Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Let's assume we have a lottery with $49$ balls. We draw six of them without putting them back and sort them afterwards. E.g. one possible result is ($1$,$4$,$7$,$8$,$30$,$49$).

Obviously we have $\frac{49!}{(49-6)!6!}$ different outcomes.

How many of them contain three pairs?(e.g. 2,3,7,8,15,16)

Our approach so far:

We tried to simplify this by transforming our result $(x_1,x_2,...,x_6), 1 <= x_1 < x_2 < ... < x_6 <= 49$

to

$(x_1,x_2+1,x_3+2,...,x_6+5), 1<=x_1<=x_2<=...<=x_6<=44$.

Example: $(3,5,6,20,36,49) \Rightarrow (3,4,4,17,32,44)$.

Now we can count a pair for each $x_i = x_j$. If we draw a simple graph now and add the probabilities, we get a result that doesn't match the provided solution of $\frac{43}{45402}$.

Are we on the right track? If so, where is our error?

share|improve this question
    
If you get $1,2,3,4,5,6$, how many pairs do you consider you have? –  Patrick Li Nov 5 '12 at 0:54
    
@PatrickLi: I would say three. As I read it, after sorting, you get three pairs iff $x_1+1=x_2, x_3+1=x_4, x_5+1=x_6$ –  Ross Millikan Nov 5 '12 at 1:12
    
A very naive approach: the number of ways to get three pairs is ${48 \choose 3}$, because you can choose three balls out of the first $48$ and then pick the one after each. This ignores problems of overlap, so is not quite right. Dividing by ${49 \choose 6}$ possibilities gives $\approx 0.124\%$ –  Ross Millikan Nov 5 '12 at 1:18
    
@PatrickLi: Zero. We only have a pair if the neighbors are not present. e.g. $1,2,3,6,7,12$ is a single pair. –  foobarfoox Nov 5 '12 at 2:15
    
Then you are looking for the number of ways to select three balls out of the first 48, with the restriction that if you pick ball $i$ you can pick neither $i+1$, nor $i+2$. Add the one above each of the three and you are there. A naive approach is there are 48 choices for the first, 43 for the next, and 38 for the last, divided by 6 for the orders. This gives $\approx 0.00935\%$. It is low because at the ends we don't lose the full 5 choices. –  Ross Millikan Nov 5 '12 at 2:46

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Browse other questions tagged or ask your own question.