A variation of an exercice from Chapter 16 (Counting and Choosing) of Liebeck's book

Question

The rules of a lottery are as follows: You select 10 numbers between 1 and 50. On lottery night, the celebrity mathematician Richard Thomas chooses at random 6 'correct' numbers. If your 10 numbers include all 6 correct ones, you win. How many ways are there to win the lottery?

From my understanding, the requested number is given by $$\binom{10}{6}\times \binom{44}{4}$$

Am I right? Thank you very much for you help.

For the sake of completeness, here is the original statement taken from the book:

The rules of a lottery are as follows: You select 10 numbers between 1 and 50. On lottery night, celebrity mathematician Richard Thomas chooses at random 6 “correct” numbers. If your 10 numbers include all 6 correct ones, you win. Work out your chance of winning the lottery.

Answer 1

A lot of controversy has erupted over the meaning of the problem. To me, it seems that looking at the epithet eminent, the mathematician has been entrusted the task of choosing $6$ numbers which are "correct" (ie winning numbers) instead of, say a computerised selection, and the question is how many ways "you" can win the lottery.

The numbers are divided into two: winning numbers $(6)$ and non-winning numbers $(44)$ You win only if you have drawn all $6$ winning numbers and $4$ non-winning numbers

Then # of ways of winning $=\binom66\times\binom{44}4$, and the upper binomial indices must add up to $50$, as I had commented.

And the probability of winning the prize would be $Pr = \dfrac{\binom66\binom{44}4}{\binom{50}6}$

Different interpretations will obviously give different answers !

Answer 2

The number of ways Thomas can choose $6$ winning numbers = $^{50} C_6= w$.

The number of possible number combinations you can end up with $^{50} C_{10} = b$.

Of the $b$ possibilities, you must have bought the one $6$ numbers of which match Thomas' winning number set, there are $w$ possibilities for that.

Answer 3

I post here an answer based on the comments. Different interpretations lead to different results.

1. $\binom{6}{6} \times \binom{44}{4}$: (see this answer)

   • In this interpretation, one first focus on selecting all 6 correct numbers out of the 6 numbers drawn by the mathematician.
   • Then, one needs to choose the remaining 4 numbers out of the 44 incorrect numbers not drawn by Richard.
   • By multiplying these two binomial coefficients together, one calculates all possible winning combinations where you've chosen all 6 correct numbers and 4 incorrect numbers out of the total pool of 50 numbers.

2. $\binom{50}{6}\binom{44}{4}=\binom{50}{10}\binom{10}{6}$: (thanks to users @Haris et @user469053)

   • Firstly, one select 6 numbers out of the 50 total numbers available.
   • Then, you're selecting the remaining 4 numbers out of the 44 numbers not chosen by Richard. This is represented by $\binom{44}{4}$.
   • Alternatively, one could first choose the 10 numbers out of the 50 total numbers, represented by $\binom{50}{10}$, and then Richard selects 6 numbers from your chosen 10, represented by $\binom{10}{6}$.