In my country, playing the lottery during the festive season is very common. I am trying to find the probability of winning the second prize and third prize in some lottery draw.

There are $49$ balls that can be selected. A person buys a ticket by buying a ticket with $7$ numbers on it. Of which, $1$ number is selected as the "Special" number.

A draw is done when $6$ "Normal" balls are drawn, with $1$ ball drawn as the "Special" ball.

Second prize in won when a ticket's "Special" number matches and any $5$ of $6$ "Normal" balls match.

Third prize is won when a ticket's "Special" number matches and any $4$ of $6$ "Normal" balls match.

What is the probability of winning second prize or third prize? I know it is not as simply as $\frac{6!}{49!}$.

  • $\begingroup$ You want the probability for each, not for both at the same time. It is a bit confusing here... $\endgroup$ – Martigan Feb 18 '15 at 13:19
  • $\begingroup$ Something left unclear about the special number. Can it be any of the $7$ numbers that match this one? $\endgroup$ – barak manos Feb 18 '15 at 13:25
  • $\begingroup$ hi @barakmanos, the set of $7$ numbers are all different. These $7$ numbers consist of the $6$ normal numbers and $1$ special number. $\endgroup$ – bryanblackbee Feb 18 '15 at 13:39
  • $\begingroup$ That wasn't my question. Usually the special number is drawn from a separate (typically smaller) set of numbers. Is it not the case here? Which one of the numbers that we choose is compared with the special number? Can it be any of the $7$ numbers that we choose? $\endgroup$ – barak manos Feb 18 '15 at 13:44
  • 1
    $\begingroup$ Your question is still unclear. Is the special number marked on the ticket as "special"? Or can one choose any of the numbers on the ticket and claim that it matches the special number (in case there exists such number of the ticket)? Also, can the special number be identical to any of the other numbers? In other words, is it drawn separately from a complete set of numbers between $1-49$, or is it drawn from the numbers that remain in the set after the previous $6$ numbers have been drawn? $\endgroup$ – barak manos Feb 19 '15 at 14:43

There are $49{48 \choose 6}$ different draws. You can win second prize in $6\cdot 42=252$ ways-choose one of the non-special numbers to miss and $42$ ways to pick the undrawn number. You can win third prize in ${6 \choose 2}{42 \choose 2}=12915$ ways. So the chances are second prize about $4.2\cdot 10^{-7},$ third prize about $2\cdot 10^{-5}$


Lottery fairness implies that "ticket-universe" consists of all the tickets differing by at least one of the number appearing on them.
Clearly, any 7-uple of (different) numbers has the same probability of being extracted and of appearing on your ticket.
The septuple from the urn is independent from that on the ticket.

Take the extracted septuple to be $(x_1,x_2,\cdots,x_7)$, with $x_7$ being the special number.

According to the described rules, a x-prize winning septuple will be still be x-winning if we permute it: the ratio "x-winning" / "total possible tickets" remain the same whether we consider or not the number arrangement on the tickets.
Thus we can consider tickets showing numbers in order.
There are $\binom{49}{7}$ different possible ordered tickets, something like $86$ million.

To any other extracted septuple $(y_1,y_2,\cdots,y_7)$ will correspond the same ratio of "x-winning / total" tickets.
In particular, the ratio will not change for a permutation of $(x_1,x_2,\cdots,x_7)$, and WLOG we can assume that the extracted sequence be $1,2,\cdots,6,7$.

Also, without altering the problem, we can assume that after the extraction you go and pick your number.

There will be:

  • only one first prize ticket, corresponding to a full match , i.e.: $1,2,\cdots,7$;

  • $6 \, \cdot \,42=252$ second prize tickets which are $$ \left( {1, \cdots ,x_{\,i} , \cdots ,7} \right)\quad \left| \matrix{ \;1 \le i \le 6 \hfill \cr \;8 \le x_{\,i} \le 49 \hfill \cr} \right. $$

  • $\binom{6}{2} \,\binom{42}{2}=12915$ third prize tickets which are $$ \left( {1, \cdots ,x_{\,i} , \cdots ,x_{\,j} , \cdots ,7} \right)\quad \left| \matrix{ \;1 \le i < j \le 6 \hfill \cr \;8 \le x_{\,i} < x_{\,j} \le 49 \hfill \cr} \right. $$

  • $\begingroup$ $49 \choose 7$ ignores the fact that one number is special. It should be $49{48 \choose 6}$, which is $7$ times larger. You can think of it as choosing seven numbers, which gives $49 \choose 7$, then choosing the special one from the group. $\endgroup$ – Ross Millikan Nov 16 '18 at 2:00
  • $\begingroup$ @RossMillikan: thanks to your comment, I revised my answer and in fact the 2nd and 3rd-prize winning tickets were underestimated. I re-casted it and would appreciate your reviewing. $\endgroup$ – G Cab Nov 16 '18 at 16:38
  • $\begingroup$ You are correct about the number of tickets but you have still ignored the special number. It is not important for the first prize, but it is for the second and third prizes. You do not commit to a denominator for second and third prizes, so there is nothing wrong. $\endgroup$ – Ross Millikan Nov 16 '18 at 16:45
  • $\begingroup$ @RossMillikan: unfortunately the winning probs. I got differ from yours.. $\endgroup$ – G Cab Nov 16 '18 at 16:47
  • $\begingroup$ @RossMillikan: well yes, actually, what I mis-interpreted is that I took the special number not to be pre-marked as such on the tickets, and any of the seven could be matched with the special. Reading better the post, it seems to understand that it is marked in advance. That reconducts to a double step lottery: a) matching the special b) matching 5 or 4 of the remaining six. $\endgroup$ – G Cab Nov 16 '18 at 18:15

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