This question came up just for fun when a friend told me about a TV show. The idea of the show is to put $10$ men and $10$ women for a couple of weeks together and each one has to find their partner (which is determined before but the people in the show don't know it of course). At the end of the week, one person from men or women (changes each week) will pick a partner and then it goes on until we end up with $10$ couples. Then the group gets told, how many "matches" they hit. That continues each week so they get more and more clues how to achieve $10$ matches.
My question now regards the first show where they have no knowledge. What is the expected value for the matches?
So to sum it up:
$10$ people group $A$
$10$ people group $B$
Pairs are only formed with $A+B$, $A+A$ or $B+B$ is not allowed. Also, each person can only belong to one pair.
Each $A$ has a perfect $B$. Each $A_1$ picks a B at random, then $A_2$ picks a $B$ at random from the remaining $B$s.
What I got so far:
Not much. For the first person the chance to hit the perfect match is $1$ out of $10$, so $10\%$. After that is splits up really fast depending on who A1 chose. If $A_1$ chose its perfect match, $A_2$ has a $1$ out of $9$ chance. If $A_1$ chose the match of $A_3$, then $A_2$ still has a $1$ in $9$ chance. If $A_1$ chose the match of $A_2$, then it's $0\%$, but $A_3$ then has $1$ in $8$ and so on.
Maybe someone knows an easy way to calculate this, it is just for fun so don't worry to much about it.
We will use linearity of expectation.
The probability that a given person in group $A$ is matched with the correct person in group $B$ is $1/10$. Since there are $10$ people in group $A$, the expected number of people in group $A$ who are matched with the correct person in group $B$ is $$10 \cdot \frac{1}{10} = 1$$