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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

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    Your solution is correct. "_Discrete time units_" probably refers that you count the number tosses, rather than the actual time when they were performed.2011-12-12

3 Answers 3

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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.

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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.

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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.