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marbles
12-06-2002, 02:21 PM
This one might be a little fun for the guys that are more of beginners in probability theory:

All numbers can be found at the "instant games" section of wilottery.com.

For a current scratch-off game, the wisconsin lottery gives the following odds of winning:

\$1 1:10
\$2 1:10
\$10 1:55
\$25 1:400
\$50 1:400
\$100 1:89,143
Grand prize 1:520,000

Question 1: If the Expected return on a \$1 ticket is 67.42 cents (or an E.V. of -32.58 cents, if you prefer), what is the payout for the grand prize?

Question 2: The site gives the odds of "winning" (payout greater than zero) as approximately 1:4.5. What are the odds of returning a profit on a \$1 ticket?

Question 3: What is the expected return on a ticket, given that you know it's a "winner" (payout greater than zero)? Hint: you'll need to solve question 1 first.

solutions later.

Homer
12-06-2002, 02:59 PM
1) .6742 = 1*(1/11) + 2*(1/11) + 10*(1/56) + 25*(1/401) + 50*(1/401) + 100*(1/89144) + x*(1/520001)

.6742 = .6394529 + x/520001

x = (.6742 - .6394529) * 520001

x = \$18068.53

2) 1/5.5 - 1/11 = 2/11 - 1/11 = 1/11 or 10:1

3) Hmmm....do you have to solve one first? Since we know the EV (given in 1) and the odds of getting a payout greater than zero (given in 2), can't we solve it based on that information, like this...

(4.5/5.5)*0 + (1/5.5)*x = .6742

x = \$3.708

-- Homer

marbles
12-06-2002, 04:10 PM
I was afraid this might happen, since I used the term "odds," which has a few conflicting definitions. In this case, they are using the term as the inverse of the probability of occurring. For example, their "odds" of getting a \$1 winner being 1:10 should be translated as a probability of 0.1, or 1/10. This can be particularly frustrating for a good poker player, who knows that 10:1 pot odds are required for an event that has a 1/11 chance of occurring.

Other than that, I like the logic you used on all three solutions... Plug them in using the definition of odds I just gave, and I'm pretty sure you'll get the right answers.

I'm particularly curious what you get for #3, as your logic seems reasonable, although it's not how I solved it. I think you're right that you don't need the jackpot solution if you already have the E.V.

PseudoPserious
12-06-2002, 04:39 PM
Here are Homer's answers using this definition of odds...
[WARNING: CUT and PASTE ERRORS MAY BE PLENTIFUL]

1) .6742 = 1*(1/10) + 2*(1/10) + 10*(1/55) + 25*(1/400) + 50*(1/400) + 100*(1/89143) + x*(1/520000)

.6742 = .67044 + x/520000

x = (.6742 - .67044) * 520000

x = \$1955.20

2) (chance of return &gt; 0) = (chance of return &gt;= 0) – (chance of return = 0)

1/4.5 - 1/10 = 20/90 - 9/90 = 11/90 or 12.22%

3) E.V. = (chance of zero payout) * (expected return on zero payout ticket) + (chance of non-zero payout) * (expected return on non-zero payout ticket)

(79/90)*0 + (11/90)*x = .6742

x = \$5.52

-- PP

"I am so smart. S-M-R-T..."

Homer
12-06-2002, 04:50 PM
Thank you sir or madam.

-- Homey

PseudoPserious
12-06-2002, 05:03 PM
I'm a boy!

marbles
12-06-2002, 05:27 PM
1. Pseudo's modified version of Homer's answer is correct. The actual answer is \$2,000, but rounding errors kick it down to \$1,955 and change. (you'll get a lot closer if you use the expected return of .674286129..., but \$1955 is close enough).

2. Taking the sum of all of the probabilities, (1/10+1/10+1/55+1/400+1/400+1/89143+1/520000), you can confirm their approximation (the sum ~ .223, or 1/4.48). Since any \$1 "winners" offer zero profit, you subtract 1/10 from this total probability to get approx 0.12319 (the probability of the ticket being profitable). Take the inverse, and you'll see that 1 out of every 8.12 tickets makes money.
3. Doh, the boss is watching... Gotta get back to work. Will post solution later.