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A fair coin is tossed four times. Let the random variable $X$ denote the number of heads in the first 3 tosses, and let the random variable $Y$ denote the number of heads in the last 3 tosses. What is the joint pmf of $X$ and $Y$?

The solution is the following table, which I am having trouble with.

$$ \begin{array}{l|llll} X \backslash Y & 0 & 1 & 2 & 3 \\ \hline 0 & 1/16 & 1/16 & 0 & 0 \\ 1 & 1/16 & 3/16 & 2/16 & 0 \\ 2 & 0 & 2/16 & 3/16 & 1/16 \\ 3 & 0 & 0 & 1/16 & 1/16 \end{array} $$ (Original image here).

I do not understand $(1,1)$. Where do they get $3/16$ from? The only possibility of only $1$ head in the first $3$ tosses and only $1$ in the last $3$ tosses is HTTH, hence it should be $1/16$?

Furthermore I do not understand $(2,2)$. Where do they get $3/16$ from? The only possibility of only $2$ heads in both the first $3$ tosses and the last $3$ tosses is THHT, hence it should also be $1/16$?

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    $\begingroup$ For (1,1) you can also have THTT and TTHT and for (2,2) HTHH and HHTH $\endgroup$ – Elsa Aug 22 '18 at 15:23
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For $(1,1)$, note that in contrary to what you are writing, $$ \{X=1, Y=1\} = \{HTTH, THTT, TTHT\} $$ and for $(2,2)$ the same trick works $$ \{X=2, Y=2\} = \{THHT, HHTH, HTHH\} $$

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For (1, 1), you may have HTTH, THTT, TTHT.

For (2, 2), you may have THHT, HTHH, HHTH

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