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Gokul43201
#3
May24-04, 09:24 PM
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The first part of this calculation is only approximately correct, but is probably as good an approximation as you'll ever need. Actually for each year the odds of failure are very slightly smaller than (19/20)^3. The second part is fine.

The reason that the above calculation is usually good enough is that the total number of tickets out there is large compared to 3. If not, you do have to calculate conditional probabilities. The outcomes are not strictly independent.

Here's an illustration of a case where your calculation would not work. Suppose you bought ALL the tickets; you have to win - in fact you'll win on multiple tickets. So the probability of failure is 0. But the "independent failure" calculation will give you (19/20)^N. This gets to be really small for large N, but never goes to 0.

anyways, mathematically...it never pays to buy lottery tickets.