I have the following question to answer:
Cards are drawn at random, without replacement, from a well-shuffled pack of 52 cards.
- 12 cards are dealt with six face down and the others being a jack, two kings, a 7, and two 3s. If a 13th card is dealt, what is the probability that it will be a 2, 3 or 4?
Since I do not know the value of the first 6 cards, I thought that the most I can do is just giving an interval of the probability of my event. However, it seems that since I haven't seen the first 6 cards, this does not change the probability. Hence,
$$\left(\frac{4}{46}\right) + \left(\frac{2}{46}\right) + \left(\frac{4}{46}\right) = \left(\frac{10}{46}\right) = \left(\frac{5}{23}\right)$$
I am now pretty sure to be wrong. And even if I were right, I would appreciate if someone could make this whole idea of "not changing the probability" more clear.
Thanks in advance for your precious help.