drawing help
August 29th, 2015 at 1:58:02 PM
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If there are 1,000 tickets in a drum and 150 of them are mine, when a ticket gets picked I have a 15% chance of winning.
If 10 people are picked, then I have a 15% chance of winning the first ticket. Since I can't win more than once, does this make sense?
#1: 0.15
#2: (1 - #1) * 0.15 = 0.85 * 0.15 = 0.1275
#3: (1 - #1 - #2) * 0.15 = = (1 - 0.15 - 0.1275) * 0.15 = 0.108375
#4...etc.
What I don't understand is, if each ticket (or place) has a different amount of winnings...how do I calculate all that? If #1 wins $1,000, #2 wins $500, #3 wins $400, etc.....?
I want to figure out the EV of such a drawing. I am not interested in the "if someone else wins then your chances improve for the next ticket because all his tickets can't get picked again" thing. But definitely want to make sure I'm doing the math right (will be doing the math right*) for the "I can only win one ticket" thing.
If 10 people are picked, then I have a 15% chance of winning the first ticket. Since I can't win more than once, does this make sense?
#1: 0.15
#2: (1 - #1) * 0.15 = 0.85 * 0.15 = 0.1275
#3: (1 - #1 - #2) * 0.15 = = (1 - 0.15 - 0.1275) * 0.15 = 0.108375
#4...etc.
What I don't understand is, if each ticket (or place) has a different amount of winnings...how do I calculate all that? If #1 wins $1,000, #2 wins $500, #3 wins $400, etc.....?
I want to figure out the EV of such a drawing. I am not interested in the "if someone else wins then your chances improve for the next ticket because all his tickets can't get picked again" thing. But definitely want to make sure I'm doing the math right (will be doing the math right*) for the "I can only win one ticket" thing.
August 29th, 2015 at 3:13:37 PM
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You're close. However, if the first ticket picked isn't yours, then there are only 999 tickets left, of which 150 are yours.
#2 is 850/1000 x 150/999
#3 is 850/1000 x 849/999 x 150/998
#4 is 850/1000 x 849/999 x 848/998 x 150/997
and so on
The EV is (the probability of winning the #1 prize) x (the amount of the #1 prize) + (the probability of winning the #2 prize) x (the amount of the #2 prize) + .... + (the probability of winning the #10 prize) x (the amount of the #10 prize).
However, this assumes that all of the other 850 tickets are bought by 850 different people. If the "nobody can win more than once" rule applies to everybody and not you, then there's no particularly easy way to figure it out as you have to take into account every possibility of particular people winning particular prizes and having their tickets removed from subsequent draws.
#2 is 850/1000 x 150/999
#3 is 850/1000 x 849/999 x 150/998
#4 is 850/1000 x 849/999 x 848/998 x 150/997
and so on
The EV is (the probability of winning the #1 prize) x (the amount of the #1 prize) + (the probability of winning the #2 prize) x (the amount of the #2 prize) + .... + (the probability of winning the #10 prize) x (the amount of the #10 prize).
However, this assumes that all of the other 850 tickets are bought by 850 different people. If the "nobody can win more than once" rule applies to everybody and not you, then there's no particularly easy way to figure it out as you have to take into account every possibility of particular people winning particular prizes and having their tickets removed from subsequent draws.
