Sec. 3.4 MATHEMATICAL EXPECTATION 71 



For this situation p = .006, n = 50,000; therefore the mathemat- 

 ically expected number of death benefits among those thirty-year- 

 olds is E(r) = .006(50,000) = 300. The reader will realize that it 

 would be unsound financially for the company to be prepared to pay 

 only 300 death benefits because this time the number of deaths might 

 be considerably higher. All that is being said is that over a period 

 of years of such calculations the average number of deaths among 

 thirty -year-olds in this same insurance class will be very close to 300. 



When chance occurrences are of the type b described above there 

 may be associated with the occurrence of E some value, say a finan- 

 cial gain or loss. Then we may wish to predict the loss or gain to 

 be expected on the average under the given conditions. For example, 

 suppose that you are going to roll a pair of unbiased dice and are 

 to be paid 60 cents if you get a sum of 7. How much should you 

 pay to play such a game if you just wish to break even? Obviously 

 you will receive either 60 cents or zero cents after each game; but 

 over many games what will be your average winnings per game? 

 That is the amount you can pay and break even. Because the prob- 

 ability of throwing a sum of 7 is 1/6 you expect, mathematically, to 

 win 60 cents on about one-sixth of your throws and to win zero cents 

 on the other five-sixths of the throws. Hence, the mathematical 

 expectation logically is 



(1/6) • (60 cents) + (5/6) • (0 cents) = 10 cents. 



Therefore you can expect to break even in the long run if you pay 

 10 cents to play each game. 



The game just described can be extended to include a reward of 

 90 cents if you throw a sum of 11 on the two dice. In this circum- 

 stance you can win in either of two mutually exclusive ways, that is, 

 you can throw a 7 or an 11. Therefore attention is centered on three 

 classes of events and the corresponding rewards: 



A sum of 7 with a reward of 60 cents, 



a sum of 11 with a reward of 90 cents, and 



a sum other than 7 and 11 with a reward of cents. 



Therefore, over a large number of games you will tend to win 60 

 cents on one-sixth of the throws, 90 cents on one-eighteenth of the 

 throws, and cents on the other seven-ninths of the throws. Hence 

 your mathematical expectation on this game is (1/6) (60 cents) + 

 (1/18) (90 cents) + (7/9) (0 cents) = 15 cents, because that is the 



