I want to calculate the probability to win in a certain lottery. These are the rules:

  1. In the lottery, participants can participate individually or as a group. Groups can have up to 200 members.

  2. Each lottery ticket has a different number. Numbers are three digits long. Each digit can be any of 0 to 9. Thus, numbers run from 000 to 999 and there are one thousand lottery tickets.

  3. When an individual participates, he wins a price when his number is drawn. When a group participate, they all get different numbers, and the group wins a price when one of their numbers is drawn.

    (The value of the price is irrelevant for the question. For the question, winning as an individual is as good as winning as a member of a group. Imagine the price being a journey for all winners.)

  4. There can be between 1 and 20 prices, all of equal value. That is, between one and twenty numbers are drawn. Numbers differ between drawings.

  5. When the numbers are drawn, each digit is drawn individually. First, the first digit of the number is drawn from an urn with balls numbered 0 to 9. After the first digit has been drawn, the drawn ball is returned to the urn and the second digit is drawn. Again the ball is put back and the third digit is drawn. If there is more than one price, the third ball is returned to the urn and the next numbers are drawn in the same way as the first.

  6. Therefore it is possible that the same number is drawn multiple times. If this happens, the drawing is counted as a blank and another number is drawn. That is, each number can only win once, even if it is drawn multiple times.

The main question I'm trying to answer is whether the probability for a group win differs between the following two cases:

A: The lottery ticket numbers of all members in a group begin with the same digit, e.g.


B: The lottery ticket numbers of all members in a group begin with different digits, e.g.


For the question I assume a group of three persons whose numbers differ in the second and third digit. That is, only the first digit is relevant for the question.

I'm interested in the answer, of course, but I'd like to understand how to calculate it. I have some basic knowledge in calculating probablities for combinations and permutations, with and without returning the ball to the urn, but I find myself unable to wrap my head around the group win problem.

  • $\begingroup$ Every number has equal probability of getting a prize, regardless of what other numbers have been drawn before, so why should these two cases be different? $\endgroup$
    – David
    Jul 5, 2019 at 10:03
  • $\begingroup$ @David For the group, when their numbers all begin with the same digit (e.g. 8), the group as a whole have a one in ten chance that this first digit is drawn. But when their first digits differ (e.g. 7, 4, and 1), the group as a whole have a three in ten chance that one of their first digits is drawn. Remember, the three digits are drawn separately. The urn doesn't contain a thousand numbers all with equal probability. The urn contains ten digits which are drawn one after another. But I may be thinking wrong, which is why I ask here. $\endgroup$
    – user252827
    Jul 5, 2019 at 10:42
  • $\begingroup$ But when the $8$ is drawn, group $A$ has now a $3$ in $10$ chance of getting the 2nd number right, while if the $7$, $4$ or $1$ are drawn instead, group $B$ has only a $1$ in $10$ chance of getting the second digit right $\endgroup$
    – David
    Jul 5, 2019 at 10:48
  • $\begingroup$ The entire "three-step extraction thing" is just a distraction. Every number has a $1$ in $1,000$ chance of getting the first prize. Every number has a $1$ in $1,000-n$ of getting a prize after $n$ prizes have been given before, regardless of which number actually took the previous prizes $\endgroup$
    – David
    Jul 5, 2019 at 10:50

1 Answer 1


Both cases are equivalent to matching any three different numbers between $0$ and $999$ with another number between $0$ and $999$, for which the probability is $3/1000.$ The process you described makes any number between $0$ and $999$ equally likely. The reason you might be confused is because the process feels different, if you're following the draw one by one. The probability of elimination at each round is different in the different cases, but the probability of elimination overall remains the same.

Consider the case with all different digits in all 3 slots. There's a $3/10$ chance of a match for the first digit. If that digit matches, however, then the next two digits must match the particular individual that the first digit matched. The other two are essentially eliminated. So the next match is $1/10$ and the following is also $1/10$, making the overall probability of matching $3/1000$, exactly as you expect for $3$ individual numbers matching with a single number out of $1000.$

Now consider the case for the same digit in the first slot, and different digits in the following two. Clearly there's a $1/10$ chance to match the first digit. However, if the first digit is matched, then all players are still in the game. Because they have different second digits, there's now a $3/10$ chance for the second digit to match. If the second digit matches someone, the other two are eliminated and the third must match that individual's third slot. So the chance for a third slot match is $1/10.$ Again, these multiply to $3/1000.$

  • $\begingroup$ Thank you for setting my head on straight :-) $\endgroup$
    – user252827
    Jul 5, 2019 at 14:49

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