(\text{Simon wins}) = \underbrace{\frac{2}{3}}_{\text{Simon > 1000}} + \underbrace{\frac{999}{6000}}_{\text{Simon ≤ 1000}}\]Let’s simplify:
\[\frac{2}{3} = \frac{4000}{6000} \quad \Rightarrow \quad \text{Total} = \frac{4000 + 999}{6000} = \frac{4999}{6000}\]Final Answer: Thus the required probability is \(\boxed{\frac{4999}{6000}}\)
💡 Did you enjoy this problem? Check out more puzzles in the Problems section!