Το καλάθι σου είναι άδειο.
Expected value (higher level courseware)
Think about it: why is it worth operating gambling slot machines or casinos? How are the respective prices for the odds determined in betting, or how do insurance companies calculate the insurance fees? It is obvious that the smaller chance, for example, a horse has winning a horse race, the greater the prize will be if it wins indeed. The more risky the thing is we are contracting an insurance for, the more we have to pay for it.
Example 1
Cathy and Paul play with a regular die. Cathy likes “curved” numbers, so that she will win if the roll is \latex{ 2;\, 3;\, 5 } or \latex{ 6; } while Paul likes the “straight” ones, and thus he is the winner when the roll is \latex{ 1 } or \latex{ 4 }. If Cathy wins, Paul “pays” her \latex{ 3 } chips, if he wins, then she “pays” him \latex{ 4 } chips. Who has an advantage in this game?
Solution
Out of the \latex{ 6 } numbers possible to be rolled, Cathy has four winning numbers, while Paul has two. Therefore if they play many times, Cathy will win in \latex{\frac{4}{2}=\frac{2}{3}} times of the rounds, while Paul will win in \latex{\frac{1}{3}} time of the rounds, since relative frequency approximates probability well for large number of rolls. This means that for n rolls, Cathy will win approx.
\latex{\frac{2}{3}\times n} times and Paul will win approx. \latex{\frac{1}{3}\times n} times, therefore Cathy gains \latex{\frac{2}{3}\times n\times 3-\frac{1}{3}\times n\times 4=\frac{2}{3}\times n} chips from Paul, thus she has an advantage in the game. We can also say that in case of many rounds of gaming, Cathy will win an average of \latex{\frac{2}{3}} chips per game.
\latex{\frac{2}{3}\times n} times and Paul will win approx. \latex{\frac{1}{3}\times n} times, therefore Cathy gains \latex{\frac{2}{3}\times n\times 3-\frac{1}{3}\times n\times 4=\frac{2}{3}\times n} chips from Paul, thus she has an advantage in the game. We can also say that in case of many rounds of gaming, Cathy will win an average of \latex{\frac{2}{3}} chips per game.
The value around which the average prize fluctuates is called the expected value of that prize. (The exact mathematical definition of the expected value requires some other concepts which we cannot discuss here.)
The expected value of Cathy's prize is found if the possible prize values are multiplied with the probability of her winning them, and these products are summed up. Therefore the expected value of Cathy's prize is
\latex{3\times \frac{2}{3}+(-4)\times \frac{1}{3}=\frac{2}{3},}
and using the same method, the expected value of Paul's prize is:
\latex{4\times \frac{1}{3}+(-3)\times \frac{2}{3}=\frac{2}{3},}
which is negative, meaning an actual loss. Therefore despite that Paul receives more chips when he wins, the game favours Cathy as she has better chances of winning.
Example 2
In roulette, the player bets on one number out of \latex{ 37 } between \latex{ 0 } and \latex{ 36 }. If the player guesses the correct number, they will receive \latex{ 36 } times the bet. In the long run, will the player win or lose?
Solution
Let us calculate supposing a \latex{ 1 } euro bet. We have a probability of \latex{\frac{1}{37}} for guessing the right number, in which case we receive \latex{ 36 } euros while our \latex{ 1 } euro deposit is given to the bank, thus we win \latex{ 35 } euros.
There is a \latex{\frac{36}{37}} probability for not guessing the right number, in which case we lose our \latex{ 1 } euro bet, so our prize is \latex{ –1 } euro. Therefore the expected value of our prize is
There is a \latex{\frac{36}{37}} probability for not guessing the right number, in which case we lose our \latex{ 1 } euro bet, so our prize is \latex{ –1 } euro. Therefore the expected value of our prize is
\latex{35\times \frac{1}{37}+(-1)\times \frac{36}{37}=-\frac{1}{37}=-0.027,}
which means that playing in a long run, in average we lose \latex{ 2.7 }% of the bet, so it is not worth playing roulette.
Example 3
Albert and Bill play with a regular die. After each roll, Albert pays Bill an amount of chips according the rolled number, while Bill pays \latex{ 3 } chips to Albert after each roll. Who has an advantage in this game?
Solution
Calculate what the possible values of Albert's prize are and the corresponding probabilities.
If 1 is rolled: the prize is \latex{3-1=2} chips, with a \latex{\frac{1}{6}} probability;
if 2 is rolled: the prize is \latex{3-1=2} chips, with a \latex{\frac{1}{6}} probability;
if 3 is rolled: the prize is \latex{3-3=0} chips, with a \latex{\frac{1}{6}} probability;
if 4 is rolled: the prize is \latex{3-4=-1} chips, with a \latex{\frac{1}{6}} probability;
if 5 is rolled: the prize is \latex{3-5=-2} chips, with a \latex{\frac{1}{6}} probability;
if 6 is rolled: the prize is \latex{3-6=-3} chips, with a \latex{\frac{1}{6}} probability;
If 1 is rolled: the prize is \latex{3-1=2} chips, with a \latex{\frac{1}{6}} probability;
if 2 is rolled: the prize is \latex{3-1=2} chips, with a \latex{\frac{1}{6}} probability;
if 3 is rolled: the prize is \latex{3-3=0} chips, with a \latex{\frac{1}{6}} probability;
if 4 is rolled: the prize is \latex{3-4=-1} chips, with a \latex{\frac{1}{6}} probability;
if 5 is rolled: the prize is \latex{3-5=-2} chips, with a \latex{\frac{1}{6}} probability;
if 6 is rolled: the prize is \latex{3-6=-3} chips, with a \latex{\frac{1}{6}} probability;
Albert's expected gain is
\latex{2\times \frac{1}{6}+1\times \frac{1}{6}+0\times \frac{1}{6}+(-1)\times \frac{1}{6}+(-2)\times \frac{1}{6}+(-3)\times \frac{1}{6}=-\frac{1}{2},}
which means the game favours Bill.
Calculate how much should Bill “pay” to create a fair game.
Expected value of the chips “paid” by Albert upon each roll is:
\latex{1\times \frac{1}{6}+2\times \frac{1}{6}+3\times \frac{1}{6}+4\times \frac{1}{6}+5\times \frac{1}{6}+6\times \frac{1}{6}=3.5.}
This means that if Bill gives \latex{ 3.5 } chips to Albert upon each roll, the game is fair and the expected value of the prize will be \latex{ 0 } for both of them.
Note that \latex{ 3.5 } is the expected value of rolling die, which shows that the expected value is not the most probable value, but the result of the average of a large number of tries.
Note that \latex{ 3.5 } is the expected value of rolling die, which shows that the expected value is not the most probable value, but the result of the average of a large number of tries.
Example 4
Before spinning the spinning top presented in Figure 12, bets can be made on the color of the region the arrow will point at, for \latex{ 5 } euros. The prize is won if the arrow points at the color for which the bet was made. If we won by betting red, we gain \latex{ 9 } euros; winning by betting blue worth \latex{ 30 } euros and green comes with a prize of \latex{ 16.5 } euros.
- Which color is worth betting for most?
- We made a bet for blue in ten subsequent game turns. What is the probability that
- we win at least \latex{ 200 } euros?
- we neither win nor lose?
- we lose?
Figure 12
Solution (a)
If we bet red, we have \latex{\frac{6}{12}} probability of winning in which case we gain \latex{ 9 } euros. Since our bet was \latex{ 5 } euros, we earn \latex{9-5= 4} euros. Similarly, we have \latex{\frac{6}{12}} probability for not spinning red, in this case we “earn” \latex{–5} euros.
Therefore the expected value of our prize is
Therefore the expected value of our prize is
\latex{4\times \frac{6}{12}+(-5)\times \frac{6}{12}=-0.5} euros.
If we make a bet on blue, we have \latex{\frac{2}{12}} probability to win, while our prize will be \latex{ 25 } euros.
Similarly, we have \latex{\frac{10}{12}} probability for not winning, in this case we “earn” \latex{–5 } euros.
When betting blue, the expected value of our prize is
When betting blue, the expected value of our prize is
\latex{25\times \frac{2}{12}+(-5)\times \frac{10}{12}=0.}
If we make a bet on green, we have \latex{\frac{4}{12}} probability to win, while our prize will be \latex{11.5} euros.
We have \latex{\frac{8}{12}} probability for not winning, in this case we “earn” \latex{–5} euros.
When betting green, the expected value of our prize is:
We have \latex{\frac{8}{12}} probability for not winning, in this case we “earn” \latex{–5} euros.
When betting green, the expected value of our prize is:
\latex{11.5\times \frac{4}{12}+(-5)\times \frac{8}{12}=0.5} euros.
Therefore the expected value of our prize is the largest when we bet on green.
Solution (b/i)
If we bet on blue, we either win \latex{ 25 } euros or lose \latex{ 5 } in one game. In order to win \latex{ 200 } euros, we have to win at least \latex{ 8 } times.
If we lose \latex{ 2 } times besides \latex{ 8 } wins, we only have \latex{ 190 } euros, so we can win at least \latex{ 200 } euros only if we win all \latex{ 10 } rounds or if we win \latex{ 9 } times and lose only once.
If we made a bet on blue, our probability of winning a round is \latex{\frac{2}{12}=\frac{1}{6}} and the probability of winning \latex{ 10 } rounds is \latex{\left(\frac{1}{6} \right)^{10}.}
If we lose \latex{ 2 } times besides \latex{ 8 } wins, we only have \latex{ 190 } euros, so we can win at least \latex{ 200 } euros only if we win all \latex{ 10 } rounds or if we win \latex{ 9 } times and lose only once.
If we made a bet on blue, our probability of winning a round is \latex{\frac{2}{12}=\frac{1}{6}} and the probability of winning \latex{ 10 } rounds is \latex{\left(\frac{1}{6} \right)^{10}.}
The chances of winning \latex{ 9 } times and losing once is \latex{\left(\begin{matrix} 10 \\ 1 \end{matrix} \right)\times \left(\frac{1}{6} \right)^{9}\times \left(\frac{5}{6} \right)^{1},} as there are \latex{\left(\begin{matrix} 10 \\ 1 \end{matrix} \right)=10} ways of choosing the losing round out of the \latex{ 10 } rounds played.
Therefore, our chances of winning at least \latex{ 200 } euros are
Therefore, our chances of winning at least \latex{ 200 } euros are
\latex{\left(\frac{1}{6} \right)^{10}+10\times \left(\frac{1}{6} \right)^{9}\times \frac{5}{6}=0.000000843.}
Solution (b/ii)
One win (\latex{ 25 } euros) is counterbalanced by \latex{ 5 } losses (\latex{ 5 } euros each). If there is \latex{ 1 } winning round within the \latex{ 10 } rounds, our losses will be larger than our earnings. If there are \latex{ 2 } wins, then the \latex{ 8 } losing rounds are not enough to lose our \latex{ 50 } euros prize, so our earnings will be larger; therefore, it is impossible to win exactly as much as we lose.
Solution (b/iii)
The above shows us that if we win only \latex{ 2 } rounds out of \latex{ 10 }, we have already won more than we have lost. Therefore we only lose money overall if we lose all \latex{ 10 } rounds or if we win only \latex{ 1 } and lose all the others. The probability of this event is
\latex{\left(\frac{5}{6} \right)^{10}+10\times \left(\frac{5}{6} \right)^{9}\times \frac{1}{6}=0.32.}
Example 5
Betty’s locker contain scarves: \latex{ 3 } are blue, \latex{ 1 } is yellow, \latex{ 1 } is grey and \latex{ 2 } are green. She is looking for a blue one, but in her hurry she just pulls out scarves from the locker at random until she finds a blue one. If the scarf drawn is not blue, she tosses it away. What is the expected value of the necessary tries?
Solution
It might happen that she finds a blue scarf upon first try; the probability of this is \latex{\frac{3}{7}.}
If the first scarf is not blue, the second one can still be blue. In this case \latex{ 2 } tries are needed, the probability of this is
If the first scarf is not blue, the second one can still be blue. In this case \latex{ 2 } tries are needed, the probability of this is
\latex{\frac{4}{7}\times \frac{3}{6}= \frac{2}{7}.}
If the second scarf is not blue either, the third one can still be blue. In this case \latex{ 3 } tries are needed, the probability of this is
\latex{\frac{4}{7}\times \frac{3}{3}\times\frac{3}{5} = \frac{6}{35}.}
If the third scarf is not blue either, the fourth one can still be blue. In this case \latex{ 4 } tries are needed, the probability of this is
\latex{\frac{4}{7}\times \frac{3}{6}\times\frac{2}{5}\times \frac{3}{4} = \frac{3}{35}.}
If even the fourth scarf is not blue, then the fifth one must be blue. In this case \latex{ 5 } tries are needed, the probability of this is
\latex{\frac{4}{7}\times \frac{3}{6}\times\frac{2}{5}\times \frac{1}{4} \times \frac{3}{3} = \frac{1}{35}.}
Therefore the expected value of the necessary tries is
\latex{1\times \frac{3}{7}+2\times \frac{2}{7}+3\times \frac{6}{35}+4\times \frac{3}{35}+5\times \frac{1}{35}=\frac{15+20+18+12+5}{35}=2.}
Notice that the probability for finding a blue scarf upon the first try is larger than those with all other number of tries. Despite this, if Betty regularly searches for a blue scarf this way, she will find a blue one after \latex{ 2 } tries on average.
Exercises
{{exercise_number}}. A horse derby offers betting for the victory of one of three horses. If Tornado wins, the bookmaker pays \latex{ 1.5 } times the amount of the bet deposited; for the victory of Lightning, the prize is ten times the deposit, and for that of Whirlwind, the prize is three times. According to secret sources, Tornado has \latex{ 60 }% chances of winning, while the chances of the victory of Lightning and Whirlwind are, respectively, \latex{ 10 }% and \latex{ 30 }%. Which horse is the best bet, and what is the expected value of the prize?
{{exercise_number}}. An insurance company offers a special insurance for a five days ski holiday, which can be bought for \latex{ 35 } euros. The insurance company reimburses \latex{ 15,000 } euros in case of death and, similarly, \latex{ 15,000 } euros in case of a permanent health damage. In the event of an injury, an average of \latex{ 2 } thousand euros is reimbursed for medical care, and \latex{ 1 } thousand euros are paid in case of damaged luggage. According to the insurance company's experts, probability of death is \latex{ 0.005 }%, that of permanent health damage is \latex{ 0.01 }%, that of the need for medical help is \latex{ 1 }% and the probability of damages to the luggage is \latex{ 0.5 }%. What is the expected value of the insurance company's profit on an insurance?
{{exercise_number}}. A ball is rolling down on the track of the arcade game machine presented on the figure. The ball has equal chances to go left or right at every obstacle, so all routes are equally possible. There are obstacles on \latex{ 4 } levels of the track, and in the end, the ball can end up in one of \latex{ 5 } slots. One must pay \latex{ 2 } euros for the launch of each ball, and we have written to each slot how much the player wins if the ball ends up there. What is the expected value of the player's profit?
\latex{ 6 }
\latex{ 1.5 }
\latex{ 1 }
\latex{ 1.5 }
\latex{ 6 }
{{exercise_number}}. What is the expected number of correct guesses on the lottery (where \latex{ 5 } numbers are picked out of \latex{ 90 })?
What do you experience if you conduct the experiment \latex{60} times or \latex{90} times?
{{exercise_number}}. Cards with the numbers \latex{ 1;\, 2;\, 3;\, 4;\, 5;\, 6;\, 7;\, 8;\, 9;\, 10 } are put in a hat. One is drawn at random. We can bet \latex{ 1 } euro that the number drawn is even, divisible by \latex{ 3 } or divisible by \latex{ 5 }. If we guessed right that it is even, we win \latex{ 1.8 } euros. If we win by guessing divisibility by \latex{ 3 }, we win \latex{ 4 } euros and by guessing right that the number is divisible by \latex{ 5 }, we gain \latex{ 5 } euros. What is worth betting?
{{exercise_number}}. A gaming machine has three panels, all of which has \latex{ 5 } stances. The panels are spun and all of them end up randomly in one of the \latex{ 5 } stances. Two stances of the first panel display an orange, \latex{ 2 } display an apple and one displays a pear. The second panel displays one orange, one apple and three pears, the third one displays two oranges, two apples and one pear. Playing a round costs \latex{ 1 } euro and the player wins if the same fruit is displayed on all three panels. Determine the expected gain if the prize is \latex{ 10 } euros.
{{exercise_number}}. Anne and Blanche roll two regular dice. Anne bets that the product of the two rolls will be even while Blanche bets on it being odd. If Anne loses, she pays \latex{ 12 } euros to Blanche. How much should Blanche pay to Anne when Anne wins so that the game is fair?