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Please explain in simple terms two questions down at the bottom? I've an exercise:

In the lottery of 100 tickets half of them are winning. How many lottery tickets need to be bought to reach a probability of wining up to 0.999?

The solution to this exercise is:

Let $X_i$ be a probability of winning of ticket i, that is $P(X_i) = p$. Where p is a probability value and equals 0.5. Then, the probability of getting AT LEAST one winning ticket in n tickets equals to sum of independent events $X_1, X_2, \cdots , X_n$, or $P(X_1, X_2, \cdots , X_n) = 1 - (1-p)^n $.


Calculations:

$1 - (1 - p)^n > \ge 0.999 $

Given half of tickets are winning, therefore p = 0.5.

$(1 - 0.5)^n \le 1 - 0.999$

Logorithmizing we get:

$n \times \log(1 - 0.5) \le \log(1 - 0.999)$

Given that $\log(1 - 0.5)$ is negative we change $\le$ to $\ge$:

$n \ge$ $\frac{\log(1 - 0.999)} {\log(1 - 0.5)}$

$ n \ge$ $\frac{\log(0.001)} {\log(0.5)}$
$n \ge 10$

2 questions I can't really wrap my head around

  1. When buying one ticket out of 100, the probability of a ticket changes, because 1st ticket's probability of winning is $\frac{50}{100}$ and the 2nd's is $\frac{49}{99}$ and so on (which is hypergeometric distribution I guess?). So how does it happen that the $p = 0.5$ remains constant when we deal with AT LEAST scenario in this question? Why p remains 0.5, whereas 5 tickets are bought?

  2. Each time we buy a lottery, each ticket is an independent event. But yet it is joint event. I assume it is joint because, perhaps of the first question, each time one ticket is purchased, the probability of winning second time changes? I don't get it.

And I might be mistakenly mixing and mashing some concepts now by saying that: if they are joint and independent should I not somehow take into account their intersection? Because if I work with a sum of independent events $X_1, X_2, \cdots , X_n$, shall we not substract their intersection?

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    $\begingroup$ The solution you gave ignores the fact that if the first ticket is losing the chance of the second ticket being losing is reduced. It assumes the chance of winning is constant at $p$. This is correct if you are flipping a coin or rolling a die, where the events are independent, but is not correct in the lottery case. $\endgroup$ Commented Feb 5, 2018 at 18:52
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    $\begingroup$ Does this answer your question? Lottery probability of winning $\endgroup$
    – user1147844
    Commented Apr 23, 2023 at 21:00

2 Answers 2

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Assuming that exactly 50 tickets are guaranteed to be winners, you are correct: the events are not independent, and the solution given is incorrect. It's not totally without merit, though:

  • You could imagine a lottery in which each ticket had a probability of $1/2$ of being a winner, with no correlation. (Though this arguably stretches the definition of the word "lottery".) In that case, the solution given would be correct.
  • Even for the problem written, though the solution is incorrect, it isn't that incorrect. Note that it suggests that you need to buy 10 tickets. By the time you've done that, the probability that the last ticket is a loser has only fallen from $1/2$ to $41/91$, or roughly $0.45$. So it's a solid way to estimate the number of tickets you need to buy. In fact, when you do the correct computation, you find that $$ P(\text{losing after $9$ tickets}) \approx 0.001317 $$ while $$ P(\text{losing after $10$ tickets}) \approx 0.000593 $$ So the "incorrect" computation was a good enough approximation in this particular case to give the correct answer!
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  • $\begingroup$ Perhaps worth pointing out that the incorrect solution becomes closer and closer to the correct solution as the sample size increases. For a pool of only 10 items, there will be a reasonable difference between sampling with and without replacement. For a pool of 10 million items, it's very unlikely to draw the same item even with replacement, so sampling with replacement becomes a very good approximation of the more-difficult-to-calculate case without replacement. $\endgroup$ Commented Feb 5, 2018 at 19:22
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It is equivalent to compute the number of tickets need to be bought so that the probability of losing is less than $0.001$. To this end, note that $$ P(\text{losing})=\frac{50}{100}\times\frac{49}{99}\times \dotsb\times \frac{50-n+1}{100-n+1}\tag{1}. $$ The first ticket bought is a loser ticket with probability $50/100$. Given that the first was a loser, there are 49 loser tickets left out of $99$ remaining tickets and so on. You can find when this probability in (1) is less than $0.001$.

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  • $\begingroup$ thanks for response John! I understand P(loosing) very well and it does make sense. But how does it come down to a formula (1 - p)^n? This is what I absolutely stuck in :( $\endgroup$
    – Vlad
    Commented Jan 17, 2018 at 22:07
  • $\begingroup$ Does the probability of a ticket winning depend on the previous ticket losing? If yes, then use the formula above. If no, then use (1 - p)^n. $\endgroup$
    – Michael
    Commented Jan 16, 2022 at 19:58

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