Expert: Paul Klarreich - 3/10/2009

Question
hi sir, i am having trouble solving this particular question, can you please help me. thankyou.

In the Lotto649, players must choose six numbers between 1 and 49. On Wednesday and Saturday nights, the lottery corporation draws six numbers. They then also draw one bonus number from the numbers remaining. It does not matter in what order the player chooses their numbers, nor does it matter in what order the numbers are drawn (except for the bonus which is drawn last). Players win prizes based on how many numbers they have correctly guessed. The probability of drawing any number is uniformly distributed so all numbers have an equal probability of being drawn. Each number drawn is independent of all other numbers drawn.
Each ticket costs $2.Find the probability of winning each dollar amount. What is your expected value if you buy one ticket? Is this a fair game?

Answer
Questioner:   Ashwati
Category:  Advanced Math
Private:  Yes
 
Subject:  sets and probability
Question:  hi sir, i am having trouble solving this particular question, can you please help me. thankyou.

In the Lotto649, players must choose six numbers between 1 and 49.

On Wednesday and Saturday nights, the lottery corporation draws six numbers.

They then also draw one bonus number from the numbers remaining.

It does not matter in what order the player chooses their numbers, nor does it matter in what order the numbers are drawn (except for the bonus which is drawn last).

Players win prizes based on how many numbers they have correctly guessed.

The probability of drawing any number is uniformly distributed so all numbers have an equal probability of being drawn. Each number drawn is independent of all other numbers drawn.

Each ticket costs $2.

Find the probability of winning each dollar amount. What is your expected value if you buy one ticket? Is this a fair game?
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The company chooses 6 numbers.  So given any number chosen by you, the probability is 6/49 that your number is a winner.

P(all six win) = C(6,6) (6/49)^6(43/49)^0

P(five win)    = C(6,5) (6/49)^5(43/49)^1

P(four win)    = C(6,4) (6/49)^4(43/49)^2

P(three win)   = C(6,3) (6/49)^3(43/49)^3

P(two  win)    = C(6,2) (6/49)^2(43/49)^4

P(one  win)    = C(6,1) (6/49)^1(43/49)^5

P(zero win)    = C(6,0) (6/49)^0(43/49)^6


Note that you get those from the binomial expansion of:

(6/49 + 43/49)^6, which = 1, of course.  The sum of these probabilities is 1, because one of these must happen.

You can compute each of those.  Each is a fraction.  Now multiply each fraction by the award, add them up and you have your expectation. (minus the $2, of course.)
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