# Expected value of lottery

If the probability of winning the lottery is $$\frac {1}{3000000}$$, and the prize is $$\9000000$$, I calculate the expected value to be $$\frac {9000000}{3000000} = 3$$

The price of each ticket is $$\2$$.

So I understand that the expected value is the average after a large number of trials/tickets purchased.

So how many tickets would I need to buy in order to get close to the expected value in the example above? If I buy 1 ticket I will stand a very low chance to win the lottery, so is there a minimum amount I could buy to get close to the expected value of $$\3$$, without buying all $$3000000$$ tickets?

Is there a huge difference in the expected value between buying say $$10000$$ tickets or $$100000$$ tickets, or even more tickets ?

• When you say get close to the expectation, do you mean your profit would be close to $3$? You haven't really said what the price of the ticket is. – Keen-ameteur Nov 6 '18 at 10:53
• I think what the OP does not take into account the price of the ticket when calculating ticket price. – Kyky Nov 6 '18 at 10:54
• Yes, that is what I mean. My profit would be close to 3. Thanks, I've edited it. The price of the ticket is $2. – Frankie139 Nov 6 '18 at 10:55 • Seems like a funny lottery. You can buy all the tickets, for$6,000,000$and you are then guaranteed to win$9,000,000$. That's not how lotteries usually work. – lulu Nov 6 '18 at 10:55 • Expected value of what? If you buy one ticket you win$8,999,998$with probability$\frac 1{3\times 10^6}$, and you lose$2$with probability$\frac {3\times 10^6-1}{3\times 10^6}$, making the expected value of a single ticket$1$. That's linear of course, so if you buy$N$tickets your expectation is$N\$. – lulu Nov 6 '18 at 11:00

## 1 Answer

Given that the question is rather vague, I will assume that you mean "How many tickets needed to have a $$95\%$$ chance that you will get at least $$99\%$$ of the expected value (i.e. $$\0.99$$)". Let us utilise Central Limit Theorem (https://en.wikipedia.org/wiki/Central_limit_theorem):

When independent random variables are added, their properly normalized sum tends toward a normal distribution (informally a "bell curve") even if the original variables themselves are not normally distributed... as $$n$$ approaches infinity then the limit, $$\sigma^2=\lim_{n\to\infty}{\frac {\operatorname {E} \left(S_{n}^{2}\right)}{n}}}$$ exists

Where $$n$$ is the number of trials and $$S_n$$ is the sample mean. So, let us assume that the distribution of every set of variables forms a bell curve. Now, we need to find the (normal) standard deviation of a distribution a variable. So we can have $$3000000$$ trials where the mean is $$1$$ and there are $$2999999$$ trials with a value of $$-1$$ and $$1$$ trial with a value of $$9000000$$. This has a standard deviation of around $$5196.15$$. We can take the square root of the above limit and show that $$\sigma=\frac {\sqrt {E(S^2_n)}}{\sqrt n}$$. As calculating the distribution of the above is the same but assuming $$n=1$$, and the left Z-Score of $$95\%$$ being $$1.65$$ according to https://socratic.org/questions/what-is-the-z-score-for-95, we can create the following formula: $$1-1.65\frac{5196.15}{\sqrt n}=0.99$$ and rearranging gives us $$\frac{5196.15}{\sqrt n}=\frac1{165}$$ According to Wolfram Alpha, $$n=7.351\cdot10^{11}$$