Assume a lottery game of the following rules:

Picking your numbers:

  • Pick a total of 6 different numbers from the lot containing 42 numbers (1 to 42).


  • Draw 7 balls, with no repetition(i.e: ball drawn is not put back in the lot) from the lot labeled from 1 to 42.


  • If the first 6 balls drawn matches your own 6 numbers (order doesn't matter): Jackpot.
  • If 5 of the first 6 balls drawn matches 5 of your numbers (order doesn't matter) and the 7th drawn ball matches your 6th number: second prize.
  • If 5 of the first 6 balls drawn matches 5 of your numbers and nothing else matches: third prize.

I'll end it here for not having many other prizes.

If I want to check my chance of winning the jackpot, it's pretty straightforward and looks like a combination $C(42,6)$, so it should be:

$$ \frac{42\cdot41\cdot40\cdot39\cdot38\cdot37}{6!} = 5,245,786. $$

So my chance of getting the jackpot is $(\frac{1}{5,245,786})$

For the third prize it's also a straightforward combination $C(42,5)$, it's equal to:

$$ \frac{42\cdot41\cdot40\cdot39\cdot38}{5!} = 850,668. $$

So third prize probability is equal to $\left(\frac{1}{850,668}\right)$

Now I am being stumbled on how to calculate the 2nd prize probability. My memories from school are not helping me enough to get my answer. I know that it should be between the two numbers I got there, however any calculations I am making end ups with a probability much higher than the first prize's.

Could you please verify that my 1st and 3rd prize probabilities are well calculated and help me calculate the 2nd prize probability?


Your logic for case number (3) does not seem correct.

First, you have to draw 6 balls with exactly 5 balls matching 5 numbers from your set of 6 picked numbers. In total There are C(42,6) ways to draw six balls. How many winning combinations do we have? From your set of 6 balls you can pick five matching balls in C(6,5) ways. The sixth number can be any number from the remaining 42-5=37, minus one (your sixth pick). So the total number of winning combinations is $C(6,5)\times36$. So the probability is:

$$p_1=\frac{36\times{6\choose 5}}{42\choose 6}$$

But you have to draw the seventh number as well from the remaining 42-6=36 numbers without hitting your last (sixth) number. That probability is:


The total proability is:

$$p=p_1\times p_2=\frac{36\times{6\choose 5}}{42\choose 6}\times\frac{35}{36}=\frac{35\times{6\choose 5}}{42\choose 6}=\frac{15}{374699}$$

You can use a similar logic for case (2).

The probability $p_1$ is the same. For the seventh ball we have 36 choices and there is 1 winning ball between them. So the probability $p_2$ is:


...and the final probability for the second prize is:

$$p=p_1\times p_2=\frac{36\times{6\choose 5}}{42\choose 6}\times\frac{1}{36}=\frac{{6\choose 5}}{42\choose 6}=\frac{3}{2622893}$$

(35 times smaller than the probability for the third prize)

  • $\begingroup$ Great! This explains where most of my mistakes where. Thank you for clarifying. $\endgroup$ – Paul Karam Oct 18 '18 at 11:03
  • $\begingroup$ I wanted to test if I did correctly understand the logic here. So I tested with rules: 4 balls matches, and another 3 ball matches. In both cases we won't have a p2 since the 7th ball changes nothing. So my answers for the 4 ball match is 1/555 and 3 ball matches is 1/36, can you please confirm? $\endgroup$ – Paul Karam Oct 19 '18 at 7:23

For $3^{rd}$ prize, you need to select which 5 numbered balls you will be getting, out of 6 in $^6C_5$ ways (get balls matching to those 5 numbers from 42); and then, select the $6^{th}$ ad $7^{th}$ ball in $^{36}C_2$ ways
(42-6=36, balls not matching any of the 6 numbers)

So, probability of winning $3^{rd}$ prize is $$\frac{^6C_5\cdot^{36}C_2}{^{42}C_7}$$

For $2^{nd}$ prize,
Among first 6, only 5 match, selected their corresponding numbers in $^6C_5$ ways.
First 5 balls drawn match these 5 numbers.
Now, $6^{th}$ ball is selected from 36 (42-6=36, it does not match any of 6 numbers); and number corresponding to last ball to be drawn is already known(one left from 6)

So, probability of winning $2^{nd}$ prize is: $$\frac{^6C_5\cdot^{36}C_1\cdot^{1}C_1}{^{42}C_7}$$

Similarly, For $1^{st}$ prize, $$\frac{^6C_6\cdot^{36}C_1}{^{42}C_7}$$

  • 2
    $\begingroup$ I think that your calculation for the third prize is also not correct. You are ignoring the fact that you have to draw the seventh ball as well and that ball must not hit your last pick. Just read the condition carefully: If 5 of the first 6 balls drawn matches 5 of your numbers and nothing else matches: third prize. "Nothing else" must include the seventh ball as well. $\endgroup$ – Oldboy Oct 18 '18 at 10:49
  • $\begingroup$ yup, now i guess its correct. $\endgroup$ – idea Oct 18 '18 at 10:55
  • $\begingroup$ omega, maybe I wasn't clear in my question, the 7th ball drawn doesn't allow you to win the first prize. Only the second. So those calculations are way off in my opinion. I may be wrong, but after looking at the answer I accepted and this one, I feel that (maybe) I wasn't very clear. $\endgroup$ – Paul Karam Oct 19 '18 at 7:37

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