Calculating probability with extra knowledge about deck Say I separate 10 cards from a standard 52 card deck. If I draw a card from these 10 cards, the probability of drawing a card less than 5 is 16/52. A friend looks at the remaining 42 cards and names off some of the cards. The friend also looks at the 10 separated cards and names off a few of those cards. He does not tell me the positions of the cards. How does this extra knowledge affect the probability?
For example, the friend tells me there is a 2 in both the remaining 42 cards and the 10 separate cards.
For the first situation (knowledge about cards in the remaining 42), I have come up with this formula:
$$\frac{baseline - known_{target}}{total - known_{total}}$$
Where $known_{target}$ is the number of cards less than 5 that we know about. And $known_{total}$ is the number of cards we know about, regardless of their value. So, for the example above, it would give us:
$$\frac{16 - 1}{52 - 1} = \frac{15}{51}$$
I believe this gives the correct probability, if we know a card is not in the set. I don't know, however, how the second situation (knowledge about cards in the separate 10) affects the probability calculation.
 A: If your friend looks at the $10$ cards and reports whether or not there is at least one $2$ (where the value $2$ is known before he looks) you have a random sample of all 10 card hands with/without $2$'s.  If there are no $2$'s, you can imagine that you start with the deck, throw away the four $2$'s, and draw a card.  The chance it is under $5$ is now $\frac {12}{48}.$  To get the chance of a card below $5$ when we are told that there is at least one $2$ we can do the following:
Chance of a card below a $5$ without information about the $2$'s is $\frac {16}{52}=\frac 4{13}$
Chance the $10$ cards do not contain a $2$ is $\frac {{48 \choose 10}}{{52 \choose 10}}=\frac {246}{595}$
Chance the $10$ cards do not contain a $2$ and we get less than $5:  \frac{12}{48}=\frac 14$
Chance the $10$ cards contain at least one $2$ and we get less than $5: \frac 4{13}-\frac{246}{595\cdot 4}=\frac {3161}{15470}\approx .20433$
Chance the $10$ cards contain a $2$ is $1-\frac {246}{595}=\frac {349}{595}$
Chance we get less than $5$ given that there is at least one $2: \frac{\frac {3161}{15470}}{\frac {349}{595}}=\frac{3161}{9074}\approx0.348$
