A lottery wait and the geometric distribution

Problem

Each draw wins with fixed probability $0.1$, and each draw spends one token. The question asks for the mean number of draws until the first win, the chance of at least one win from a finite number of tokens, whether ten straight misses are less likely than a win on the first draw, and the least number of tokens that pushes that chance strictly above nine-tenths. The full question and the options are on the CEEC 113 AST Mathematics I paper.

Where it goes wrong

The official item analysis records a score rate of $56\%$ and a discrimination of $0.42$. A common slip is to write “at least one win in two draws” as $0.1+0.1$, or to turn “the probability approaches $1$” into “some finite number of draws guarantees that it equals $1$”. Another trap is missing that “greater than nine-tenths” is a strict inequality.

Idea

Let $X$ be the number of draws until the first win. Then

This is the geometric distribution counted from $1$: trials until the first success. Its expected value is

With $n$ tokens, the chance of at least one win comes from the complement:

Two draws give $1-0.9^2=0.19$, not $0.2$. Ten straight misses have probability $0.9^{10}\approx 0.349$, which is larger than the $0.1$ chance of winning on the first draw. A chance strictly above $0.9$ needs $0.9^n<0.1$, and the least such integer is $n=22$. Every finite $n$ still has $0.9^n>0$, so the chance only approaches $1$ and is never guaranteed to equal $1$.

What the figure shows

• Drag At most n draws and compare $1-0.9^n$ with the nine-tenths threshold.
• The histogram piles up ten thousand waits. Each wait is the number of trials until the first win, and the rightmost bin pools every $X\ge 24$.
• The line underneath tracks the sample mean. As the sample grows, the line settles near the expected value $10$.