buy 1000 lottery tickets

buy 1000 lottery tickets

Likelihood – Expectation?

A lottery offers a prize of $ 1,000, a $ 500 prize and $ 5100 prizes. 1000 tickets were sold for $ 3 each. Find the expectation if a person buys two tickets. Suppose player ticket is replaced after each draw and the same ticket can win more than one prize. Please show me how to fix this, thanks

Dear ShortShot78, The fastest and easiest way to solve this problem is to add the award amounts to obtain the total value of the awards, which is $ 2000. Given no ticket is more likely to win than another, expects earnings received for each ticket is then $ 2000 / 1000 = $ 2 per ticket. Since each ticket is $ 3, the expectation of each ticket is $ 2 – $ 3 = – $ 1. Therefore, a person who purchases two tickets have a loss Expected $ 2. If you choose to solve this problem with a longer, more difficult path, then here is one possible approach. Find the respective probabilities either win a prize or two awards. For a prize, you can find the probability of winning a prize in the first draw, the second draw, the draw for third and so on, until the draw meeting. During the first call probability is given as p1 = P (winning a prize in the draw only the first) = (2 / 1000) (998/999) (997 / 998) (996/997) (995/996) (994/995) (993/994) = (2 / 1000) (993/999) = 331/166500. You can verify that the probability is the same to win a prize on the second draw only, or third party only, etc (call these p2, p3 ,…, p7), and since they are all equal, it simply can be identified by p. No loss of generality, we can think of more prizes to be awarded first, followed by superiors, so that payment is expected to win just one prize is 100 pp + 100 + 100 + 100 + 100 + 500 ppp + p p = 2000 p. 1000 Consider now won two awards in the first two draws, the first and the third is based, the first and the fourth is based, and still, so to win the sixth and seventh draws. Take the prize were awarded lower amounts for the first time, as before. Q12 = P (winning two awards in the first two draws) = (2 / 1000) (1 / 999) (998/998) (997/997) (996/996) (995/995) (994/994) = (2 / 1000) (1 / 999) = 1 / 499 500. Again, you can verify that each of the twenty-one chances to win two awards is equally likely (ie Q13 = Q14 Q12 = =… = q67), so it can only be identified by q. Payment is expected to win just two awards is (100 + 100) (10 q) + (100 + 500) (5 q) + (100 + 1000) (5 q) + (500 + 1000) = 2000 + 3000 qq + 5500 + 1500 = 12000 qqq q. The total projected payments for a person with two inputs is the sum of expected payments to win one or two awards: 2000 p + q = 12000 (2000) (331 / 166500) + (12000) (1 / 499500) = (2000) (993 / 499500) + (12000) (1 / 499500) = [(2000) (993) + 12000] / 499500 = 1998000 / 499500 = 4. By Therefore, as with the "quick and easy solution, since each ticket costs $ 3, the total expectation is $ 4 – ($ 3) (2) = – $ 2. It ie a person who purchases two tickets have expected a loss of $ 2.

1000 Dollar Mania and Diamond Dash Scratch Off Tickets