Suppose Y is a random variable with probability distribution function f(y). The mathematical expectation, or expected value, E(Y) is defined as: E n ( Y) y f( ) µ i i y i The sum is taken over all values of the random variable. Toss a balanced coin once. Let Y denote the number of heads. Find the expected value of Y. 0 ½ 0 ½ ½ E(Y) ½ Toss a balanced coin twice. Let Y denote the number of heads. Find the expected value of Y. 0 ¼ 0 ½ ½ ¼ ½ E(Y)
The variance of the probability mass function f(y) is σ y µ f y ( ) ( ) The standard deviation is ( y µ ) f( ) σ y An equivalent formula for the variance σ y f y µ ( ) y-µ (y-µ) (y-µ) f(y) 0 0.5 0 -.5.5 0.85 0.375 0.375-0.5 0.5 0.09375 0.375 0.750 +0.5 0.5 0.09375 3 0.5 0.375 +.5.5 0.85 µ.5 σ 0.75 σ 0.866054
Suppose the random variable Y represents the monetary profit or loss from an economic decision. Then E(Y) is called the expected monetary value. Repeatedly toss a die until the number six comes up. Let Y denote the number of tosses; the player wins Y dollars. Find the expected monetary value of the game. 5 f ( y) 6 6 y,,... y 0.667 0.667 0.0043 0.093 4 0.000 0.0046 0.389 0.778 0.0036 0.0797 4 0.000 0.0040 3 0.57 0.347 3 0.0030 0.0694 43 0.000 0.0034 4 0.0965 0.3858 4 0.005 0.0604 44 0.000 0.009 5 0.0804 0.409 5 0.00 0.054 45 0.000 0.005 6 0.0670 0.409 6 0.007 0.0454 46 0.0000 0.00 7 0.0558 0.3907 7 0.005 0.0393 47 0.0000 0.008 8 0.0465 0.37 8 0.00 0.0340 48 0.0000 0.005 9 0.0388 0.3489 9 0.000 0.093 49 0.0000 0.003 0 0.033 0.330 30 0.0008 0.053 50 0.0000 0.00 0.069 0.96 3 0.0007 0.08 5 0.0000 0.0009 0.04 0.69 3 0.0006 0.087 5 0.0000 0.0008 3 0.087 0.430 33 0.0005 0.06 53 0.0000 0.0007 4 0.056 0.8 34 0.0004 0.038 54 0.0000 0.0006 5 0.030 0.947 35 0.0003 0.09 55 0.0000 0.0005 6 0.008 0.73 36 0.0003 0.00 56 0.0000 0.0004 7 0.0090 0.53 37 0.000 0.0087 57 0.0000 0.0003 8 0.0075 0.35 38 0.000 0.0074 58 0.0000 0.0003 9 0.0063 0.89 39 0.000 0.0064 59 0.0000 0.0003 0 0.005 0.043 40 0.000 0.0054 60 0.0000 0.000 E(Y) 5.9988
y 0 y ( p) y y p p y ( p) E ( Y ) yp( p) y p p p y For p /6, we have Expected Monetary Value E(Y) /(/6) 6 dollars. With interest rate r per period, the Present Value of an infinite number of periodic cash payments of one dollar is /r dollars. y 0 ( r) y r With the above formula, we see the present value of an infinite number of periodic cash payments of one dollar when the interest rate is /6 per period is 6 dollars. Therefore, this is monetarily equivalent to playing a game that can terminate with probability of p r, and that pays one dollar for each round played.
! " A small lumber company is considering building a new sawmill. It is faced with uncertain future economic conditions and it has three alternatives: build a large facility, build a small facility or do nothing. However, the company knows how much money can be made depending on what the economic conditions may be. The various payoffs are displayed in the following table. What should the company do? Alternatives The Future Favorable Market Unfavorable Market Probability 0.5 0.5 Payoffs EMV Large Facility 00,000-80,000 0,000 Small Facility 00,000-0,000 40,000 Do Nothing 0 0 0 The company has no real information about the future economy so it assigns a subjective probability 0.5 to each of two possibilities. Using the probability distribution function for the payoffs for each of the alternative decisions, it determined the expected monetary value (EMV) of each set of payoffs. It turns out that building a small facility yields the largest EMV of 40,000 dollars.! "#$ A gambling game is said to be a fair or equitable game if the expected amount won or lost is equal to zero. # Take ten fair coins; repeatedly toss them until all tails come up. Let Y denote the number of tosses and get back Y dollars. What wager amount would make this a fair game? First must find the probability that a game will terminate. In this case, this is the probability of getting all tails. By the special multiplication law of probability, the probability, p, that the game will terminate in a given round is P(Tail, Tail, Tail, Tail, Tail, Tail, Tail, Tail, Tail, Tail) P(Tail) P(Tail) P(Tail) P(Tail) P(Tail) P(Tail) P(Tail) P(Tail) P(Tail) P(Tail) (/)(/)(/)(/)(/) (/)(/)(/)(/)(/)/04
Thus p /04, and E(Y) /p 04. Therefore, the fair wager amount should be,04 dollars. $ % Bet 7 dollars and toss a pair of dice. Get back the amount of dollars equal to the sum of numbers on the two dice. Does the wager amount of 7 dollars make this a fair game? Yes. Let Y denote the amount won or lost after playing the game once. The expected monetary value is E(Y) 0. -5 /36 (-5 X ) / 36-5/36-4 /36 (-4 X ) / 36-8/36-3 3/36 (-3 X 3) / 36-9/36-4/36 (- X 4) / 36-8/36-5/36 (- X 5) / 36-5/36 0 6/36 (0 X 6) / 36 0 5/36 ( X 5) / 36 5/36 4/36 ( X 4) / 36 8/36 3 3/36 (3 X 3) / 36 9/36 4 /36 (4 X ) / 36 8/36 5 /36 (5 X ) / 36 5/36 E(Y) 0 %& A gambling game pays W to one and the probability of winning is p. What should W be to make this a fair game? As stated, a gambler bets one dollar in order to win W dollars with probability p. Let Y denote the amount won or lost. The probability distribution function and expected value of Y is: W p Wp - -p -(-p) E(Y) Wp - (-p) For a fair game, the expected monetary value must be zero: Wp - (-p) 0 W (-p)/p
That is, W-to-one are the odds of loosing. In addition, one-to-w are the odds of winning. & '( Suppose a horse pays 0 to, which means upon winning it will pay 0 dollars for each one-dollar bet in addition to the original bet. The pay out is determined by the number of people that bet on the same horse. For a fair bet, the odds of winning are one-to-0; i.e. the horse must win with probability /. For probabilities this small, research has shown that people severely underestimate the true probability of a horse winning due to choosing horses that are more attractive too often. Therefore, in fact, long-shot bets are in favor of the gambler, giving an expected monetary value significantly larger than the bet. Suppose you went to the horse races and consistently bet the 0-to-one horses, one per race. The apparent probability of wining is /, assume the true probability is /0. The expected monetary value is /0 0.05 dollars. This is better than any casino game. 0 /0-9/0-9/0 E(Y) /0 However, it takes too long win: The expected number of races to the first win is 0 races, about two days worth.
) *+" Blaise Pascal may have invented the original roulette game. In American-style roulette, there are 36 numbers, - 36, and the 0. Half the numbers are red and half are black. A straight up bet is on a single number and pays 35 to. Is this a fair game? No. The expected monetary value of the game is -/37 dollars for a one dollar wager. '&() 35 /37 35/37-36/37-36/37 E(Y) -/37 The red or black bet pays to. What is its monetary value? The line bet is on six numbers, pays 5 to. What is its monetary value? Find out one other roulette bet. What is its monetary value?
, + % In many gambling games, trusting your intuition about probability can be disastrous. A simple betting game with three cards proves it. Three cards are manufactured to special specifications. The first card has a spade on both sides. The second card has a diamond on both sides. The third card has a spade on one side and a diamond on the other. The banker shakes the cards in a hat and lets you draw one card randomly, putting it on the table. He then bets even money that the underside suit is the same as the top. To con you into thinking it is a fair game, the banker tells you that your card cannot be the spade-spade card. Therefore, it is either the spadediamond card or the diamond-diamond card, so you and he have equal chances of winning. What is the expected monetary value of the game? Suppose the bet is one dollar and the diamond is the showing suit. Let Y denote the money you win including your bet. Label the sides of the diamond-diamond card A or B. In the following the first symbol is the side that is on top. y Elementary Events f(y) yf(y) - (, ) ( A, B) ( B, A) /3 /3 /3 -/3 E(Y) -/3
-+*+.- % An old carnival and casino game dating back to the early 800 s, chucka-luck appears to be a fair game. To play the game, players bet one dollar and choose a number from one to six; three dice are tossed and the bankers pays a dollar for each die showing the players number. If there were only one die, a number has winning probability /6. However, with two dice, it seems the number has winning probability /6. Even more so, with three dice, it seems the number has winning probability of 3/6, so since the game pays one to one, this seems to be a fair game. However, the banker has the advantage when there are doubles or triples since he must pay back the original bet only once even though the winning number is showing two or three times. When all numbers are different, the banker breaks even. Suppose there are six players, each choosing a different number. Y represents the banker s wins and losses. For a triplet, say, the banker gives back three to the winning player plus her original bet, for a net profit of two dollars. For a double, say, the banker gives back two dollars to the double plus one, and one dollar plus one to the single, for a net profit of one dollar. For all singles, say 4, the banker gives back one dollar plus one to each winning player for a net profit of zero. The expected monetary value for the banker is 0/6 0.47 dollars for six players, or 0.078 dollars per player. The following table demonstrates the calculations.
Type of event Triplet Double Elementary Events 6 6 333 444 555 666 6 6 6 6 5 3 6 90 6 90 6 3 3 3 4 4 4 5 5 5 6 6 6 3 3 3 4 4 4 5 5 5 6 6 6 33 33 33 33 33 33 334 343 433 335 353 533 336 363 633 44 44 44 44 44 44 443 434 344 445 454 544 446 464 644 55 55 55 55 55 55 553 535 355 553 535 355 554 545 455 66 66 66 66 66 66 663 636 366 664 646 466 665 656 566 All Singles 0 5 4 6 0 6 0 0 3 4 5 6 6 3 34 35 36 4 43 45 46 5 53 54 56 6 63 64 65 3 4 5 6 3 33 34 35 4 43 45 46 5 53 54 56 6 63 64 65 3 34 35 36 3 34 35 36 34 34 345 346 35 35 354 356 36 36 364 365 4 43 45 46 4 43 45 46 43 43 435 436 45 45 453 456 46 46 463 465 5 53 54 56 5 53 54 56 53 53 534 536 54 54 543 546 56 56 563 564 6 63 64 65 6 63 64 65 63 63 634 635 64 64 643 645 65 65 653 654 E(Y) 0 6