Question

Question: Suppose that we roll a fair die until a\(6\)comes up or we have rolled it\(10\)times. What is the expected number of times we roll the die?

Short answer

Answer:

The expected number of times we roll the dice is\(5.03\).

Step-by-step solution

Given information

We roll a fair die until a six comes up or we have rolled it ten times

Step 2: Definition

The expected value, also called the expectation or mean, of the random variable\(X\) on a sample space is equal to

\(E(x) = \sum {X.P(X)} \)

Formula used:

\(E(x) = \sum {X.P(X)} \)

Step 4: Compute the outcome for ten rolls

We need to compute\(\sum\limits_{i = 1}^{10} {i*p(X = i)} \)

\({1^{st}}roll\)

\(1*\frac{1}{6} = \frac{1}{6}\)

\({2^{nd}}roll\)

\(2*\frac{1}{6}*\frac{5}{6} = \frac{{10}}{{36}}\)

\({3^{rd}}roll\)

\(3*\frac{1}{6}*{\left( {\frac{5}{6}} \right)^2} = \frac{{75}}{{217}}\)

\({4^{th}}roll\)

\(4*\frac{1}{6}*{\left( {\frac{5}{6}} \right)^3} = \frac{{500}}{{1296}}\)

\({5^{th}}roll\)

\(5*\frac{1}{6}*{\left( {\frac{5}{6}} \right)^4} = \frac{{3125}}{{7776}}\)

\({6^{th}}roll\)

\(6*\frac{1}{6}*{\left( {\frac{5}{6}} \right)^5} = \frac{{18750}}{{46656}}\)

\({7^{th}}roll\)

\(7*\frac{1}{6}*{\left( {\frac{5}{6}} \right)^6} = \frac{{109375}}{{279936}}\)

\({8^{th}}roll\)

\(8*\frac{1}{6}*{\left( {\frac{5}{6}} \right)^7} = \frac{{625000}}{{1679616}}\)

\({9^{th}}roll\)

\(9*\frac{1}{6}*{\left( {\frac{5}{6}} \right)^8} = \frac{{3515625}}{{10077696}}\)

\({10^{th}}roll\)

\(10*\frac{1}{6}*{\left( {\frac{5}{6}} \right)^9} = \frac{{19531250}}{{10077696}}\)

After adding these values, we get\(\frac{{50700551}}{{10077696}}\)which is approximately\(5.03\).

So, the expected number of times we roll the dice is\(5.03\).