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In a random drawing with 50 entries are the odds of winning better if I:

A. Enter 1 contest 3 times
B. Enter 3 contests 1 time

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Context? Do all drawings have the same odds? – Arturo Magidin Feb 28 '11 at 20:34
@Arturo Magidin - Assume all drawings have 1/50 odds. – P.Brian.Mackey Feb 28 '11 at 20:36
And I ask again: Context? Is this homework? Idle curiosity? Actual practical application? – Arturo Magidin Feb 28 '11 at 20:41
@Arturo Magidin Practical - work contest. I took prob and stats, but only remember enough to know that I do not know the answer to this question. – P.Brian.Mackey Feb 28 '11 at 20:42
up vote 3 down vote accepted

In the first case, you have three entries out of 50, so your chance of winning assuming all drawings have equal chance of being picked is $$\frac{3}{50}=0.06.$$

If you enter 3 contests 1 time, and I assume each contest is the same, then your chance of winning at least one of them is the complement to losing all of them. This probability is given by $$ 1-\left(\frac{49}{50}\right)^3=\frac{7351}{125000}\approx 0.058808. $$ So it seems you have a slightly better chance of winning at least once if put all three of your bets in one contest.

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Nice, do you know the name of the second formula you used? That sure seems like a handy one to know, me being a programmer and all. – P.Brian.Mackey Feb 28 '11 at 20:52
@P.Brian: It's just an application of three facts: (i) if events are independent, then the odds of "$x$ in event $1$ and $y$ in event $2$" is the product of the odds of $x$ in $1$ and of $y$ in even $2$; (ii) the odds of "not($x$)" are $1$ minus the odds of $x$; (3) De Morgan's laws. It's easier to compute odds of multiple independent events all happening than the odds of at least one of many independent events happening, so if you want to know the odds that at least one of independent events $x_1,\ldots,x_n$ happen, it's easier to compute the odds that none happen and taking the complement. – Arturo Magidin Feb 28 '11 at 20:58

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