The showcase showdown. On the popular television game show The Price Is Right, contestants can play “The Showcase Showdown.” The game involves a large wheel with 20 nickel values, 5, 10, 15, 20,…, 95, 100, marked on it. Contestants spin the wheel once or twice, with the objective of obtaining the highest total score without going over a dollar (100). [According to the American Statistician (August 1995), the optimal strategy for the first spinner in a three-player game is to spin a second time only if the value of the initial spin is 65 or less.] Let x represent the total score for a single contestant playing “The Showcase Showdown.” Assume a “fair” wheel (i.e., a wheel with equally likely outcomes). If the total of the player’s spins exceeds 100, the total score is set to 0.

a. If the player is permitted only one spin of the wheel, find the probability distribution for x.

b. Refer to part a. Find E(x) and interpret this value

c. Refer to part a. Give a range of values within which x is likely to fall.

d. Suppose the player will spin the wheel twice, no matter what the outcome of the first spin. Find the probability distribution for x.

e. What assumption did you make to obtain the probability distribution, part d? Is it a reasonable assumption?

f. Find$\mathit{\mu }$ and$\mathit{\sigma }$ for the probability distribution, part d. and interpret the results.

g. Refer to part d. What is the probability that in two spins the player’s total score exceeds a dollar (i.e., is set to 0)?

h. Suppose the player obtains a 20 on the first spin and decides to spin again. Find the probability distribution forx.

i. Refer to part h. What is the probability that the player’s total score exceeds a dollar?

j. Given the player obtains a 65 on the first spin and decides to spin again, find the probability that the player’s total score exceeds a dollar.

k. Repeat part j for different first-spin outcomes. Use this information to suggest a strategy for the one-player game.

From the information, it is clear that a game involves large wheel with 20 nickel values.

That is,$\left\{5,10,15,20,25,30,35,40,45,50,55,60,65,70,75,80,85,90,95,100\right\}$

Let x represents the total score for a single contestant playing.

Since, the 20 possible outcomes are equally likely, the probability of choosing any of the outcomes is clearly $\frac{1}{20}$.

The probability distribution of x is obtained below:

That is,

$$\begin{gathered} P\left(x=5\right)=\frac{1}{20} \\ P=\left(x=10\right)=\frac{1}{20} \end{gathered}$$

Similarly for the remaining values are shown in the probability distribution table.

Hence p(x) =0.05 for all the values of ‘x’.

The formula for the mean is given below

$$\begin{gathered} \mathit{\mu }\mathbf{=}\mathit{E}\left(x\right) \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{=}\mathbf{\sum }_{\mathbf{}}\mathit{x}\mathbf{}\mathit{p}\left(x\right) \end{gathered}$$

$$\begin{gathered} =\left\{\begin{array}{c}5\left(\frac{1}{20}\right)+10\left(\frac{1}{20}\right)+15\left(\frac{1}{20}\right)20\left(\frac{1}{20}\right)+25\left(\frac{1}{20}\right)\\ +30\left(\frac{1}{20}\right)+35\left(\frac{1}{20}\right)+40\left(\frac{1}{20}\right)45\left(\frac{1}{20}\right)+50\left(\frac{1}{20}\right)\\ +55\left(\frac{1}{20}\right)+60\left(\frac{1}{20}\right)+65\left(\frac{1}{20}\right)+70\left(\frac{1}{20}\right)+75\left(\frac{1}{20}\right)\\ +80\left(\frac{1}{20}\right)+85\left(\frac{1}{20}\right)+90\left(\frac{1}{20}\right)+95\left(\frac{1}{20}\right)+100\left(\frac{1}{20}\right)\end{array}\right\} \\ =52.5 \end{gathered}$$

Thus, the mean of x is 52.5

On an average total score for a single contestant playing is 52.5.

c.

The formula for the variance of x is given below:

${\mathit{\sigma }}^{\mathbf{2}}\mathbf{=}\mathit{E}{\left(x-\mu \right)}^{\mathbf{2}}\mathbf{}\mathit{P}\left(x\right)$

Therefore,

${\sigma }^2=\left\{\begin{array}{c}(5-52.5{)}^2(0.5)+(10-52.5{)}^2(0.5)+(15-52.5{)}^2(0.5)\\ +(20-52.5{)}^2(0.5)+(25-52.5{)}^2(0.5)+(30-52.5{)}^2(0.5)\\ +(35-52.5{)}^2(0.5)+(40-52.5{)}^2(0.5)+(45-52.5{)}^2(0.5)\\ +(50-52.5{)}^2(0.5)+(55-52.5{)}^2(0.5)+(60-52.5{)}^2(0.5)\\ +(65-52.5{)}^2(0.5)+(70-52.5{)}^2(0.5)+(75-52.5{)}^2(0.5)\\ +(80-52.5{)}^2(0.5)+(85-52.5{)}^2(0.5)+(90-52.5{)}^2(0.5)\\ +(95-52.5{)}^2(0.5)+(100-52.5{)}^2(0.5)\end{array}\right\}$

The standard deviation is,

$$\begin{gathered} \sigma =\sqrt{{\sigma }^2} \\ =\sqrt{831.25} \\ =28.83 \end{gathered}$$

Thus, the standard deviation of x is 28.83.

Here, the wheel is assumed to be “fair”, that is the outcomes of the wheel are equally likely. Therefore, the distribution is uniform distribution because the continuous random variables appear to have the equally likely outcomes over the range of values. Since, the uniform distribution is not bell or mound shaped. Chebychev’s rule can be applied to describe the data.

Thus, the Chebychev’s rule is used in order to find the range of values.

At least$\frac{3}{4}$ of the observations are lies between two standard deviations $\left(\mu \pm 2\sigma \right)$.

At least of the observations are lies between two standard deviations $\left(\mu \pm 3\sigma \right)$.

The observations will fall within two standard deviations of the mean is,

$$\begin{gathered} \left(\mu \pm 2\sigma \right)=\left(52.5\pm 2\left(28.83\right)\right) \\ =\left(52.3-2\left(28.83\right),4.655+2\left(28.83\right)\right) \\ =\left(52.3-57.66+52.5+57.66\right) \\ =\left(-5.16,110.16\right) \end{gathered}$$

Thus, at least$\frac{3}{4}$ of the data the range of values for wheel falls in the interval is (-5.16,110.16).

Here, all observations are lie between two standard deviations of the mean. Hence, it is clear that, x is likely to fall in any one interval with the equal length.

d.

If the player spins the wheel twice, the total outcomes are as follows:

Total outcomes=400

The sample space is as follows:

$N\left(S\right)=\left\{\begin{array}{c}(5,5)\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}(10,5)\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}(15,5)\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\cdots \hspace{0.33em}\hspace{0.33em}(100,5)\\ (5,10)\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}(10,10)\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}(15,10)\hspace{0.33em}\hspace{0.33em}\cdots \hspace{0.33em}\hspace{0.33em}(100,10)\\ ⋮\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}⋮\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}⋮\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\cdots \hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}⋮\\ (5,100)\hspace{0.33em}(10,100)\hspace{0.33em}\hspace{0.33em}(15,100)\hspace{0.33em}\hspace{0.33em}\cdots \hspace{0.33em}\hspace{0.33em}(100,100)\end{array}\right\}$

Since, there are 400 possible outcomes which are equally likely, the probability of choosing any of the outcome is clearly$\frac{1}{400}=0.0025$.

The probability distribution of x is obtained below:

Let x be the sum of the two numbers in each sample.

There is one sample in the sample space with the sum of 10.

$$\begin{gathered} P\left(x=10\right)=\frac{1}{400} \\ =0.0025 \end{gathered}$$

There are two samples in the sample space with the sum of 15.

$$\begin{gathered} P\left(x=15\right)=\frac{2}{400} \\ =0.005 \end{gathered}$$

There are three samples in the sample space with the sum of 15.

$$\begin{gathered} P\left(x=15\right)=\frac{3}{400} \\ =0.0075 \end{gathered}$$

Assume the wheel to be fair. If the total of the player’s spin exceeds 100, then set total score to 0. Then the probability for x=0 is obtained by adding all the probabilities and subtract it from the total probability 1.

That is,

$$\begin{gathered} P\left(x=0\right)=1-0.475 \\ =0.5250 \end{gathered}$$

Similarly, for the remaining values is shown in the probability distribution table.

e.

Here, the wheel is assumed to be fair. That is, the outcomes of the wheel are equally likely.

f.

Since,

$\mathit{\mu }\mathbf{=}\mathbf{\sum }_{\mathbf{}}\mathit{x}\mathbf{}\mathit{P}\left(x\right)$

$$\begin{gathered} =\left\{\begin{array}{c}0\left(0.5250\right)+10\left(0.0025\right)+15\left(0.0050\right)+20\left(0.0075\right)+25\left(0.0100\right)+\\ 30\left(0.0125\right)+35\left(0.0150\right)+40\left(0.0175\right)+45\left(0.0200\right)+50\left(0.0225\right)+\\ 55\left(0.0250\right)+60\left(0.0275\right)+65\left(0.0300\right)+70\left(0.0325\right)+75\left(0.0350\right)+\\ 80\left(0.0375\right)+85\left(0.0400\right)+90\left(0.0425\right)+95\left(0.0450\right)+100\left(0.0475\right)\end{array}\right\} \\ =33.25 \end{gathered}$$

Thus, the mean of x is 33.25.

On an average total score for a single contestant playing is 33.25.

Obtain the value of the standard deviation.

The formula for the variance of x is,

${\mathit{\sigma }}^{\mathbf{2}}\mathbf{=}\mathit{E}{\left(x-\mu \right)}^{\mathbf{2}}\mathit{P}\left(x\right)$

Consider,

$$\begin{gathered} {\sigma }^2=E{\left(x-\mu \right)}^{2}P\left(x\right) \\ =\left\{{\left(0.-33.25\right)}^2\left(0.5250\right)+...+{\left(100-33.25\right)}^2\left(0.0425\right)\right\} \\ =1471.31 \end{gathered}$$

The standard deviation is,

$$\begin{gathered} \sigma =\sqrt{{\sigma }^2} \\ =\sqrt{1471.3125} \\ =38.3577 \end{gathered}$$

Thus, the standard deviation of x is 38.3577.

g.

If the sum of the player’s spin exceeds 100, then set the probability of the score to 0.

Thus,

$$\begin{gathered} P\left(x=0\right)=1-0.475 \\ =0.5250 \end{gathered}$$

Hence, the probability that in two spins the player’s total score exceeds a dollar is 0.5250.

h.

If the player spins the wheel twice, obtain 20 on first spin and decides to spin again. Then,

The total outcomes are 20 and the sample space is as follows:

$N\left(S\right)=\left\{\begin{array}{c}(20,5)(20,10)(20,15)(20,20)(20,25)(20,30)\\ (20,35)(20,40)(20,45)(20,50)(20,55)(20,60)\\ (20,65)(20,70)(20,75)(20,80)(20,85)(20,90)\\ (20,95)(20,100)\end{array}\right\}$

Since, there are 20 possible outcomes which are equally likely, the probability of choosing any of the outcome from the sample space is clearly$\frac{1}{20}=0.05$

The probability distribution of x is obtained below:

Let x be the sum of the two numbers in each sample.

There is one sample in the sample space with the sum of 25.

$$\begin{gathered} P\left(x=25\right)=\frac{1}{20} \\ =0.05 \end{gathered}$$

There is one sample in the sample space with the sum of 30.

$$\begin{gathered} P\left(x=30\right)=\frac{1}{20} \\ =0.05 \end{gathered}$$

There is one sample in the sample space with the sum of 35.

$$\begin{gathered} P\left(x=35\right)=\frac{1}{20} \\ =0.05 \end{gathered}$$

Assume the wheel to be fair. If the total of the player’s spin exceeds 100, then set total score to 0. Then the probability for x=0 is obtained if the sum of the player’s spin exceeds 100. That is,

$$\begin{gathered} P\left(x=0\right)=\frac{1}{20} \\ =0.20 \end{gathered}$$

Similarly for the remaining values is shown in the probability distribution table.

If the sum of the player’s spins exceeds 100, then set the probability of the score to 0.

Thus,

$$\begin{gathered} P\left(x=0\right)=\frac{4}{20} \\ =0.20 \end{gathered}$$

Hence, the probability that in two spins the player’s total score exceeds a dollar is 0.20.

j.

If the player spins the wheel twice, obtains 65 on first spin and decides to spin again. Then the total outcomes are as follows:

$N\left(S\right)=\left\{\begin{array}{c}(65,5)(65,10)(65,15)(65,25)(65,30)\\ (65,35)(65,40)(65,45)(65,50)(65,55)\\ (65,60)(65,65)(65,70)(65,75)(65,80)\\ (65,85)(65,90)(65,95)(65,100)\end{array}\right\}$

Since, there are 20 possible outcomes which are equally likely, the probability of choosing any of the outcome is clearly$\frac{1}{20}$.

Let x be the sum of the two numbers in each sample.

The probability distribution of x is obtained below:

There is one sample in the sample space with the sum of 70.

$$\begin{gathered} P\left(x=70\right)=\frac{1}{20} \\ =0.05 \end{gathered}$$

There is one sample in the sample space with the sum of 75.

$$\begin{gathered} P\left(x=75\right)=\frac{1}{20} \\ =0.05 \end{gathered}$$

There is one sample in the sample space with the sum of 80.

$$\begin{gathered} P\left(x=80\right)=\frac{1}{20} \\ =0.05 \end{gathered}$$

Similarly for the remaining values is shown in the probability table.

Also assume the wheel to be fair. Moreover if the total of the player’s spin exceeds 100, then set total score to 0.

Then the probability for x=0 is obtained if the sum of the player’s spin exceeds 100. Here, there are 13 samples that exceeds total score 100 from the sample space that is as follows:

$$\begin{gathered} P\left(x=0\right)=\frac{13}{20} \\ =0.65 \end{gathered}$$

k.

If the player spins the wheel twice, obtain 70 on first spin and decides to spin again. Then the total outcomes are 20 and the sample space is as follows:

$N\left(S\right)=\left\{\begin{array}{c}(70,5)\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}(70,10)\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}(70,15)\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}(70,20)\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}(70,25)\\ (70,45)\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}(70,50)\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}(70,55)\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}(70,60)\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}(70,65)\\ (70,70)\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}(70,75)\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}(70,80)\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}(70,85)\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}(70,90)\\ (70,95)\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}(70,100)\hspace{0.33em}\end{array}\right\}$

Since, there are 20 possible outcomes which are equally likely, the probability of choosing any of the outcome is clearly$\frac{1}{20}$.

Let x be the sum of the two numbers in each sample.

The probability distribution of x is obtained below:

There is one sample in the sample space with the sum of 75.

Consider,

$$\begin{gathered} P\left(x=75\right)=\frac{1}{20} \\ =0.05 \end{gathered}$$

There is one sample in the sample space with the sum of 80.

$$\begin{gathered} P\left(x=80\right)=\frac{1}{20} \\ =0.05 \end{gathered}$$

There is one sample in the sample space with the sum of 85.

$$\begin{gathered} P\left(x=85\right)=\frac{1}{20} \\ =0.05 \end{gathered}$$

Similarly for the remaining values is shown in the probability distribution table.

Also assume the wheel to be fair. Moreover if the total of the player’s spin exceeds 100, then set total score to 0.

Then the probability for x=0 is obtained if the sum of the player’ spin exceeds 100. Here, there are 14 samples that exceeds total score 100 from the sample space that is as follows:

$$\begin{gathered} P\left(x=0\right)=\frac{14}{20} \\ =0.70 \end{gathered}$$

If the player spins the wheel twice, obtains 75 on first spin and decides to spin again. Then the total outcomes are 20 and the sample space is as follows:

$N\left(S\right)=\left\{\begin{array}{c}(75,5)\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}(75,10)\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}(75,15)\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}(75,20)\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}(75,25)\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}(75,30)\\ (75,35)\hspace{0.33em}\hspace{0.33em}(75,40)\hspace{0.33em}\hspace{0.33em}(75,45)\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}(75,50)\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}(75,55)\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}(75,60)\\ (75,65)\hspace{0.33em}\hspace{0.33em}(75,70)\hspace{0.33em}\hspace{0.33em}(75,75)\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}(75,80)\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}(75,80)\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}(75,85)\\ (75,90)\hspace{0.33em}\hspace{0.33em}(75,95)\hspace{0.33em}\hspace{0.33em}(75,100)\end{array}\right\}$

Since, there are 20 possible outcomes which are equally likely, the probability of choosing any of the outcome is clearly$\frac{1}{20}$.

Let xbe the sum of the two numbers in each sample.

The probability distribution of x is obtained below:

There is one sample in the sample space with the sum of 80.

Consider,

$$\begin{gathered} P\left(x=80\right)=\frac{1}{20} \\ =0.05 \end{gathered}$$

There is one sample in the sample space with the sum of 85.

$$\begin{gathered} P\left(x=85\right)=\frac{1}{20} \\ =0.05 \end{gathered}$$

There is one sample in the sample space with the sum of 90.

$$\begin{gathered} P\left(x=90\right)=\frac{1}{20} \\ =0.05 \end{gathered}$$

Similarly for the remaining values are shown in the probability distribution table.

Also assume the wheel to be fair. Moreover if the total of the player’s spin exceeds 100, then set total score to 0.

Then the probability for x=0 is obtained if the sum of the player’ spin exceeds 100. Here, there are 15 samples that exceeds total score 100 from the sample space that is as follows:

$$\begin{gathered} P\left(x=0\right)=\frac{15}{20} \\ =0.75 \end{gathered}$$

Hence,

If the player spins the wheel twice and obtains 65 on first spin, then the probability if the players score exceeds dollar on the sum of the two spins is 0.65.

If the player spins the wheel twice and obtains 70 on first spin, then the probability if the players score exceeds dollar on the sum of the two spins is 0.70.

If the player spins the wheel twice and obtains 75 on first spin, then the probability if the players score exceeds dollar on the sum of the two spins is 0.75.

Hence, it is clear that as the score on the first spin increases, then the probability of the players score exceeds dollar also increases.