## Question:

A player in Ankara plays a roulette game in a casino by betting on a single number each time. Because the wheel has 38 numbers, the probability that the player will win in a single play is 1/38. Note that each play of the game is independent of all previous plays.

a. Find the probability that the player will win for the first time on the 5th play.

b. Find the probability that it takes the player more than 80 plays to win for the first time.

c. The gambler claims that because he has 1 chance in 38 of winning each time he plays, he is certain to win at least once if he plays 38 times. Does this sound reasonable to you? Find the probability that he will win at least once in 38 plays.

## Probability :

Probability is a well-used approach in the field of mathematics and statistics; it is used to predict whether winning in a casino is also used in card games to predict the opponent’s next move. Probability is all about making an educated guess for the upcoming events.

## Answer and Explanation:

**Given Information**

- The wheel has 37 numbers.
- Probability that the player will win in a single play is 1/38.

**(a)** The probability that the player will win for the first time on the 5th play is computed as;

Here, the probability of not winning a single play is {eq}1 – dfrac{1}{{38}} = dfrac{{37}}{{38}}{/eq}

Then the probability that the player will win for the first time on the 5th play means that on the first 4 times he losses and then wins at the 5th time, hence;

{eq}begin{align*}

&= {left( {dfrac{{37}}{{38}}} right)^4} times dfrac{1}{{38}}

&= 0.0237

end{align*}{/eq}

Hence, the probability that the player will win for the first time on the 5th play is **0.0237**.

**(b)** The probability that it takes the player more than 80 plays to win for the first time means he losses the first 80 plays;

{eq}begin{align*}

&= {left( {dfrac{{37}}{{38}}} right)^{80}}

&= 0.1184

end{align*}{/eq}

Hence the probability that it takes the player more than 80 plays to win for the first time is **0.1184**.

**(c)**

No the gambler is incorrect. In order to compute the probability that he wins at-least once in 38 plays, let us first compute that he loses 38 times in a row, that is equal to {eq}{left( {dfrac{{37}}{{38}}} right)^{38}}{/eq}.

Now to obtain that he wins at least once, subtract the above probability from 1.

{eq}begin{align*}

&= 1 – {left( {dfrac{{37}}{{38}}} right)^{38}}

&= 0.637

end{align*}{/eq}

Which is distant from a certainty.