Odd Die, Even Result After Subtraction
A fair $n$-sided die, where $n$ is odd, has faces numbered $1$ through $n$. You roll the die, record the outcome, and subtract $1$.
What is the probability that the result is even? Compute the answer for $n = 11$.
Hints
- Subtracting 1 flips parity: an odd outcome becomes even, and an even outcome becomes odd. So identify which die faces produce an even result after subtraction.
- Among $\{1, 2, \ldots, n\}$ with $n$ odd, count how many values are odd -- there is one more odd value than even.
- The number of odd values in $\{1, \ldots, n\}$ when $n$ is odd is $\frac{n+1}{2}$, giving probability $\frac{n+1}{2n}$.
Worked Solution
How to Think About It: Subtracting 1 flips the parity of the outcome: even becomes odd, odd becomes even. So asking "what is the probability the result is even" is exactly the same as asking "what is the probability the die shows an odd number?" Now count: among $\{1, 2, \ldots, n\}$ with $n$ odd, there are $\frac{n+1}{2}$ odd numbers and $\frac{n-1}{2}$ even numbers -- the odd numbers outnumber the even numbers by exactly one, since $n$ itself is odd and goes into the odd bucket.
Quick Estimate: For $n = 11$: faces are $\{1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11\}$. Odd faces: $\{1, 3, 5, 7, 9, 11\}$ -- that is 6 faces. Total: 11 faces. So probability $= 6/11 \approx 0.545$. Slightly more than half, as expected.
Formal Solution:
Let $X$ be the outcome of the die roll, $X \sim \text{Uniform}\{1, 2, \ldots, n\}$.
We want $P(X - 1 \text{ is even}) = P(X \text{ is odd})$.
Among $\{1, 2, \ldots, n\}$ with $n$ odd: - Odd values: $1, 3, 5, \ldots, n$ -- these are $\frac{n+1}{2}$ values. - Even values: $2, 4, 6, \ldots, n-1$ -- these are $\frac{n-1}{2}$ values.
Since the die is fair: $$P(X \text{ is odd}) = \frac{(n+1)/2}{n} = \frac{n+1}{2n}$$
For $n = 11$: $$P = \frac{11+1}{2 \times 11} = \frac{12}{22} = \frac{6}{11}$$
Answer: $\dfrac{n+1}{2n}$, which equals $\dfrac{6}{11}$ for $n = 11$.
Intuition
The key move here is recognizing that a simple operation on the outcome (subtract 1) just relabels the outcomes -- it does not change the probability structure, only which outcomes are in the "even" bucket. Once you see that subtracting 1 flips parity, the problem reduces to a simple counting exercise.
The slight asymmetry ($\frac{n+1}{2n} > \frac{1}{2}$) comes entirely from the fact that $n$ is odd, so the set $\{1, \ldots, n\}$ has one more odd element than even. If $n$ were even, the probability would be exactly $\frac{1}{2}$. This kind of parity argument -- tracking how many elements in $\{1, \ldots, n\}$ are odd versus even depending on whether $n$ is odd or even -- appears frequently in discrete probability problems.