I have a friend who plays the lottery by buying one line that gets played in several games. So if he buys $20 he plays one set of numbers for 20 games. I think he would have better chance at buying 20 different lines for one game. I am not good at the specific math behind the odds. Am I right in this assumption or is my friend?

I talked with a friend who explains it like this: If you have a standard six sided die and bet that you would either get a 1 or 2 your chances are 2/6. If you bet that in two rolls at least one will be a 1 your chances are 11/36. There are 36 possible outcomes and 11 outcomes where at least one die is a 1. So 1/6=33.3% which is better than 11/36=30.5%.


closed as off-topic by John B, Lord Shark the Unknown, amWhy, Christopher, Don Thousand Nov 9 '18 at 16:53

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Well, let's work an example. Say there are $N$ possible lines in a single lottery, so that your chance of winning with a single line in a single lottery is $p(N)=\frac 1N$.

Let's say $N=1000$ to be explicit.

If we choose $20$ lines out of one lottery, our chances of winning are $$\frac {20}{1000}=\frac 1{50}=.02$$

If we choose one line in each of $20$ lotteries, our chance of winning at least one is the complement of our chance of winning none, hence $$1-\left( \frac {999}{1000}\right)^{20}=.019811114$$

Which, of course, is slightly worse. Of course, you might win more than one with the second strategy. But the probability of winning at least two is $$1-\left( \frac {999}{1000}\right)^{20}-\binom {20}1\times \left( \frac 1{1000}\right)\times \left(\frac {999}{1000}\right)^{19}=.0001873$$ so it is fairly safe to neglect that possibility.

  • $\begingroup$ If we assume an equal prize each time then your expected winnings will be the same in each case. $\endgroup$ – badjohn Nov 8 '18 at 20:47

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