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

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?

• 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. Commented Feb 5, 2018 at 18:52
– user1147844
Commented Apr 23, 2023 at 21:00

• 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!
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 $0.001$.