Tag Archives: Probability and Statistics

Curing the Compulsive Gambler: Challenging Probability Problem


Mr. Brown always bets a dollar on the number 13 at roulette against the advice of Kind Friend. To help cure Mr. Brown of playing roulette, Kind Friend always bets Brown $20 at even money that Brown will be behind at the end of 36 plays. How is the cure working?

(Most American roulette wheels have 38 equally likely numbers. If the player’s number comes up, he is paid 35 times his stake and gets his original stake back; otherwise, he loses his stake)


Solution:

If Mr. Brown wins once in 36 turns, he is even with the casino. His probability of losing all 36 times is \displaystyle \left( \frac{37}{38} \right)^{36} \approx 0.383 . In a single turn, his expectation is

35\left( \frac{1}{38} \right)-1\left( \frac{37}{38} \right) = - \frac{2}{38}\ \text{dollars}

and in 36 turns

-\frac{2}{38}\left( 36 \right) = -1.89 \ \text{dollars}

Against Kind Friend, Mr. Brown has an expectation of

+20\left( 0.617 \right)-20\left( 0.383 \right)\approx +4.68 \ \text{dollars}

And so, all told, Mr. Brown gains +4.68 – 1.89 = + 2.79 dollars per 36 trials; he is finally making money at roulette. Possibly Kind Friend will be cured first. Of course, when Brown loses all 36, he is out $56, which may jolt him a bit.


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Chuck-a-Luck: Challenging Probability Problem


Chuck-a-Luck is a gambling game often played at carnivals and gambling houses. A player may bet on anyone of the numbers 1, 2, 3, 4, 5, 6. Three dice are rolled. If the player’s number appears on one, two, or three of the dice, he receives respectively one, two, or three times his original stake plus his own money back; otherwise, he loses his stake. What is the player’s expected loss per unit stake? (Actually, the player may distribute stakes on several numbers, but each such stake can be regarded as a separate bet.)


Solution:

Let us compute the losses incurred (a) when the numbers on the three dice are different, (b) when exactly two are alike, and (c) when all three are alike. An easy attack is to suppose that you place a unit stake on each of the six numbers, thus betting six units in all. Suppose the roll produces three different numbers, say 1, 2, 3. Then the house takes the three unit stakes on the losing numbers 4, 5, 6 and pays off the three winning numbers 1, 2, 3. The house won nothing, and you won nothing. That result would be the same for any roll of three different numbers.

Next suppose the roll of the dice results in two of one number and one of a second, say 1, 1, 2. Then the house can use the stakes on numbers 3 and 4 to payoff the stake on number 1, and the stake on number 5 to payoff that on number 2. This leaves the stake on number 6 for the house. The house won one unit, you lost one unit, or per unit stake you lost 1/6.

Suppose the three dice roll the same number, for example, 1, 1, 1. Then the house can pay the triple odds from the stakes placed on 2, 3, 4 leaving those on 5 and 6 as house winnings. The loss per unit stake then is 2/6. Note that when a roll produces a multiple payoff the players are losing the most on the average.

To find the expected loss per unit stake in the whole game, we need to weight the three kinds of outcomes by their probabilities. If we regard the three dice as distinguishable –say red, green, and blue — there are 6 \times 6 \times 6= 216 ways for them to fall.

In how many ways do we get three different numbers? If we take them in order, 6 possibilities for the red, then for each of these, 5 for the green since it must not match the red, and for each red-green pair, 4 ways for the blue since it must not match either of the others, we get 6 \times 5 \times 4 = 120 ways.

For a moment skip the case where exactly two dice are alike and go on to three alike. There are just 6 ways because there are 6 ways for the red to fall and only 1 way for each of the others since they must match the red.

This means that there are 216 - 126 = 90 ways for them to fall two alike and one different. Let us check that directly. There are three main patterns that give two alike: red-green alike, red-blue alike, or green-blue alike. Count the number of ways for one of these, say red-green alike, and then multiply by three. The red can be thrown 6 ways, then the green 1 way to match, and the blue 5 ways to fail to match, or 30 ways. All told then we have 3 \times 30 = 90 ways, checking the result we got by subtraction.

We get the expected loss by weighting each loss by its probability and summing as follows:

\underbrace{\frac{120}{216}\times 0}_\text{none alike} + \underbrace{\frac{90}{216}\times \frac{1}{6}}_\text{2 alike}+\underbrace{\frac{6}{216}\times \frac{2}{6}}_\text{3 alike} = \frac{17}{216} \approx 0.079

Thus, you lose about 8% per play. Considering that a play might take half a minute and that government bonds pay you less than 4% interest for a year, the attrition can be regarded as fierce.

This calculation is for regular dice. Sometimes a spinning wheel with a pointer is used with sets of three numbers painted in segments around the edge of the wheel. The sets do not correspond perfectly to the frequencies given by the dice. In such wheels I have observed that the multiple payoffs are more frequent than for the dice, and therefore the expected loss to the bettor greater.


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Coin in Square: Challenging Probability Problem


In a common carnival game, a player tosses a penny from a distance of about 5 feet onto the surface of a table ruled in 1-inch squares. If the penny (3/4 inch in diameter) falls entirely inside a square, the player receives 5 cents but does not get his penny back; otherwise, he loses his penny. If the penny lands on the table, what is his chance to win?


Solution:

When we toss the coin onto the table, some positions for the center of the coin are more likely than others, but over a very small square we can regard the probability distribution as uniform. This means that the proba­bility that the center falls into any
region of a square is proportional to the area of the region, indeed, is the area of the region divided by the area of the square. Since the coin is 3/8 inch in radius, its center must not land within 3/8 inch of any edge if the player is to win. This restriction generates a square of side 1/4 inch within which the center of the coin must lie for the coin to be in the square. Since the proba­bilities are proportional to areas, the probability of winning is \displaystyle \left( \frac{1}{4} \right)^2 = \frac{1}{16}. Of course, since there is a chance that the coin falls off the table altogether, the total probability of winning is smaller still. Also, the squares can be made smaller by merely thickening the lines. If the lines are 1/16 inch wide, the winning central area reduces the probability to \displaystyle \left( \frac{3}{16} \right)^{2} = \frac{9}{256} or less than \displaystyle \frac{1}{28}.


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Trials until First Success: Challenging Probability Problem


On the average, how many times must a die be thrown until one gets a 6?


Solution:

Let p be the probability of a 6 on a given trial. Then the probabilities of success for the first time on each trial are (let q = 1 - p):

TrialProbability of success on trial
1 p
2 pq
3 pq ^2
..
..
..

The sum of the probabilities is

\begin{align*}
p+pq+pq^2+\ldots & = p\left( 1+q+q^2+\ldots \right) \\ \\
 & = \frac{p}{1-q} \\ \\
 & = \frac{p}{p} \\ \\
 & = 1
\end{align*}

The mean number of trials, m, is by definition,

m = p + 2pq + 3pq^2 + 4pq^3+ \ldots

Note that our usual trick for summing a geometric series works:

qm = pq + 2pq^2+3pq^3 + \ldots

Subtracting the second expression from the first gives

m-qm=p+pq+pq^2+\ldots

or

m\left( 1-q \right) = 1

Consequently,

mp=1

and

m=1/p

We see that p=1/6, and so m=6.

On the average, a die must be thrown 6 times until one gets a 6.


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Probability & Statistics for Engineers & Scientists Ninth Edition by Walpole Exercise 1.2


According to the journal Chemical Engineering, an important property of a fiber is its water absorbency. A random sample of 20 pieces of cotton fiber was taken and the absorbency on each piece was measured. The following are the absorbency values:

18.7121.4120.7221.8119.2922.4320.17
23.7119.4420.5018.9220.3323.0022.85
19.2521.7722.1119.7718.0421.12

(a) Calculate the sample mean and median for the above sample values.
(b) Compute the 10% trimmed mean.
(c) Do a dot plot of the absorbency data.
(d) Using only the values of the mean, median, and trimmed mean, do you have evidence of outliers in the data?


Solution:

Part A. The sample mean is computed as follows:

\begin{align*}
\bar x & = \frac{\Sigma x_{i}}{n} \\
\bar x & = \frac{18.71+21.41+20.72+\cdots +21.12}{20} \\
\bar x & = 20.77 \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)
\end{align*}

We can solve for the sample median by arranging the data in increasing order first.

18.04, \
18.71, \
18.92, \
19.25, \
19.29, \
19.44, \
19.77, \
20.17, \
20.33, \
20.50 \\
20.72, \
21.12, \
21.41, \
21.77, \
21.81, \
22.11, \
22.43, \
22.85, \
23.00, \
23.71

Since there are 20 measurements (even), the middle measurements are the (20/2) 10th and the (20/2 + 1) 11th measurement. The 10th measurement is 20.50 and the 11th measurement is 20.72. The median is the average of these two measurements.

\begin{align*}
\tilde x & = \frac{20.50+20.72}{2} \\
\tilde x & = 20.61 \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)
\end{align*}

Part B. The 10% trimmed mean is calculated by removing 10% of the lowest data and 10% of the highest data. That is, removing the 2 lowest and 2 highest data. We are left with the following:

18.92, \
19.25, \
19.29, \
19.44, \
19.77, \
20.17, \
20.33, \
20.50 \\
20.72, \
21.12, \
21.41, \
21.77, \
21.81, \
22.11, \
22.43, \
22.85

The 10% trimmed mean, \bar x _{tr10} is

\begin{align*}
\bar x _{tr10} & = \frac{\Sigma x_{i}}{n} \\
\bar x _{tr10} & = \frac{18.92+19.25+\cdots+22.85}{16} \\
\bar x _{tr10} & = 20.74 \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)
\end{align*}

Part C. The dot plot is shown

dot plot for Probability & Statistics for Engineers & Scientists Ninth Edition by Walpole Exercise 1.2

Part D. Since the values of the mean, median, and trimmed mean are not actually far from each other, we can conclude that there are no outliers in the given measurements.


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Probability & Statistics for Engineers & Scientists Ninth Edition by Walpole Exercise 1.1


The following measurements were recorded for the drying time, in hours, of a certain brand of latex paint.

3.42.54.82.93.6
2.83.35.63.72.8
4.44.05.23.04.8

Assume that the measurements are a simple random sample.
(a) What is the sample size for the above sample?
(b) Calculate the sample mean for these data.
(c) Calculate the sample median.
(d) Plot the data by way of a dot plot.
(e) Compute the 20% trimmed mean for the above data set.
(f) Is the sample mean for these data more or less descriptive as a center of location than the trimmed mean?


Solution:

Part A. Sample size, n is the total number of measurements.

n=15 \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)

Part B. The sample mean, \bar x is computed as follows:

\begin{align*}
\bar x & = \sum_{i=1}^{n}\frac{x_{i}}{n} \\
& = \frac{3.4+2.5+4.8+2.9+3.6+2.8+3.3+5.6+3.7+2.8+4.4+4.0+5.2+3.0+4.8}{15} \\
& = \frac{56.8}{15} \\
& = 3.79 \ \text{hours} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)
\end{align*}

Part C. The sample median, \tilde x is the number at the middle of the arranged measurements in increasing magnitude. There are 15 measurements, n=15. If we arranged the data in increasing magnitude, the median is the measurement in the middle.

2.5, \ 2.8, \ 2.8, \ 2.9, \ 3.0, \ 3.3, \ 3.4, \ \underset{\color{Blue} \text{middle number}}{3.6}, \ 3.7, \ 4.0, \ 4.4, \ 4.8, \ 4.8, \ 5.2, \ 5.6

The middle number is 3.6. That is

\tilde x= 3.6 \ \text{hours} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)

Part D. This dot plot has been created using the Statistical Software Rguroo

a dot plot for the data: 3.4, 2.5, 4.8, 2.9, 3.6, 2.8, 3.3, 5.6, 3.7, 2.8, 4.4, 4.0, 5.2, 3.0, 4.8. this dot plot was made possible through Rgurro at https://www.rguroo.com/

Part E. The 20% trimmed mean means the average of the measurements left after removing 20% highest and 20% lowest data. This means we remove the 3 highest and 3 lowest numbers. Therefore, the data becomes

\ 2.9, \ 3.0, \ 3.3, \ 3.4, \ 3.6,  \ 3.7, \ 4.0, \ 4.4, \ 4.8,

The 20% trimmed mean, \bar x _{tr20} is

\begin{align*}
\bar x _{tr20} & = \frac{2.9+3.0+ \cdots+4.8}{9} \\
\bar x _{tr20} & = \frac{33.1}{9} \\
\bar x _{tr20} & = 3.678 \ \text{hours} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)
\end{align*}

Part F. The sample mean for these data is \bar x = 3.79 \ \text{hours} while the 20% trimmed mean is \bar x _{tr20} = 3.678 \ \text{hours}. Seems like the two means are not really far from each other, but because of the elimination of the extreme values, we can treat the trimmed mean as a better descriptive mean.


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Probability & Statistics for Scientists & Engineers Ninth Edition by Walpole, Myers, Myers, and Ye

Chapter 2: Probability

Chapter 3: Random Variables and Probability Distributions

Chapter 4: Mathematical Expectation

Chapter 5: Some Discrete Probability Distribution

Chapter 6: Some Continuous Probability Distribution

Chapter 7: Functions of Random Variables

Chapter 8: Fundamental Sampling Distributions and Data Descriptions

Chapter 9: One- and Two-Sample Estimation Problems

Chapter 10: One- and Two-Sample Tests of Hypotheses

Chapter 11: Simple Linear Regression and Correlation

Chapter 12: Multiple Linear Regression and Certain Nonlinear Regression Models

Chapter 13: One-Factor Experiments: General

Chapter 14: Factorial Experiments (Two or More Factors)

Chapter 15: 2k Factorial Experiments and Fractions

Chapter 16: Nonparametric Statistics

Chapter 17: Statistical Quality Control

Chapter 18: Bayesian Statistics


Textbook Solutions for Probability and Statistics



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Successive Wins: Challenging Problem in Probability


To encourage Elmer’s promising tennis career, his father offers him a prize if he wins (at least) two tennis sets in a row in a three-set series to be played with his father and the club champion alternately: father-champion-father or champion-father-champion, according to Elmer’s choice. The champion is a better player than Elmer’s father. Which series should Elmer choose?


Solution:

Since the champion plays better than the father, it seems reasonable that fewer sets should be played with the champion. On the other hand, the middle set is the key one, because Elmer cannot have two wins in a row without winning the middle one. Let C stand for the champion, F for father, and W and L for win and loss by Elmer. Let f be the probability of Elmer’s winning any set from his father, c the corresponding probability of winning from the champion. The table shows only possible prize-winning sequences together with their probabilities, given independence between sets, for the two choices.

Set with:

Father First

FCFProbability
WWWfcf
WWLfc(1-f)
LWW(1-f)cf
Totalfc(2-f)

Champion First

CFCProbability
WWWcfc
WWLcf(1-c)
LWW(1-c)fc
Totalfc(2-c)

Since Elmer is more likely to best his father than to beat the champion, f is larger than c, and 2-f is smaller than 2-c, and so Elmer should choose CFC. For example, for f=0.8, c=0.4, the chance of winning the prize with FCF is 0.384, that for CFC is 0.512. Thus, the importance of winning the middle game outweighs the disadvantage of playing the champion twice.

Many of us have a tendency to suppose that the higher the expected number of successes, the higher the probability of winning a prize, and often this supposition is useful. But occasionally a problem has special conditions that destroy this reasoning by analogy. In our problem, the expected number of wins under CFC is 2c+f, which is less than the expected number of wins for FCF, 2f+c. In our example with f=0.8 and c=0.4, these means are 1.6 and 2.0 in that order. This opposition of answers gives the problem flavor.


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