GAM470 Name_____Answers________

April 19, 2005 Homework
  1. On the Super Bowl you find the following prop.

    2. Coin toss is heads -102

    Coin toss is tails -102

    Assuming a fair coin, what is the expected value of either side of this bet?

    -102 is equivalent to 1/1.02 to 1. So the expected value is .5*(1/1.02) - .5 = -0.98%.

  2. What probability of winning is required of a handicapper to break even over the long run against the normal –110 line?

    3. As usual the formula for expected value is pr(win)*odd – pr(loss) = expected value. Here we know the expected value is 0, but we don’t know the probability. Let’s let p stand for the probability then…

    p*(1/1.1) – (1-p) = 0

    p*(1/1.1 + 1) = 1

    p*(2.1/1.1) = 1

    p = 1.1/2.1 = 11/21 = 52.38%.

  3. Your probability of winning against the point spread is 54%. What is your expected value when laying –105?

    4. -105 is equivalent to 1/1.05 to 1. So the expected value is .54*(1/1.05) - .46 = +5.43%.

  4. In today’s baseball game the Orioles to win pays +175. You feel the probability they will win 40%. What is the expected value of this bet?

    5. +175 is equivalent to 1.75 to 1. So the expected value is 40%*(1.75) – 60% = 10%.

  5. In another baseball game the line on the Padres is –300. You feel this bet has a 5% advantage. What is the probability of the Padres winning?

    6. -300 means the bettor must risk $300 to win $100, which is equivalent to 1/3 to 1. Using the usual expected value formula and solving for p…

    p*(1/3) – (1-p) = .05

    p*(1/3 + 1) –1 = .05

    p*(4/3) = 1.05

    p=1.05*(3/4) =

    p=78.75%

    Problems 6,7:

    You see the following prop on the Super Bowl

    Total Solo & Assisted Tackles by Michael Lewis

    Over 5.5 -130

    Under 5.5 EV

    You feel the expected number of tackles is 5.2.

  6. Using the Poisson distribution, what is the probability of less than 5.5 tackles? Please use the table below.

    7. Tackles

    Formula

    Probability

    0

    exp(-5.2)*5.2^0/0!

    0.005517

    1

    exp(-5.2)*5.2^1/1!

    0.028686

    2

    exp(-5.2)*5.2^2/2!

    0.074584

    3

    exp(-5.2)*5.2^3/3!

    0.129279

    4

    exp(-5.2)*5.2^4/4!

    0.168063

    5

    exp(-5.2)*5.2^5/5!

    0.174785

    Total

    .

    0.580913
  7. What is the expected value of both bets?

    8. The table above shows the probability of under 5.5 is 58.09%, so the probability of over 5.5 must be 41.91%.

    Under 5.5: 58.09%*1 – 41.91% = 16.18%.

    Over 5.5: 41.91%*(1/1.3) – 58.09% = -25.85%.

    Problem 8,9:

    You see the following prop on the Super Bowl

    Will there be a safety?

    Yes +600

    No -850

    You feel the average number of safties per game is 0.05.

  8. What is the probability of zero safties?

    9. exp(-0.05)*0.05^0/0! = exp(-0.05) = 95.12%.

  9. What is the expected value of both bets?

    10. The probability on the no is 95.12% so the probability on the yes is 4.88%. +600 is equivalent to 6 to 1 and –850 is equivalent to 1/8.5 to 1.

    Yes: 4.88%*6 – 95.12% = -65.86%

    No: 95.12%*(1/8.5) – 4.88% = +6.31%

    Problem 10,11:

    You see the following prop on the Super Bowl

    Total number of different Eagles to score:

    Over 3 -115

    Under 3 -115

    You feel the average number of Eagles to score is 2.5.

  10. Use the following table to determine the probability of under 3, exactly 3, and over 3 Eagles to score.
    11.

    Eagles to Score

    Formula

    Probability

    0

    exp(-2.5)*2.5^0/0!

    0.082085

    1

    exp(-2.5)*2.5^1/1!

    0.205212

    2

    exp(-2.5)*2.5^2/2!

    0.256516

    3

    exp(-2.5)*2.5^3/3!

    0.213763

    Total

    .

    0.757576

    Under 3 = 0.082085 + 0.205212 + 0.256516 = 0.543813

    Exactly 3 = 0.213763

    Over 3 = 1-0.757576 = 0.242424

  11. What is the expected value of both bets?

    12. -115 is equivalent to paying 1/1.15 to 1. As usual the expected value is pr(win)*pays – pr(lose).

    Over 3: 0.242424*(1/1.15) – 0.543813 = -0.33301

    Under 3: 0.543813*(1/1.15) – 0.242424 = +0.230457

  12. Your sock drawer has 8 white socks and 5 black socks. You pull out 4 of them at random and without replacement. What is the probability you get a three of a kind (3 of one color and 1 of the other)?

3 whites and 1 black combinations: combin(8,3)*5 = 56*5 = 280

1 white and 3 blacks combinations: combin(5,3)*8 = 10*8 = 80

Total successful combinations: 280+80 = 360

Total all combinations: combin(13,4) = 715

Probability = 360/715 = 50.35%.