from 34 pick 5, repeat 11 times. Expected number of matches between the 11 trials? Let's assume we're using a deck of cards with $34$ cards. Shuffle the deck, and draw $5$ cards. Record the hand. After returning the hand to the deck, shuffle and repeat the trial. When looking over the records, what is the expected number of matches?
The question I'm asking is similar to having people in a room and checking birthdays. With $23$ people in a room there's a $50\%$ chance that $2$ will share a birthday. Above, with $11$ trials, how many matches would any $2$ expect?
I've worked on this, and I'm steering myself in circles. 
EDIT:
I'm defining a match as: two trials sharing a single card. If $2$ players share $2$ cards, that's $2$ matches. I'm trying to find the expected value for matches.
For the $11$ trials: Above I said shuffle the deck and repeat the trial. I should've said "repeat $11$ times"
We could restate the problem as $11$ players:
$11$ players choose $5$ cards each from independent decks of $34$. How many cards should any given player expect to share with someone else in the group?
 A: It is not totally clear what you are looking for:


*

*Any particular card has a probability $\frac{5}{34}$ of being picked by a particular player. Thus any particular card has a probability $\frac{5^2}{34^2}$ of being picked by both of two  particular players. The expected number of cards picked by both of two particular players is the expected number of cards a particular player shares with another particular player and can be calculated as $34 \times \frac{5^2}{34^2}$ or as $5 \times \frac{5}{34}$ so is $\frac{25}{34}\approx 0.735$. So repeating this in $11$ trials, you multiply by $11$ and get that you would expect a total such pairs between the two players across the $11$ rounds of $\frac{275}{34}\approx 8.088$ 

*With $10$ players and a single round, the probability that a particular card is not picked by any of them is $\frac{29^{10}}{34^{10}}$ so the probability a particular card is picked by at least one of them is $\frac{34^{10}-29^{10}}{34^{10}} \approx 0.796$. So with $11$ players, the number of cards a particular player expects to share with at least one of the other $10$ is $5$ times this, about $3.981$

*With $11$ players and a single round, the probability that a particular card is not picked by any of them is $\frac{29^{11}}{34^{11}}$ and the probability a particular card is picked by exactly one of them is $11 \times \frac{5}{34}\times \frac{29^{10}}{34^{10}}$ so the probability a particular card is picked by at least two of eleven is $\frac{34^{11}-29^{11}-55\times 29^{10}}{34^{11}} \approx 0.497$. So the expected number of cards picked by at least two of eleven is $34$ times this, about $16.881$


The first result differs from the second for various reasons, mainly because in the first result the second player effectively chooses $55$ cards for comparison with the first player's cards while in the second result the $10$ other players choose an expected $34 \times  \frac{34^{10}-29^{10}}{34^{10}} \approx 27.071$ distinct cards for comparison.  So your restatement leads to a different question  
