Probability in poker In his book "The poker face of Wall Street", A.Brown wrote this:
"If two hands have two different cards ten or higher, there is a $62\%$ chance they share a card of the same rank.
If two hands are pairs or suited connectors, there is only a $7\%$ probability they share a card of the same rank.
If they are both suited hands, there is only a $19\%$ chance they share the same suit."
How did he find these numbers?
 A: For the first one, say:  There are $5$ ranks at issue here $\{10,\;J,\;Q,\;K,\,A\}$.  And of course there are $4$ possible suits for each rank.  We are assured that player $1$ has cards from exactly two of these (doesn't matter which).  That leaves $6$ cards from those two ranks and $4$ each from the other three, $18$ in all.  Look at the two relevant cards from the second player's hand, call them $X,Y$.  The probability that $X$ is not of one of the ranks covered by the first player is $\frac {12}{18}$.  The probability that $Y$ has none of the ranks covered so far (that is, the two covered by the first player and the one covered by $X$) is then $\frac 8{14}$. As we need both events we multiply to see that the probability that the players do not share the same rank is $$\frac {12}{18}\times \frac 8{14}=\frac 8{21}\sim .381$$
The desired probability is, of course $1$ minus this, or about $.619$.
I don't know what a connector is, so will not touch the second one.
I get a different answer for the last one.  Say the first player has Spades (it doesn't matter).  As $5$ spades are needed for that hand, there are only $8$ spades left in the deck.  There are $\binom 85$ ways to make a spade flush, then...and of course $\binom {13}5$ ways to make a flush out of any of the other suits.  Thus the probability that the second player also has a spade flush (given that the hand is a flush) is:  $$\frac {\binom 85}{\binom 85+\binom {13}5+\binom {13}5+\binom {13}5}\sim .143$$  Perhaps I am misunderstanding that question.
EDIT:  On rethinking, I will make a guess and say that these were meant to be hands of two cards only.  Of course, my calculation assumed five card hands.  It is easy to modify the calculation and this assumption does match the stated answer.
