Texas Hold 'Em textbook question "Texas Hold-em is a poker game in which players are each dealt two cards face
down (called your hole or pocket cards), from a standard deck of 52 cards, followed by a round
of betting, and then five cards are dealt face up on the table with various breaks to permit players
to bet the farm. These are communal cards that anyone can use in combination with their two
pocket cards to form a poker hand. Players can use any five of the face-up cards and their two
cards to form a five card poker hand. Probability calculations for this game are not only required
at the end, but also at intermediate steps and are quite complicated so that usually simulation is
used to determine the odds that you will win given your current information, so consider a simple
example. Suppose we were dealt 2 Jacks in the first round.
(a) What is the probability that the next three cards (face up) include at least one Jack?
(b) Given that there was no Jack among these next three cards, what is the probability that there
is at least one among the last two cards dealt face-up?"
I don't know how to do (b). Uhh...I don't really think I got the rules of Texas Hold 'Em from this short description. (a) and (c) were questions that weren't really about the game rules itself, but what does "last two cards dealt" mean? Aren't we already told that the last two cards dealt are all jacks?
 A: (a) It’s easier to calculate the probability that none of the first three face up cards is a jack. You’ve seen two cards $-$ your hole cards $-$ and both of them are jacks. Thus, there are $50$ cards that you’ve not seen, and $2$ of them are jacks. The probability that the first face-up card is not a jack is therefore $\frac{48}{50}$. Assuming that it’s not a jack, there are still $2$ jacks left among the $49$ cards that you’ve not seen, so the probability that the next face-up card is not a jack is $\frac{47}{49}$. Repeating the reasoning one more time, we see that if neither of the first two face-up cards is a jack, the probability that the third face-up card is also not a jack is $\frac{46}{48}$. Thus, the probability that none of the first three face-up cards is a jack is
$$\frac{48}{50}\cdot\frac{47}{49}\cdot\frac{46}{48}=\frac{47\cdot46}{50\cdot49}=\frac{1081}{1225}\approx0.88245\;,$$
and the probability that at least one is a jack is therefore 
$$1-\frac{48}{50}\cdot\frac{47}{49}\cdot\frac{46}{48}=\frac{144}{1225}\approx0.11755\;.$$
(b) Given that there is no jack in the first three face-up cards, there must still be $2$ jacks unaccounted for among the $52-4=47$ cards that you’ve not seen. The reasoning used in (a) still applies: the probability that the fourth face-up card is not a jack is $\frac{45}{47}$, and if it’s not a jack, the probability that the fifth and last face-up card is not a jack is $\frac{44}{46}$. The probability that neither is a jack is therefore
$$\frac{45}{47}\cdot\frac{44}{46}=\frac{990}{1081}\approx0.91582\;,$$
and the probability that at least one is a jack is 
$$1-\frac{45}{47}\cdot\frac{44}{46}=\frac{91}{1081}\approx0.08418\;.$$
