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#1
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simple probability exercise
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. |
#2
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My answers
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 |
#3
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Re: My answers
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. |
#4
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Re: My answers
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 > 0) = (chance of return >= 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..." |
#5
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Re: My answers
Thank you sir or madam.
-- Homey |
#6
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Re: My answers
I'm a boy!
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#7
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Solutions
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. |
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