0

The following question asks what is the probability that in a 7/45 lottery, that the exact same seven numbers are drawn in two consecutive draws:

Probability of 'same lottery numbers drawn twice'?

My question is similar but I'm interested in looking backwards. Let's say we have a set of 5000 previous draws. What are the chances that a selection of 3, 4, 5, 6, or 7 numbers have previously appeared in the set of 5000?

Edit for more clarity: The seven numbers are drawn without repetition and in any order. I'm wondering what is the probability that 3, 4, 5, 6, or 7 of the numbers has appeared (in any combination or order) at least once among the previous set of 5000 draws. For example, let's say I draw 3, 15, 16, 32, 34, 40, 45. What are the chances that the numbers 15, 16, 32, 34 have appeared at least once in the previous 5000 draws?

Additional edit for clarity: For the above example, 15, 16, 32 and 34 would have to appear in the same draw at least once in the set of previous 5000 draws.

I know each draw is independent but according to the accepted answer in the linked question, there is apparently a much smaller probability.

jslmsca
  • 113
  • you need to provide more details. for instance, are the numbers drawn with or without repetition? are you asking what the chances are of x numbers previously appearing in the set of 5000 exactly twice, extactly three times, or 2 or more, or ...? do you want to know the chances of them being drawn consecutively or in any order among the 5000 times? – RyRy the Fly Guy Apr 11 '23 at 20:20
  • 1
    It doesn't matter whether you look forwards or backwards. Each draw is (assumed to be) independent, so the chance of a match between any pair of draws is the same. – Ross Millikan Apr 11 '23 at 21:02
  • Does this answer your question? Odds of Winning the Lottery Using the Same Numbers Repeatedly Better/Worse? –  May 21 '23 at 20:58
  • @user Thanks but I don't think it's the same question as I was not interested about the odds of "pet numbers" or using the same numbers in each draw. I was interested in backtesting over a known number of draws, trying to find out how many times 2, 3, 4, 5 numbers appear. – jslmsca May 28 '23 at 16:06

1 Answers1

1

The probability is practically $100\%$.

Let us find the complementary probability, that none of the previous $5000$ tickets shared three or more numbers with the current ticket. In order for this to happen, each previous ticket needs to have at most two numbers in common with the current ticket. The probability that a particular ticket has at most two numbers in common with the current ticket is $$ P(\text{two tickets have at most $2$ numbers in common})=\frac{\binom{38}{7}}{\binom{45}{7}} +\frac{\binom{38}{6}\binom71}{\binom{45}{7}} +\frac{\binom{38}{5}\binom72}{\binom{45}{7}}\approx 0.9362 $$ To find the probability that this occurs for none of the previous $5000$ tickets, we raise the above probability to the $5000^\text{th}$ power. This is because the events are independent. Therefore, the probability that we fail to find a ticket which overlaps is $$ (0.9362)^{5000}\approx 8.22\times 10^{-144} $$ This number is so small that it is essentially an impossibility. Therefore, the you can say there will certainly exist at least one ticket in the past five thousand which shares at least $3$ numbers with your ticket.

Mike Earnest
  • 84,902