## Dart Game

A game consists of throwing darts onto a target. The target is divided into four sectors as shown in the figure below

Assume that the throws are independent from one another and that the players hit the target on every throw.

1. A player throws a dart. Let $p_0$ represent the probability of getting $0$ point, $p_3$ the probability of getting $3$ points, and $p_5$ the probability of getting $5$ points.
Thus, $p_0+p_3+p_5=1$. Given $p_5=\frac{1}{2}p_3$ and $p_5=\frac{1}{3}p_0$, find the values of $p_0,p_3$ and $p_5$.

2. One of the games consists of throwing a maximum of three darts. The player wins the game if she obtains a total (for the three throws) of $8$ or more points. If she has a total of $8$ or more points after two throws, she does not throw the third dart.
Let $W_2$ represent the event “the player wins the game in $2$ throws”,  $W_3$ the event “the player wins the game in $3$ throws”, and $L$ the event “the player loses the game”. Let $p\left (A\right )$ represent the probability of an event $A$.
(a) Show that $p\left (W_2\right )=\frac{5}{36}$ by using a possibility tree
(b) Derive $p\left (L\right )$ knowing that $p\left (W_3\right )=\frac{7}{36}$

3. A player plays six games according to the rules given in Question 2. What is the probability that she will win at least one game?

4. The price  for one game is fixed at $2$ euros. If the player wins the game in two throws, she receives $5$ euros. If she wins in three throws, she receives $3$ euros. If she loses, she receives nothing. Let $X$ be the random variable that represents the algebraic gain by the player for one game. The possible values of $X$ are therefore: $-2, 1$ and $3$.
(a) Give the probability distribution of $X$
(b) Determine the expected value of $X$. Is the game favorable to the player?
Source: Baccalaureat General, Session Avril 2011, Pondichery, Serie Scientifique, www.ilemaths.net

SOLUTION

1. Find the values of $p_0, p_3$ and $p_5$.
We have three unknown variables $p_0,p_3, p_5$ and three equations
$p_0+p_3+p_5=1\quad\quad,\left (1\right )$
$p_5=\frac{1}{2}p_3\quad\quad\quad\qquad,\left (2\right )$
$p_5=\frac{1}{3}p_0\quad\quad\quad\qquad,\left (3\right )$
Compare Eq. $\left (2\right )$ and $\left (3\right )$
$\frac{1}{2}p_3=\frac{1}{3}p_0$
Multiply both sides by $2$
$p_3=\frac{2}{3}p_0\quad\quad\quad\qquad,\left (4\right )$
Substitute the values of $p_3$ and $p_5$ into Eq. $\left (1\right )$
$p_0+\frac{2}{3}p_0+\frac{1}{3}p_0=1$
$\frac{3p_0+2p_0+p_0}{3}=1$
$\frac{6p_0}{3}=1$
$2p_0=1$
$p_0=\frac{1}{2}$
Substitute the value of $p_0$ into Eq. $\left (4\right )$
$p_3=\frac{2}{3}\times\frac{1}{2}$
$=\frac{1}{3}$
Substitute the value of $p_0$ into Eq. $\left (3\right )$
$p_5=\frac{1}{3}\times \frac{1}{2}$
$=\frac{1}{6}$

2. (a) Show that $p\left (W_2\right )=\frac{5}{36}$ by using a possibility tree
We will draw the tree symbolically by writing the scores of a $W_2$ event as follows
$5-5$, first throw 5 points; second throw 5 points
$5-3$, first throw 5 points; second throw 3 points
$3-5$, first throw 3 points; second throw 5 points
$p\left (W_2\right )=p_5p_5+p_5p_3+p_3p_5$
$=\frac{1}{6}\cdot\frac{1}{6}+\frac{1}{6}\cdot\frac{1}{3}+\frac{1}{3}\cdot\frac{1}{6}$
$=\frac{1}{36}+\frac{1}{18}+\frac{1}{18}$
$=\frac{1+2+2}{36}$
$=\frac{5}{36}$

2. (b) Derive $p\left (L\right )$ knowing that $p\left (W_3\right )=\frac{7}{36}$
$p\left (L\right )+p\left (W_2\right )+p\left (W_3\right )=1$
$p\left (L\right )+\frac{5}{36}+\frac{7}{36}=1$
$p\left (L\right )=1-\frac{5}{36}-\frac{7}{36}$
$=\frac{36-5-7}{36}$
$=\frac{24}{36}$
$=\frac{2}{3}$

3. A player plays six games according to the rules given in question 2.
What is the probability that she will win at least one game?
Winning at least one game means winning $1,2,3,4,5$ or $6$ games. In terms of losing that means losing $5,4,3,2,1$ or $0$ games.
Probability of losing all six games
$=\left [p\left (L\right )\right ]^6$
$=\left (\frac{2}{3}\right )^6$
$\therefore$ Probability of winning at least one game
$=1-\left (\frac{2}{3}\right )^6$
$\approx .91$

4. (a) Give the probability distribution of $X$
$X=-2$ when the player loses the game
$p\left (X=-2\right )=p\left (L\right )$
$=\frac{2}{3}$
$X=1$ when the player wins the game in three throws
$p\left (X=1\right )=p\left (W_3\right )$
$=\frac{7}{36}$
$X=3$ when the player wins the game in two throws
$p\left (X=3\right )=p\left (W_2\right )$
$=\frac{5}{36}$

4. (b) Determine the expected value of $X$. Is the game favorable to the player?
$E\left (X\right )=\left (-2\right )p\left (X=-2\right )+1p\left (X=1\right )+3p\left (X=3\right )$
$=\left (-2\right )\times \frac{2}{3}+1\times \frac{7}{36}+3\times \frac{5}{36}$
$\approx -.72$
The negative expected value indicates that the player is expected to lose money over the long run. The game is not favorable to the player.

1. $p_0=\frac{1}{2};\;p_3=\frac{1}{3};\;p_5=\frac{1}{6}$
2. (b) $p\left (L\right )=\frac{2}{3}$
3. Probability of winning at least one game $\approx .91$
4. (a) $p\left (X=-2\right )=\frac{2}{3};\;p\left (X=1\right )=\frac{7}{36};\;p\left (X=3\right )=\frac{5}{36}$
4. (b) $E\left (X\right )\approx -.72$; game not favorable to player