Lottery probability help Lottery probability help
Think lottery (1-49) and the fact that you have 1 in 57 (1 in 56.66) chance of matching 3 numbers of 6 being chosen and 6 drawn.
NOW?! HERE is the question...
What are the odds of STILL matching 3 when 10 are chosen (ie just on paper) BUT still 6 are drawn??
So, in short
What are the odds of matching 3 numbers out of 10 (simply written on paper before draw) via a 6 numbers drawn from 1-49 lottery draw?
 A: Think hypergeometric distribution. For the population size $N$, the chosen numbers $K$ and number of draws $n$, the probability of matching $k$ equals:
$$
     \frac{\binom{K}{k} \binom{N-K}{n-k}}{\binom{N}{n}}
$$
The first case: The population size $N=49$, the number of chosen numbers $K=6$, the number of drawn numbers $n=6$. The probability of matching $k=3$:
$$
    \frac{\binom{6}{3} \binom{43}{3}}{\binom{49}{6}} \approx \frac{1}{56.6559}
$$
For the second case: the population size $N=49$, the number of chosen numbers $K=10$, the number of drawn numbers $n=6$. The probability of matching $k=3$:
$$
    \frac{\binom{10}{3} \binom{39}{3}}{\binom{49}{6}} = \frac{45695}{582659} \approx \frac{1}{12.751}
$$
A: For a general solution, let:
\begin{eqnarray*}
N &=& \mbox{Number of numbers in total} \\
D &=& \mbox{Number of numbers drawn} \\
S &=& \mbox{Number of number selected (by the player)} \\
M &=& \mbox{Number of matches.}
\end{eqnarray*}
There are $\binom{D}{M}$ ways of choosing $M$ numbers from $D$ to match with.
There are $\binom{N-D}{S-M}$ ways of choosing the other $S-M$ numbers from $N-D$ to not match with.
There are $\binom{N}{S}$ total possible selections of $S$ numbers from $N$.
Therefore, the probability of getting exactly $M$ matches is:
$$\dfrac{\binom{D}{M} \binom{N-D}{S-M}}{\binom{N}{S}}.$$
Your specific question has:
\begin{eqnarray*}
N &=& 49 \\
D &=& 6 \\
S &=& 10 \\
M &=& 3
\end{eqnarray*}
So the probability is
$$\dfrac{\binom{6}{3} \binom{43}{7}}{\binom{49}{10}} \approx 0.0784\;\; (1 \mbox{ in } 12.75).$$
