Drawing pair of cards. Did my brother play fair or unfair? The Game
There is a pool of $n$ cards that are marked either by a A or by a B. There is a proportion $p$ of the cards that are marked by a A and a proportion $1-p$ of the cards that are marked by a B. The game is the following. I randomly draw a card and put it back in the pool. I then randomly draw a second card. These two first cards form a pair of cards. A pair can be either AA, AB or BB. I take note of this pair on my spreadsheet and replace the second card in the pool as well. I repeat the draws up to the time I drew $n$ pairs of cards to get $n$ entries on my spreadsheet.
The Probabilities for each pair to be AA, AB or BB are $p^2$, $2p(1-p)$ and $(1-p)^2$  respectively. Therefore, the probability that at the end of the game I drew $x$ AA is $p^{2x} + (1-p)^{n-x}$. The probability that at the end of the game I drew $y$ AB is $2p(1-p)^y + (1-2p(1-p))^{n-y}$. The probability that at the end of the game I drew $z$ BB is $(1-p)^{2x} + p^{n-x}$.
Everything seems correct up to this point?
Question
I ask my little brother to play this game and he can basically do two things. He either plays fair and chose to draw the cards randomly or he can play unfair and chose to draw the first card randomly and then draw the second card non-randomly so that he may end up with lots of AA and BB or with lots of AB for examples. If he plays unfair he may, for example, chose that every time he draws an A at first hand, his probability to draw another A to complete the pair becomes $p+\Delta p$ resulting to an excess of AA and a lack of AB.
So he plays and he observes $x$ AA, $y$ AB and $z$ BB, where $x+y+z=n$ by definition. The values $x$, $y$, $z$ differ from the expectation of $p^2$, $2p(1-p)$ and $(1-p)^2$. How can I tell with a 95% confidence interval (using the distributions of my second paragraph) whether my brother played fair or unfair?
 A: Those probabilities for $ x, y, z $ don't look right.  You basically have a trinomial sampling distribution (or multinomial with dimension 3).  That is $(x, y, z)\sim Trinomial (n, p^2,2p (1-p), (1-p)^2) $
You can use this to get 95% marginal CIs for each of $ x, y, z $ by using the binomial distribution.  These have the form
$$ lower_x = n(p^2) - 1.96\sqrt{n (p^2 (1-p^2))}$$
$$ upper_x = n(p^2) + 1.96\sqrt {n (p^2 (1-p^2))}$$
$$ lower_y = n(2p(1-p)) - 1.96\sqrt{n (2p (1-p) (1-2p (1-p)))} $$
$$ upper_y = n(2p (1-p)) + 1.96\sqrt{n (2p (1-p) (1-2p (1-p)))}$$
$$ lower_z = n((1-p)^2) - 1.96\sqrt{n ((1-p)^2 (1-(1-p)^2))}$$
$$ upper_z = n((1-p)^2) + 1.96\sqrt{n ((1-p)^2 (1-(1-p)^2))} $$
Of course these are all based on the large $ n $ scenario, so that the normal approximation to the binomial can be applied.
You could also use the chi-square or deviance test.
$$ X^2=\sum_k \frac {(O_k-E_k)^2}{E_k} $$
$$ D =2\sum_k O_k\log\left(\frac {O_k}{E_k}\right) $$
The $ O_k $ are the observed numbers $(x, y, z) $ and the $ E_k $ are the expected numbers $(np^2,2np (1-p),n(1-p)^2) $.  If $ n $ is large then the two numbers should be about the same, and both would have an approximate chi-square distribution with $2$ degrees of freedom (as there are $3$ categories).
A: I suggest a chi-square test that compares the observed little brother frequences to the theoretical frequencies (in the ratio of 1:2:1 for AA, AB, and BB, respectively).
