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In the UK the lottery uses numbers $1$ to $49$ and a total of six numbers are picked. It has been said may times that there is as much chance of numbers $1, 2 ,3 ,4 ,5 ,6$ to be picked as any other random combination.

My question is this:

Let's say that the first $3$ numbers to come out are $1,2$ and $3$. What are the chances of a number between $1$ and $10$ coming out next Vs a number between $11$ and $20$?

There are obviously less numbers between $1$ and $10$ now that we already lost $1$ to $3$, so surely the probability is that a number between $11$ and $20$ is more likely? In which case, the chances of a lottery selection of $1,2,3,4,5,6$ is less likely than $2,12,21,28,32,47$ for example...

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  • $\begingroup$ That is true, but a priori to the first three drawings the probability is the same. Once you drew three balls it becomes a conditional probability. $\endgroup$
    – Asaf Karagila
    Oct 9 '10 at 14:10
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    $\begingroup$ If you consider 'numbers with digit 2 in them', by your logic, 1,2,3,4,5,6 has higher probability than 2,12,21,28,32,42... $\endgroup$
    – Aryabhata
    Oct 9 '10 at 14:28
  • $\begingroup$ On a related note, if you want to maximise your expected winnings (and still participate) then you are better served by choosing combinations of numbers unlikely to be chosen by others, so as to minimise your chances of sharing your possible winnings with others. $\endgroup$ Dec 14 '17 at 5:25
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The chances of the next number being between 1 and 10 is $\frac{7}{49}$, as opposed to the probability of it being between 11 and 20 being $\frac{10}{49}$. So, among other things, it is less likely that a lottery ticket will have only numbers between 1 and 10, as opposed to numbers between 1 and 20. However, that does not mean that a given ticket with numbers between 1 and 10 is less likely then a given ticket with numbers between 1 and 20. The fact that there are more tickets with numbers between 1 and 20 exactly cancels with the higher chance of getting such a ticket.

Example: You have a bag of pens; a red pen, and 99 blue pens numbered 1-99. The chance of you getting a red pen is only .01 whereas the chance of oyu getting a blue pen is .99, but the chance of getting any given blue pen (say, #42) is .99 / 99 = .01, which is the same as the chance of getting the red pen.

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Having chosen 1,2,3 all possibilities (there are 46*45*44/6) that include 1,2,3 are equally likely and all possibilities that do not include 1,2,3 have probability zero. The psychological error comes from grouping "numbers less than 10" together. As Moron commented you could also group "numbers containing a digit 2". Each of these are legal groupings and the probability of selecting all numbers from within them can be calculated. But it is still true that any particular combination is equally likely at the outset. A draw of 13,18,29,33,36,45 looks "random" to our eye and we group it with all other "random looking" combinations. From this point of view, a "random looking" combination is highly likely.

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It is CORRECT to say that the probability of drawing numbers 1 2 3 4 5 6 is identical to the probability of drawing any 'given' combination of 6 numbers. Proof: The probability of the first number being drawn as either 1,2,3,4,5 or 6 is 1/49.
Likewise the probability of the first number drawn to correspond with any of the numbers in a given combination is also 1/49. Assuming the first number drawn is '1' then the probability of the second number being drawn as either 2,3,4,5 or 6 is 1/48. Likewise, if the first number drawn corresponds to any one of the numbers in the given combination, then the probability of the second number being drawn corresponding to any one of the 5 remaining numbers in the given combination is also 1/48. This pattern will hold true for the remaining 4 balls, where the probability of a 3rd corresponding ball being 1/47, 4th ball being 1/46, 5th ball being 1/45 and the last ball being 1/44.

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