0
$\begingroup$

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 ?

$\endgroup$
  • 1
    $\begingroup$ 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. $\endgroup$ – Keen-ameteur Nov 6 '18 at 10:53
  • $\begingroup$ I think what the OP does not take into account the price of the ticket when calculating ticket price. $\endgroup$ – Kyky Nov 6 '18 at 10:54
  • $\begingroup$ 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. $\endgroup$ – Frankie139 Nov 6 '18 at 10:55
  • $\begingroup$ 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. $\endgroup$ – lulu Nov 6 '18 at 10:55
  • 1
    $\begingroup$ 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 $N$. $\endgroup$ – lulu Nov 6 '18 at 11:00
1
$\begingroup$

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, ${\displaystyle \sigma^2=\lim_{n\to\infty}{\frac {\operatorname {E} \left(S_{n}^{2}\right)}{n}}}$ 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, $n=7.351\cdot10^{11}$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.