- Can we predict what will happen in the long run?
- If there is a cost associated with winning and losing, we can.
- This is called expected value.
- I really like his first example.
An insurance policy costs $50 a year. If you get sick and can't
continue college it pays $200,000. If the probability of
getting sick and not continuing college is 1/5000 is this
a good deal for the insurance company?
P(sick) * cost(Sick) + P(Not Sick) * price(Not Sick)
1/5000 * 200,000 = $40
- Expected value is the amount expected if a game is played over and over again.
- It is the sum of probability times cost for all outcomes.
- What is the expected value of a die toss
Outcome Probability
1 1/6
2 1/6
3 1/6
4 1/6
5 1/6
6 1/6
E = 1*1/6 + 2*1/6 + 3*1/6 + 4*1/6+5*1/6+6*1/6 = 21/6 = 3.5
- What if I roll two four sided die and find the sum?
Outcome Probability
2 1/16
3 2/16
4 3/16
5 4/16
6 3/16
7 2/16
8 1/16
E = 2/16 + 6/16 + 12/16 + 20/16 + 18/16 + 14/16 + 8/16 = 82/16
82/16 = 5.125
- Do problem 2, page 668
E = 50,000 x .2 + 100,000 * .1 + 150,000 * .03 + 200,000 * .01 + 250,000 * .01
= 29,000
They will spend an average of $29,000 on each customer.
They need to charge $29,050 to make $50 on each customer.
- Build a table if you can.
- If a profit or loss is involved, we multiply by that.
- If I charge $1 for a bet on a dice game, I pay $5 if you toss the die and guess the number. Is this a good game?
Outcome Probability Profit/loss
right 1/6 4
wrong 5/6 -1
1/6*4 - 1 * 5/6 = -1/6
On average, each time you will lose 17 cents.
- The only tricky question is what is the profit or loss.
- Sometimes you get to keep your cost plus the jackpot
- Sometimes you don't.
- Look at problem 6.
Event P(E) C(E)
win 1/5 38,500
lose 4/5 -1,500
E = 38,500 * 1/5 - 4/5 * 1,500
= 6,500
- Look at problem 16
Event P(E) C(E)
1 1/6 -.25
2 1/6 .75
3 1/6 1.75
other 1/2 -1.25
E = -.25/6 + .75/6 + 1.75/6 - 1/2 = -.04