# Lottery: Finding the probability of winning respective prizes

I got this question as my country is conducting a lottery this weekend.

So my country's lottery works like this:

A person buys a ticket with $6$ numbers from a pool of $45$ numbers, $1 - 45$. During the draw, the gamemaster will draw $6$ numbers. Afterwhich, he will draw 1 more number from the remaining $(45 - 6 = ) \space 39$ numbers and calls this the additional number.

The top $5$ prizes are as follows:

1. 1st Prize - Ticket matches all $6$ numbers
2. 2nd Prize - Ticket matches only $5 \space \text{of} \space 6$ numbers AND the additional number
3. 3rd Prize - Ticket matches only $5 \space \text{of} \space 6$ numbers
4. 4th Prize - Ticket matches $4 \space \text{of} \space 6$ numbers AND the additional number
5. 5th Prize - Ticket matches $4 \space \text{of} \space 6$ numbers

The question is: What is the respective probability of the prizes?

I am guessing for the 1st, 3rd & 5th prize respectively

$$P(\text{1st prize}) = \frac{6 \choose 6}{45 \choose 6} \\ P(\text{3rd prize}) = \frac{{6 \choose 5}{39 \choose 1}}{45 \choose 6} \\ P(\text{5th prize}) = \frac{{6 \choose 4}{39 \choose 2}}{45 \choose 6} \\$$ Are these calculations correct, and also what is the respective probability of the rest of the prizes?

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I think you got $P(1st prize)$ correctly. But, to me it seems the other two answers are wrong. Let me share my views.

For the $3^{rd}$ prize winner, he/she should not get the additional number picked by the master, otherwise he/she won $2^{nd}$ prize, but not $3^{rd}$.

I would suggest you to divide the 45 numbers into 3 groups.

Group-1 : the 6 numbers picked by master
Group-2 : the additional number picked by master
Group-3 : remaining numbers( 38 )


Now, for a person to win $4^{th}$ prize, he should choose 4 numbers from Group-1 and 1 number from Group-2. So, the probability to win $4^{th}$ prize = $\frac{\binom{6}{4} * \binom{1}{1} * \binom{38}{1}}{\binom{45}{6}}$

Think along these lines.

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Nice clear explanation. – André Nicolas Feb 21 '13 at 12:29

$$p(1)=\binom 66 / \binom{45} 6$$

$$P(2)=\binom 65 * \binom {39}1/ \binom{45} 6 * \binom 1 1/ \binom{39} 1$$ $$P(3)=\binom 65* \binom{39} 1 /\binom{45}6$$ $$P(4)=\binom 64 * \binom {39} 2 / \binom{45} 6 * \binom 1 1/ \binom{39} 1$$ $$P(5)=\binom 64 * \binom {39} 2 / \binom{45} 6$$

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