Question

Roulette Marti decides to keep placing a 1$ bet on number 15 in consecutive spins of a roulette wheel until she wins. On any spin, there's a 1-in-38 chance that the ball will land in the 15 slot.

a. How many spins do you expect it to take for Marti to win?

b. Would you be surprised if Marti won in 3 or fewer spins? Compute an appropriate probability to support your answer.

Short answer

(a) Mean of X $\mathrm{E}\left(\mathrm{X}\right)=\frac{1}{p}=\frac{1}{1/38}=38$

(b) It can be seen that the probability of 3 or fewer is too low So, it is surprising if Martin won in 30 or fewer spins.

Step-by-step solution

Part (a) Step 1: Given Information

Number of spins $=15$

Probability of the ball landing in any slot $=\frac{1}{38}$.

Concept used:

If the trials are repeated until a success occurs and the probability of success is same in each trial, then it is the case of geometric distribution.

In this case, a bet is placed until a win and the probability of winning in each chance is same and was independent. So, this follows geometric distribution.

Part (a) Step 2: Simplificaiton

Let X be the number of spins for a success

So, for Geometric with $\mathrm{p}=\left(\frac{1}{38}\right)$

$P(X=k)=(1-p{)}^{k-1}\times p$

Mean of X,

$\mathrm{E}\left(\mathrm{X}\right)=\frac{1}{p}=\frac{1}{1/3\mathrm{s}}=38$

Part (b) Step 1: Given information

Number of spins is 15.

Probability of the ball landing in any slot is$1/38$

Part (b) step 2: Calculation

The likelihood of winning in three or fewer spins

$$\begin{gathered} P(X\le 3)=0+{\left(1-\frac{1}{38}\right)}^{1-1}\times \frac{1}{38}+{\left(1-\frac{1}{38}\right)}^{2-1}\times \frac{1}{38} \\ +{\left(1-\frac{1}{38}\right)}^{3-1}\times \frac{1}{38} \\ P(X\le 3)=0.07689 \end{gathered}$$

It is clear that the likelihood of three or fewer is excessively low.

So it's surprising Martin didn't win in 30 or less spins.