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# Probability: jackpots

Posted on 2010-11-25
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I'm stumped by a probability issue I'm trying to work out whilst coding.  My math isn't up to the task!  In simplest terms it could be worked out to this fictional scenario:

Jack bought raffle tickets for three different draws.  Due to the total number of tickets sold for each draw he knows that:
He has a 1/4 chance of winning Game 1
He has a 1/8 chance of winning Game 2
He has a 1/16 chance of winning Game 3

What is his overall chance of winning something (at least one draw)?

When I did 1/2 chance for each draw and played with some numbers, I figured out through trial and error it was around 87.5%, but I can't arrive at the number elegantly, and I certainly haven't figured it out when each draw has a different chance.  Help!
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Question by:tknudsen-qec
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LVL 58

Expert Comment

ID: 34214051
You need to draw a Venn diagram to see this, there will be 7 sections.
The ONLY section that doesn't win anything is
doesn't win game1 AND doesn't win game2 AND doesn't win game3
This is easily calculated as
3/4 * 7/8 * 15/16
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LVL 58

Accepted Solution

cyberkiwi earned 2000 total points
ID: 34214056
Oh, and the complement of that (1- the above) will be the probability of winning something.
For reference, your 1/2, 1/2, 1/2 is solved the same way

P(win something) = 1 - (1 - 1/2) x (1- 1/2) x (1 - 1/2) = 1 - 0.125 = 0.875 (or 87.5%)
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LVL 59

Expert Comment

ID: 34214081
He has a 1/4 chance of winning Game 1 = 3/4 losing
He has a 1/8 chance of winning Game 2 = 7/8 loosing
He has a 1/16 chance of winning Game 3 = 15/16 loosing

(3/4)*(7/8)*(15/16) = 0.6152

so he will not get anything %61.52 and get 1- 0.6152 = 0.3847, %39.47 somehing...
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LVL 3

Author Closing Comment

ID: 34214088
Thank you so very much, this is extremely helpful and appreciated.
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LVL 3

Expert Comment

ID: 34214098
As cyberkiwi said.

P(win something) = 1 - (1 - 1/4) x (1- 1/8) x (1 - 1/16) = 1 - 0.615234375 = 0.38476525
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