# 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?

• These exposed formulas are wrong, and also are easier 2 and 4 than 3 and 5. See my full answer. – Angel Moreno Oct 8 '18 at 16:49

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.

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

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

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

$$p(1) = \frac 1 {45 \choose 6}$$

$$p(2) = \frac {6}{45} \cdot\frac {5}{44} \cdot\frac {4}{43} \cdot\frac {3}{42} \cdot\frac {2}{41} \cdot\frac {38}{40} \cdot {{6}\choose{1}} \cdot\frac {1}{39}$$

$$p(3) = \frac {6}{45} \cdot\frac {5}{44} \cdot\frac {4}{43} \cdot\frac {3}{42} \cdot\frac {2}{41} \cdot {{6}\choose{1}} \cdot ( \frac {38}{40} \cdot\frac {38}{39} + \frac {1}{40} )$$

$$p(4) = \frac {6}{45} \cdot\frac {5}{44} \cdot\frac {4}{43} \cdot\frac {3}{42} \cdot\frac {38}{41} \cdot\frac {37}{40} \cdot {{6}\choose{2}} \cdot\frac {1}{39}$$

$$p(5) = \frac {6}{45} \cdot\frac {5}{44} \cdot\frac {4}{43} \cdot\frac {3}{42} \cdot\frac {38}{41} \cdot ( {{6}\choose{2}} \cdot\frac {37}{40} \cdot\frac {38}{39} + \frac {6 !}{4!\cdot1!\cdot1!} \cdot \frac {1}{40} \cdot \frac {39}{39})$$

I will explain the probability of $$p(5)$$ which is the most difficult:

p (5) = Probability (4 white numbers + 2 black numbers in the 6 holes) multiply Probability (red number is not in the seventh hole, knowing that it is not in the first 6 holes) + P (4 white numbers + 1 black number + red number in the 6 holes)

Let k be the number: $$k = \frac {1} {{{45}\choose{6}} \cdot 39}$$ Then: $$p(1) = 39k$$ $$p(2) = 6\cdot 38 \cdot k = 228 k$$ $$p(3) = 6\cdot (38^2+39) \cdot k = 8898k$$ $$p(4) = \frac {38 \cdot 37 \cdot 6 \cdot 5}{2 \cdot 2} \cdot k = 10545k$$ $$p(5) = \frac {38}{2}\cdot (15 \cdot 37 \cdot 38 + 30 \cdot 39) = 422940k$$

And then the interpretation of those numbers, and seen as another way to solve the problem, is:

The inverse of k is the possible number of different ballots.

The 39 of p (1) are the 38 numbers that have not come out in the raffle added to the number "special" make 39. Someone who has the first 6 holes OK and one of the 39 remaining numbers in the seventh hole takes prize of category 1.

In p (2) we choose one of the first 6 holes and one of the 38 numbers that have not left: 6 * 38.

In p (4) we chose two holes of the first 6 holes and two of the numbers of the 38 that have not come out in the draw. And then we exchange them, and that's the total number of different type 4 ballots. $$p(4) = k * {{6}\choose{2}} * {{38}\choose{2}}$$

In p (3). Choose one of the first 6 holes. Then, a) Choose one of the 38 numbers for the hole, and another of the 38 numbers (37 + 1 (removed from the hole) = 38) for the seventh hole. b) Put the "special" number in the hole, and put some of the 39 numbers (38 + 1 (removed from the hole)) in the hole 7.

That is, the problem can be solved also and much easier by calculating the number of different ballots of each prize class.

And calculating the total number of possible ballots (this is 1 / k) and taking into account that by symmetry of the problem each one of them has the same probability that it is the one that comes out exactly (except order of the first 6 holes) in the draw. (this is k).