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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 ?

lulu
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Frankie139
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    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 06 '18 at 10:53
  • I think what the OP does not take into account the price of the ticket when calculating ticket price. – Kyan Cheung Nov 06 '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 06 '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 06 '18 at 10:55
  • The ticket price is $2 in this example above – Frankie139 Nov 06 '18 at 10:55
  • Not a real lottery, I'm just thinking about how expected values would work. – Frankie139 Nov 06 '18 at 10:56
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    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 06 '18 at 11:00
  • Does this answer your question? Lottery probability of winning –  Apr 08 '23 at 22:53

1 Answers1

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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, $n=7.351\cdot10^{11}$

Kyan Cheung
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