Are LFSRs enough for this?
Here is an attack that I think will, with excellent odds, allow certain determination of which value is on the ticket of colluding players; and consequently give an advantage on guessing the tickets of these colluding players, and a (typically lesser) advantage on guessing the tickets of honest players. I'm assuming:


The adversary knows the value on every ticket (as confirmed by a comment), with no ambiguity about that under orientation of the ticket.
It is not known the initial state of the LFSR (as now explicit).
All players press their button haphazardly, which when the exact instant is unknown sorts of partially reseed the LFSR.
The adversary knows exactly when $m$ consecutive players pressed the button (because these $m$ players collude with the adversary), thus exactly how many steps of the LFSR there has been between theses presses; this allows the adversary to express the state of the LFSR at the key-presses of the colluding players as a linear function of the (assumed unknown) state when the first colluding player pressed the button.
$n$ is low enough that the adversary can perform some processing for each of the (at most) $n!/(n-m)!$ distinct orders of values for colluding players ($n=15$, $m=10$ means that about $11\cdot10^9$ orders are to be considered, and would be very feasible).
$t$ is large enough that it is likely that only a single of these ticket order matches the linear functions; there are $b$ unknown bits, $(n-1)\cdot t$ known bits, thus by an entropy argument (based on the vague and unproven hypothesis that the haphazard behavior of players makes the system of linear equations unremarkable), there should typically be a single possible order of values when
$$m\cdot t\gg b+\log_2(n!/(n-m)!)$$


It is then easy to check a candidate order, and deduce the $b$ bits of initial state. Just plug $m={2\over3}n$ for $2\over3$ of the players colluding.

Also, if the uncertainty on when the honest players pressed (assumed to have uniform distribution in an interval with $u$ steps, with that the product of time per frequency) is such that $u\ll2^t/(n-m)$, we can often guess the (value on the) tickets of other players immediately before and after the $m$ colluding players; even all of them if $u\ll2^t/(n-m)!$ or something on that tune.

Further (as the saying attributed to the NSA goes, and illustrated by a sizable improvement in version 5 of this answer): attacks only get better; they never get worse. In particular, I have not fully exploited the information that can be gained from the known value on the tickets of the honest players, which is significant if $u\ll2^t$; or speedups possible for many parameters to dramatically cut down on $n!/(n-m)!$ explorations; or what improvements could be achieved when colluding players press at a chosen instant, rather than at a random but known instant.



The above does not give a condition that would insure an adversary can't succeed. Here are some thoughts towards that. I'll consider that the LFSR is maximum-length (has period $2^b-1$, or equivalently uses a primitive polynomial and is never all-zero); and that a key-press occurring when the LFSR is all-one prints an all-one ticket, but leaves the LFSR as if no key-press had occurred.

Assume a player stays exactly $u=2^b-1$ clock periods in the room, and consider the 4 variables


LFSR state $I$ on entrance
LFSR state $P$ as sampled by key-press (of which $t$ out of $b$ bits are printed)
LFSR state $O$ on exit
number of clock periods $w$ between entrance and key-press, with $0\le w&lt;2^w$


It holds that $P=M^w\cdot I$ and $\hat P=M^w\cdot O$, where $\hat P$ is the bitwise complement $\bar P$ of $P$ except that all-ones is unchanged; and $M$ is the $b\times b$ sparse matrix corresponding to advancing the LFSR by one step.
When any of $I$ $P$ $O$ $w$ is fixed, these relations defines a one-one mapping from any of the other three variables to either of the two last variables. Note: when $w$ is unknown, finding it (or $P$ if also unknown) seems to be a discrete logarithm problem; I consider an adversary able to solve that.

It follows that when an honest player stays in the room for much longer than $2^b-1$ clock periods, to an adversary knowing only one of the LFSR states on entry or exit,


the other of these states is indistinguishable from a uniformly random non-zero variable,
the value printed on the ticket has the same distribution as for a random sampling (all-zero with odds $2^{b-t}-1\over2^b-1$, any of the other $2^t-1$ values with higher odds $2^{b-t}\over2^b-1$).


The above two unknowns are however strongly related.

If we see the LFSR as an entropy pool,


an honest player staying in the room for much longer than $2^b-1$ clock periods refills that entropy pool to its full capacity $c=\log_2(2^b-1)$ bit
however knowledge of the value of a ticket reveals at most $r=\log_2(2^b-1)-\log_2(2^{b-t}-1)$ bit of information when $t&lt;b$ (this occurs when the value printed is all-zero), or $r=\log_2(2^b-1)$ bit when $t=b$ (the value printed can't be all-zero).


From this I tentatively conjecture that with all but very small $b$, the adversary has negligible advantage if at least $h$ honest players, who each stay in the room for much longer than $2^b-1$ periods, stand between any two players colluding with the adversary, and
$$h\cdot c\ge (h+1)\cdot r \text{ , that is } t&lt;b \text{ and } (h+1)\cdot\log_2(2^{b-t}-1)\ge\log_2(2^b-1)$$

The condition is constructed as: in a sequence of $h+1$ players, at least as much entropy is injected as revealing all the ticket values consumes. My intuition is that this condition is enough that the adversary can't learn more about the state of the LFSR when a player presses a key than knowledge of the corresponding ticket value would reveal, and/thus is unable to tell if an hypothesis on which ticket belongs to who is more or less likely.