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I'm trying to calculate some probabilites about the lottery and this question showed up:

There are $20$ balls(numbers) - all different - and we pick $5$ of them with no replacement. Let's pick $firstdraw=(1,2,3,4,5)$

Within a $100$ draws, what is the probability in two consecutive draws have the same numbers?

Since it's just saying "the same numbers" I don't have to calculate the order: $$p=\frac{5}{20}.\frac{4}{19}.\frac{3}{18}.\frac{2}{17}.\frac{1}{16}$$ How can I calculate the probability of getting consecutive. I don't know how to approach $x$ times. Can I say $p^2$ as we roll $2$ dices and getting the same number?

And also how can I calculate getting the same same numbers in $(firstdraw)$ at least twice in $100$ draws. Thanks.

RChamuel
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2 Answers2

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I'm assuming the question is this: in each draw, there are $20$ balls (each marked by something different e.g. a number from $1$ to $20$) and $5$ are chosen without replacement. $100$ draws are performed. What is the probability that some two consecutive draws are equal?

You have correctly calculated the probability of one draw yielding a particular set of five balls. Let's call it $p$.

There are $99$ opportunities for two consecutive draws to be the same: the first could equal the second, the second could equal the third, ..., the $99$th could equal the $100$th.

Each event has probability $p$, because you can take the first draw in each pair to be the fixed set of balls that we are comparing the second draw in the pair to. Asking "what is the probability that the two draws is the same?" is the same as asking "what is the probability that the second draw is equal to some particular outcome, i.e. the first draw?". We know the answer is $p$. This works only because each five-ball outcome is equally likely.

Each of these events is completely unconnected to the others (because no particular draw is any more likely than any other), so the probability it doesn't happen in any of the $99$ possible places is $(1-p)^{99}$, and the probability that it does happen is $1-(1-p)^{99}$.

A.M.
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  • Thanks a lot. The word "consecutive" really bugs me out. For the second part that getting same set at least twice in $100$ draws, can we approach like it's a birthday probability:? $\frac{20!}{(20^5)(20-5)!}$ – RChamuel Oct 29 '21 at 17:14
  • This would give one or more consecutive matches, not just one as stated. – Vaekor Oct 29 '21 at 17:15
  • @Vaekor The question was "what is the probability in two consecutive draws have the same numbers". I see no implication there of "exactly one", and the most natural assumption is at least one. – A.M. Oct 29 '21 at 17:18
  • @RChamuel this is the binomial distribution. It is easier to calculate it as "what is the probability of it not happening $0$ or $1$ times?", rather than "at least twice". The probability of it happening $0$ times is $(1-p)^{99}$. The probability of it happening once is $99p(1-p)^{98}$. Subtract the sum of these from $1$. – A.M. Oct 29 '21 at 17:23
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This is a binomial distribution problem. There are a fixed number of trials (effectively 99, since we're looking at consecutive draws), there are two outcomes (they match or they don't), the trials are independent, and all trials have the same probability $p$ of success that you calculated already.

We are interested in exactly one match (1 success), and we will use the formula $\binom{n}{x}p^{x}(1-p)^{n-x}$, where $x$ is the number of trials, $x$ is the number of successes, and $p$ is the probability of success.

The probability is therefore given as $\binom{99}{1}\cdot(p)^{1}(1-p)^{98}\approx 0.006345$.

For the later question of two or more, we can subtract the probabilities of 1 success and 0 successes from 100%, such that:

$\binom{99}{0}\cdot(p)^{0}(1-p)^{99} \approx .993635$

$P(X\ge 2) \approx 1-.994635 -0.006345 = 0.00002$

Vaekor
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