# Combinatorics - Probability of drawing AT LEAST these three cards from a set of six cards with no replacement.

(Okay, brief background: I am designing a board game and developed the following problem to make sure I understand combinatorics properly in order to balance the number of cards in my game.)

There are six cards in a deck, labeled A$_1$, A$_2$, B$_1$, B$_2$, B$_3$, and C. Assuming this deck is well-shuffled, what is the probability of drawing at least one 'A' and at least one 'B' card in a 3-card draw with no replacement?

My strategy so far has been twofold:

I first wrote out all of the possible combinations of a three-card draw from this deck and counted how many of the combinations have at least one 'A' and at least one 'B.' 15 of the 20 possible draws meet this criteria, suggesting that there is a $\frac{3}{4}$ probability.

My second strategy attempted to reproduce this result more mathematically, using combinatorics (i.e. $\left(\begin{smallmatrix} n \\ k \end{smallmatrix}\right)=\frac{n!}{k!(n-k)!}$). My logic is that there are $\left(\begin{smallmatrix} 2 \\ 1 \end{smallmatrix}\right)$ combinations to draw one 'A,' $\left(\begin{smallmatrix} 3 \\ 1 \end{smallmatrix}\right)$ combinations to draw one 'B,' and the last draw can be any of the remaining four cards so $\left(\begin{smallmatrix} 4 \\ 1 \end{smallmatrix}\right)$. The total number of combinations is $\left(\begin{smallmatrix} 6 \\ 3 \end{smallmatrix}\right)$. However, when I ran the calculation:

$\frac{\left(\begin{smallmatrix} 2 \\ 1 \end{smallmatrix}\right)*\left(\begin{smallmatrix} 3 \\ 1 \end{smallmatrix}\right)*\left(\begin{smallmatrix} 4 \\ 1 \end{smallmatrix}\right)}{\left(\begin{smallmatrix} 6 \\ 3 \end{smallmatrix}\right)}=\frac{2*3*4}{20}=\frac{24}{20}$

which is both not the result I initially came up with and a probability over 1, which is impossible.

You are counting certain instances multiple times. Say you select $A_1$, then $B_1$, and then $A_2$. This exact same selection happens if you select $A_2$, then $B_1$, and then $A_1$. (See edit below for more details).

Another approach would be to count how many combinations there are without having at least one $A$ and one $B$, and then subtracting from the total. There are $\binom{4}{3}$ ways to not have any $A$'s, $\binom{3}{3}$ ways to not have any $B$'s, and it's impossible to not have any of either, so we get $$\frac{\binom{6}{3}-\binom{4}{3}-\binom{3}{3}}{\binom{6}{3}} = \frac{15}{20} = \frac{3}{4}.$$

EDIT: In your count, you choose one $A$, then one $B$, and then one card remaining. Enumerating, we obtain:

$A_1B_1A_2$

$A_1B_1B_2$

$A_1B_1B_3$

$A_1B_1C$

$A_1B_2A_2$

$A_1B_2B_1$

$A_1B_2B_3$

$A_1B_2C$

...and so forth. However we have included $A_1B_1B_2=A_1B_2B_1$ twice. Continuing in this way, we will double count all counts of the form $A_iB_jA_k$ and all counts of the form $A_iB_jB_k$. There are $\binom{2}{1}\binom{3}{1}\binom{1}{1}=6$ of the first type and $\binom{2}{1}\binom{3}{1}\binom{2}{1}=12$ second type. Thus we have $6/2+12/2=9$ extra counts. This is consistent because $24-9=15$.

• There is a typo in the number of draws without an $A$: should be $\binom{4}{3}$ as you correctly wrote below. – Evargalo Aug 7 '17 at 14:51
• Thank you for the alternative approach, that's helpful. However, it was my understanding that the $k!$ in the denominator of $\frac{n!}{k!(n-k)!}$ was used in unordered combinations to avoid overcounting...is that the case, and if so, why is my approach flawed? – user470062 Aug 7 '17 at 14:59
• @user470062 : ​ ​ ​ Your approach is flawed because you're not doing a correction for your otherwise-overcounting. ​ (The k! is a correction for the combination-formula's otherwise-overcounting; not for your overcounting.) ​ ​ ​ ​ ​ ​ ​ ​ – user57159 Aug 7 '17 at 15:24
• @user470062 It is true that the $k!$ prevents overcounting in combinations. However your count consists of multiple combinations, so things can possibly go wrong. See my latest edit for a detailed explanation as to why your count doesn't work. – John Griffin Aug 7 '17 at 18:47