I have the following problem:
So my main question is how am I supposed to choose the rank of the cards when trying to calculate the probability of one pair? I don't have the usual 52 deck of cards so I don't have joker, queen and king (I assume the digit "1" is supposed to represent ace).
If it would be a 52 deck of cards it would turn out that you can choose 13 ranks, but I have a trouble of deciding how to choose the rank? Should I focus on all the matching digits with the four different colors? In that case I can only match up to the digit 7.
Some guidence would be really appreciated!
Here's one approach, considering 16 cases. \begin{align} &a,a,8,9,10, \text{ with $a\le 7$}:&& \binom{7}{1}\binom{4}{2}\binom{3}{1}\binom{2}{1}\binom{1}{1} = 252\\ &a,a,b,8,9, \text{ with $a\le 7, b\le 7$}:&& \binom{7}{1}\binom{4}{2}\binom{6}{1}\binom{4}{1}\binom{3}{1}\binom{2}{1} = 6048\\ &a,a,b,8,10, \text{ with $a\le 7, b\le 7$}:&& \binom{7}{1}\binom{4}{2}\binom{6}{1}\binom{4}{1}\binom{3}{1}\binom{1}{1} = 3024\\ &a,a,b,9,10, \text{ with $a\le 7, b\le 7$}:&& \binom{7}{1}\binom{4}{2}\binom{6}{1}\binom{4}{1}\binom{2}{1}\binom{1}{1} = 2016\\ &a,a,b,c,8, \text{ with $a\le 7, b\le 7, c\le 7$}:&& \binom{7}{1}\binom{4}{2}\binom{6}{2}\binom{4}{1}^2\binom{3}{1} = 30240\\ &a,a,b,c,9, \text{ with $a\le 7, b\le 7, c\le 7$}:&& \binom{7}{1}\binom{4}{2}\binom{6}{2}\binom{4}{1}^2\binom{2}{1} = 20160\\ &a,a,b,c,10, \text{ with $a\le 7, b\le 7, c\le 7$}:&& \binom{7}{1}\binom{4}{2}\binom{6}{2}\binom{4}{1}^2\binom{1}{1} = 10080\\ &a,a,b,c,d, \text{ with $a\le 7, b\le 7, c\le 7, d\le 7$}:&& \binom{7}{1}\binom{4}{2}\binom{6}{3}\binom{4}{1}^3 = 53760\\ &8,8,a,9,10, \text{ with $a\le 7$}:&& \binom{3}{2}\binom{7}{1}\binom{4}{1}\binom{2}{1}\binom{1}{1} = 168\\ &8,8,a,b,9, \text{ with $a\le 7, b\le 7$}:&& \binom{3}{2}\binom{7}{2}\binom{4}{1}^2\binom{2}{1} = 2016\\ &8,8,a,b,10, \text{ with $a\le 7, b\le 7$}:&& \binom{3}{2}\binom{7}{2}\binom{4}{1}^2\binom{1}{1} = 1008\\ &8,8,a,b,c, \text{ with $a\le 7, b\le 7, c\le 7$}:&& \binom{3}{2}\binom{7}{3}\binom{4}{1}^3 = 6720\\ &9,9,a,8,10, \text{ with $a\le 7$}:&& \binom{2}{2}\binom{7}{1}\binom{4}{1}\binom{3}{1}\binom{1}{1} = 84\\ &9,9,a,b,8, \text{ with $a\le 7, b\le 7$}:&& \binom{2}{2}\binom{7}{2}\binom{4}{1}^2\binom{3}{1} = 1008\\ &9,9,a,b,10, \text{ with $a\le 7, b\le 7$}:&& \binom{2}{2}\binom{7}{2}\binom{4}{1}^2\binom{1}{1} = 336\\ &9,9,a,b,c, \text{ with $a\le 7, b\le 7, c\le 7$}:&& \binom{2}{2}\binom{7}{3}\binom{4}{1}^3 = 2240 \end{align} Adding these up yields $139160$ hands. Now divide by $\binom{34}{5}=278256$ to get the probability.
Alternatively, you can use the principle of inclusion-exclusion.
Overcount of one pair ($aa$ with $a \le 7$, $88$, $99$): $$\binom{7}{1}\binom{4}{2}\binom{34-4}{3} +\binom{3}{2}\binom{34-3}{3} +\binom{2}{2}\binom{34-2}{3}=188965 $$
Two pairs ($aabb$ with $a \le 7$ and $b \le 7$, $aa88$, $aa99$, $8899$): $$\binom{7}{2}\binom{4}{2}^2\binom{34-8}{1} +\binom{7}{1}\binom{4}{2}\binom{3}{2}\binom{34-7}{1} +\binom{7}{1}\binom{4}{2}\binom{2}{2}\binom{34-6}{1} +\binom{3}{2}\binom{2}{2}\binom{34-5}{1}=24321 $$
Full houses ($aaabb$ with $a \le 7$ and $b \le 7$, $aaa88$, $aaa99$, $88899$): $$\binom{7}{1}\binom{4}{3}\binom{6}{1}\binom{4}{2} +\binom{7}{1}\binom{4}{3}\binom{3}{2} +\binom{7}{1}\binom{4}{3}\binom{2}{2} +\binom{3}{3}\binom{2}{2} =1163 $$
Pairs minus twice two-pairs minus full houses: $$188965-2(24321)-1163=139160$$