# Something strange?

Discussion in 'Coffee Break and Community Discussion Forum' started by Ronen E, Mar 19, 2016.

1. ### Ronen EWell-Known Member

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Hi,

This is a puzzle for all you probability experts. I think I can find the answer but wanted additional perspectives.

I noticed something that seemed a little strange in our lottery results. Several times I noticed that the same number (out of six) was drawn at the same spot in the sequence (eg 3rd out of six), 2 weeks in a row. That didn't trigger my concern until this week I noticed that the last 2 numbers drawn were identical to those in last week's draw (and in the same order). It's like they forgot to draw the last two and just left the ones from last week...

So, the question is: what's the probability of occurrence of 2 numbers out of 6 being identical, and in the same positions, 2 weeks in a row? The draw involves the numbers 1-45 inclusive.

Do the positions in the sequence where the above occurs matter? Ie is it more striking that it occurred on the last 2?

2. ### MarkMeerWell-Known Member

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I'm no expert, but from what I recall, the draws are treated independently. Therefore:

The probability that the 3rd draw (out of 6) is the same two weeks in a row is: 1/45
(NOTE: if you are assuming all other draws are all DIFFERENT from the previous week, then the probability would be (44/45)*(44/45)*(1/45)*(44/45)*(44/45)*(44/45))

The probability that the last two draws are the same is: 1/45*1/45 = 1/2025
(or, if the other 4 have to be different: (44/45)*(44/45)*(44/45)*(44/45)*(1/45)*(1/45))

NOTE: this may be simplifying, as it's assuming that withdrawing a number does not exclude it from subsequent draws.

Last edited: Mar 21, 2016
3. ### Bev DModeratorStaff Member

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I always work on the assumption that probability is WAY more complex than I remember (sort of like the pain of child birth)
So in determining the probability we must incorporate all of the actual conditions of the event. In this case the number that is drawn is NOT available to be redrawn. This is sampling without replacement.

Even though the weeks are independent, conditional probability will apply for this question.

4. ### Ronen EWell-Known Member

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Thanks. Would actually determining the probability in this case be too time consuming?

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