Probability of getting different results when tossing a coin Here's a question I got for homework:

In every single time unit, Jack and John are tossing two different
  coins with P1 and P2 chances for heads. They
  keep doing so until they get different results. Let X be the number of
  tosses. Find the pmf of X (in discrete time units). What kind of
  distribution is it?

Here's what I have so far:
In every round (time unit) the possible results 
HH - p1p2 
TT - q1q2
TH - q1p2
HT - q2p1

and so P(X=k) = ((p1p2 + q1q2)^(k-1))*(q1p2+q2p1)
Which means we're dealing with a geometric distribution. 
What doesn't feel right is that the question mentions 'discrete time units'. That makes me think about a Poisson distribution, BUT - Poisson is all about number of successes in a time unit, while here we only have one round in every time unit. 
If I'm not too clear its only because I'm a little confused myself. Any hint would be perfect. Thanks in advance
 A: You may feel some discomfort, but you clearly understood the problem, and arrived at the correct answer using correct reasoning.   
You should be explicit about what $k$ ranges over.  Possibly you should say explicitly what you mean by $q_1$ and $q_2$. Maybe you should explicitly let $r=q_1p_2+q_2p_1$ and observe that $p_1p_2+q_1q_2=1-r$.   
For the sake of completeness, you should separate out the cases $p_1=p_2=1$ and $p_1=p_2=0$, for which the problem (and answer) do not make sense.
A: You can work out the probability that they get different results on the first toss, namely $p_1 (1-p_2)+ (1-p_1)p_2 = p_1+p_2 - 2p_1 p_2$.  
If they have not had different results up to the $n$th toss, then the conditional probability they get different results on the next toss is the same; this is the memoryless property and so (since the number of tosses is a positive integer, i.e. discrete) you have a geometric distribution, as you spotted.
A: The discrete time units refer to the series of flips.  You can't have the process end after $1.5$ flips, only 1, 2, etc.  Maybe that exponent $k-1$ will give you a hint to the name of the distribution.
