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:
- 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.
Results:
- 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?
Best Answer
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:
$$p_2=\frac{35}{36}$$
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:
$$p_2=\frac{1}{36}$$
...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)