Lottery probability, picking 10 numbers to get 6 winning numbers

Question

What are your chances of winning in the following scenario?

Consider a lottery game with 49 numbers and 6 are chosen as the winning numbers. You're allowed to select 10.

What is the probability that your selection contains all 6 winning numbers? (the remaining 4 can be anything else)

Please confirm if this is correct:

${43 \choose 4}$ / ${49 \choose 10}$

Reasoning:

From a selection of 10 numbers, there are ${43 \choose 4}$ combinations which contain the winning 6; i.e. ${6 \choose 6}$ * ${43 \choose 4}$

divided by the total number of combinations with 10 numbers: ${49 \choose 10}$

Answer 1

Going by what your question says, the remaining $4$ tickets don't matter, what is important is the winning $6$ tickets. And since you have all $6$ tickets, you are definitely going to win. So probability is $1$.

Hope this helps you.

Answer 2

Your logic of finding the probability of winning the lottery by choosing $10$ numbers from $49$ if the ten contain all six winning numbers is correct, and it is surprising that none of the answers previously posted have agreed to it !

P(win) = $\frac{\text{(Get all 6 winning numbers)*(Any 4 others)}}{\text{(Get any 10 numbers)}}$

which in full form gives $P(win) =\dfrac{\binom66\binom{43}4}{\binom{49}{10}}$

which can be abbreviated to

$P(win) =\dfrac{\binom{43}4}{\binom{49}{10}}$

Answer 3

In this case your winning probability is 1 as you have all 6 winning numbers.