Chapter 9, Problem 1CE

To determine

To calculate: The expected winnings for the 6-spot game.

Answer to Problem 1CE

Solution:

The expected number of winning is $\underset{\_}{+0.20312}$.

Explanation of Solution

Given:

The probability distribution for 6-spot game is shown below,

$x$	$P\left(x\right)$
0	0.16660
1	0.36349
2	0.30832
3	0.12982
4	0.02854
5	0.00310
6	0.00013

Formula used:

$E\left(x\right)={\displaystyle \sum x_i{p}_i}$

Here, $E$ is the expected value of the random variable $x$ and $p$ is the cross-ponding probability.

Calculation:

Let x be the number of matches when playing 6-spot. The outcomes of x are 0,1,2,3,4,5,6.

Therefore, the probability distribution for 6-spot game is shown below,

$x$	$P\left(x\right)$
0	0.16660
1	0.36349
2	0.30832
3	0.12982
4	0.02854
5	0.00310
6	0.00013

So, the winning for each number of matches and subtract our $\$1$ initial payment fee. Thus, the table is,

$x$	$P\left(x\right)$	Winnings
0	0.16660	0
1	0.36349	0
2	0.30832	0
3	0.12982	2
4	0.02854	12
5	0.00310	110
6	0.00013	2000

The expected winning are,

$\begin{array}{ccc}\hfill E\left(\text{winnings}\right)& =& 0\times 0.16660+0\times 0.36349+0\times 0.30832+2\times 0.12982+12\times 0.02854\hfill \\ \hfill & +& 110\times 0.00310+2000\times 0.00013\hfill \\ \hfill & =& 0+0+0+0.25964+0.34248+0.341+0.26\hfill \\ \hfill & =& 1.20312\hfill \end{array}$

Thus, the expected number of winning is $+0.20312$.