Problem Statement
Abd continually rolls a fair \(6\)-sided die until he obtains his first \(6\).
What’s the expected number of times Abd rolls a \(5\) before he stops?
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Solution
Let \(X\) be the event representing number of times we get a \(5\) before hitting our first \(6\).
Let’s try to break this down using conditional expectation.
From any given state, every die roll gives us 6 possible outcomes:
- With probability \(\frac{1}{6}\), we get a \(5\).
- With probability \(\frac{1}{6}\), we get a \(6\) → game stops.
- With probability \(\frac{4}{6}\), we get something else (\(1\), \(2\), \(3\), or \(4\)).
Let’s write the expected value in terms of these outcomes:
\[E[X] = \frac{1}{6} \cdot E[X \mid \text{5}] + \frac{1}{6} \cdot E[X \mid \text{6}] + \frac{4}{6} \cdot E[X \mid \text{Other}]\]Now, analyze each term:
- \(E[X \mid \text{6}] = 0\) → because we stop immediately, no more \(5\)s possible.
- \(E[X \mid \text{5}] = 1 + E[X]\) → we got one \(5\), and then we’re back to square one, since the process restarts.
- \(E[X \mid \text{Other}] = E[X]\) → getting \(1\), \(2\), \(3\), or \(4\) doesn’t change the structure. We’re still in the exact same state with same probabilities.
Substituting all this in:
\[E[X] = \frac{1}{6}(1 + E[X]) + \frac{1}{6}(0) + \frac{4}{6}(E[X])\] \[E[X] = \frac{1}{6} + \frac{1}{6}E[X] + \frac{4}{6}E[X]\] \[E[X] = \frac{1}{6} + \frac{5}{6}E[X]\]Bring terms together:
\[E[X] - \frac{5}{6}E[X] = \frac{1}{6} \Rightarrow \frac{1}{6}E[X] = \frac{1}{6} \Rightarrow E[X] = 1\]So the expected number of times Abd rolls a \(5\) before his first \(6\) is:
\[\boxed{1}\]💡 Did you enjoy this problem? Check out more puzzles in the Problems section!