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Suppose I want to enter a lottery where my odds of winning are p=1/625. The lottery occurs daily. If I enter this lottery every day, my odds of winning on at least one of the days that I played become better than 50% on day 433, since 1 - ((1 - p)^432) = 0.4993, but 1 - ((1 - p)^433) = 0.5001.

Just to make things slightly more complicated, say there are actually six independent lotteries daily, and each has the same p odds of winning. Now if I enter all of them every day, my odds of winning any of the lotteries on any given day rise to 1 - (1 - p)^6, and I should exceed even odds of winning any of lottery entries on any of the days I played on day 73.

Please correct me if I've flubbed yet at this point.

Now, suppose further that you can get a ticket to each of these six daily lotteries in a bundle, and they are always free to enter, but you can only play each at most once per day (bundling all six tickets doesn't change that). The payout for each of the lotteries is $100.

Here's my question: How do I calculate the expected value of a month's supply of 6-ticket bundles, given exactly 30 days in a month.

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  • $\begingroup$ You can't compute the expected value without knowing the prize. I think you are still interested in how many plays are needed to win. It sounds like you get $30$ plays in a month, regardless of the bundling. Is that true? If so, your chance of winning in the month is just like you are calculating, $1-(\frac {624}{625})^{30}$ $\endgroup$ Nov 6 '19 at 14:31
  • $\begingroup$ Edited to include the prize amount. And technically you would get 30 * 6 plays in a month. $\endgroup$
    – mttpgn
    Nov 6 '19 at 14:33
  • $\begingroup$ The expected value of $n$ tickets is $n$ times the expected value of of one ticket, by linearity of expectation. The expected value of one ticket is $\frac{100}{625}$ minus the cost of the ticket. $\endgroup$
    – saulspatz
    Nov 6 '19 at 14:40
  • $\begingroup$ I saw the once per day, but now I see that was each ticket. If you play $180$ times in a month, the exponent is $180$ $\endgroup$ Nov 6 '19 at 14:41
  • $\begingroup$ The price of each ticket is included in the question. The answer I'm hoping for will have a step-by-step walk through of how to compute expected value in this case. $\endgroup$
    – mttpgn
    Nov 6 '19 at 14:47
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How do I calculate the expected value of a month's supply of 6-ticket bundles, given exactly 30 days in a month?

So from what I understand, the expected value of a month's supply of 6-ticket bundles is the same as the expected value of 180 tickets (because the lotteries and days are independent).

How many times we win/lose the lottery is stochastic, and we notate $X$ for that variable. Since this question is about how many time we win or lose in 180(=n) trials with chance $p$ for each trial, we know that $X \sim$bin$(n,p)$. Now the expectation of the random variable $X$ is known (do you know what the expectation of a binomial distribution is?).

A more intuitive approach is weighing each possible outcome with is probability of occurrence. So if we win three times, the winning would be 300 and the probability of occurrence would be $\binom{180}{3}p^3(1-p)^{177}$, do you understand why?. Thus the weight of winning three times(X=3) is $300 * \binom{180}{3}p^3(1-p)^{177}$. Now if you do this for each possible amount of winnings $(\{0,1,2,...,180\})$ and sum them all up, this will be you expectation of $X$.

So to answer your question;

If you can determine the distribution and parameters of your random variable, then use this in order to 'know' the corresponding expected value.

An alternative is weighing each possible outcome with is probability of occurrence.

Edit on your question: So we were talking about calculating the probability of some event, in this case 177 losses and 3 wins. But there are a lot of different ways that this event can occur! We could for example win the first 3 and lose the last 177, but we could also win, lose win, lose win and lose the last 175 lotteries.

There are a lot of different combinations that would satisfy our event of 3 wins and 177 losses. Each of these events can occur with probability $p^3(1-p)^{177}.$ So therefore we multiply the amount of combinations that satisfy our event, with $p^3(1-p)^{177}$.

So how many combinations are there for 3 wins out of 180 lotteries? Well the first win could be on one of the 180 lotteries, the second on one of the 179 (because it can't win on the same day as the previous ticket) and the third on one of the 178 left.

This gives us $180*179*178$ groups of 3 out of 180. But wait, we counted each combination more then once. We have counted each combination $3*2*1$ times, do you understand why?

We divide the $180*179*178$ by $3*2*1$ to get the amount of unique combinations. We can write $\frac{180*179*178}{3*2*1}$ as $\frac{180!}{3!(180-177)!}=\binom{180}{3}$, the '!' is called a factorial.

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  • $\begingroup$ Several points elude me. I'm not familiar with binomial distribution -- I looked for some resources on it after you posted this, but I'm not seeing the connection here. For the intuitive approach, I see that (1-p)^177 represents the odds of losing for all but three plays, while p^3 represents the odds of winning 3 plays all of which are independent, but can you explain in English how the combinatorial got involved? $\endgroup$
    – mttpgn
    Nov 8 '19 at 16:02
  • $\begingroup$ Yes ofcourse! I'll edit it in the answer. $\endgroup$ Nov 8 '19 at 23:03

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