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In some lottery one can buy a ticket by choosing seven distinct numbers each of them from numbers ${\{1, 2, \dots, 45}\}$ (so $1/(45379620)$ is the probability to win the first prize).

Let this week the numbers draws be $a_1,a_2,\dots, a_7$. What is the probability that the next draw would be the same numbers, regardless of their order?

Thank you.

EDIT - Balls of lottery do not have memory of course, but why appearance of different 7-number-s happen rather than same 7-number? Suppose that we throw a coin 1000 times; is the probability of 1000 times only tail same as probability of any other mode?

callculus42
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3 Answers3

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Given that this week's draw has already happened and that it doesn't influence the next one, the probability of the numbers being the same will be (again)

$$\frac{1}{45\choose 7}=\frac{1}{45379620}$$.

man_in_green_shirt
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  • Yes, but intuitively it seems to be much more difficult appearance of same numbers. http://www.sbs.com.au/news/article/2009/09/17/identical-lottery-numbers-drawn-twice-row. Thank you –  Jul 12 '15 at 15:57
  • It's just as improbable as any other 7 numbers appearing :) – man_in_green_shirt Jul 12 '15 at 15:59
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    No - that would be different. 500 heads and 500 tails (without specifying order) would be more probable than 1000 tails. However, once 999 coins have been tossed, then regardless of what came out, the probability of the 1000th one being tails is 50%. – man_in_green_shirt Jul 12 '15 at 16:11
  • @man_in_green_shirt If X is the random variable, which expresses the number of appearing tails, Then X follows the binomial distribution. In this case the probabilities are mostly different. Only $P(X=x)$ and $P(X=n-x)$ are equal, But in the case of lottery you have a uniform distribution. Each combination is equally like. – callculus42 Jul 12 '15 at 16:23
  • @user200918 - your link is broken but the same story is at http://news.bbc.co.uk/1/hi/8259801.stm - the same numbers drawn in consecutive draws, with no top prize winners the first time and 18 people sharing the top prize the second time so receiving less each then was usual. – Henry May 22 '25 at 00:26
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As man_in_green_shirt has explained,

probability remains $\dfrac{1}{45\choose 7}=\dfrac{1}{45379620}$

You have remarked "..intuitively it seems to be much more difficult appearance of same numbers."

Please understand that what the result means is:

the same set of #s is expected to come up only once in 45379620 further lotteries !

true blue anil
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As man_in_green_shirt notes, the probability remain the same. But the probability of picking two sets $a_1, a_2, \ldots, a_7$ and $a_1', a_2', \ldots, a_7'$ that are identical over two draws is equal \begin{equation} \frac{1}{45379620}\frac{1}{45379620} = \frac{1}{2059309911344400}, \end{equation} a much smaller probability. The first situation concerns picking one unique set of integers, the probability above is repeating them twice, given that they are independent.

Forgottenscience
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    That is the probability a given set of seven numbers are picked in each of the two draws. But you have to multiply this by the possible number of sets of seven numbers, which is $ 45379620$ – Henry Jul 12 '15 at 17:06
  • "That is the probability a given set of seven numbers are picked in each of the two draws." I agree, that is what I answered, but I am also aware that this is a different calculation then the question OP answered. I just wished to point out the difference between the probability I calculated (which intuitively corresponds to what the OP wanted) and the correct probability, that the draws are independent and that $P(A|B) = P(A)$ in that case. – Forgottenscience Jul 13 '15 at 09:46