Suppose that two cards are dealt from a standard 52-deck poker deck...

Question

Suppose that two cards are dealt from a standard 52-deck poker deck. Let A be the event that the sum of the two cards is 8 (assume aces have a numerical value of 1). How many outcomes are in A?

I don't know how to do this without writing out each combination. I'm sure there must be an easier way or formula to do this.

So i just started counting the various events that could lead to this outcome. {1,7}, {2,6}, {3,5}, {4,4}

the first three if we assume they are all in the same house, and we have four different houses will lead to 4x3 outcomes here. Then the fourth pair obiously cannot happen within the same house since each house only has one 4 card. So we have 1 x 3 = 3 outcomes here. so far 12 + 3 = 15, but now we have to consider that the first three pairs can happen with numbers from different houses. This is where it is starting to get complicated for me and not sure what is the best process to follow to get this result. Anybody know the best way to do this question please?

Answer 1

So, you are good to start with trying to identify what cases there are. You correctly identified that you get a sum of 8 by having 1+7, 2+6, 3+5, as well as 4+4 (and if you consider order relevant, 5+3, 6+2, and 7+1 might be considered different cases as well). I would not have bothered with breaking into cases based on suits (the common word here in English is suits, not "houses") and instead gone straight ahead with rule of product.

So, for the case of $1+7$, we pick which Ace it was (there are four options) and then we pick which seven it was (there are four options). Applying the rule of product, that makes $4\times 4 = 16$ options here.

The same calculation applies for the remaining cases of $2+6$ and $3+5$ (as well as $5+3,6+2,7+1$)

For the case of two cards both with value of $4$... here we choose two of them (heavy emphasis on the phrase "choose two") to be used. To do that, we use binomial coefficients. There are $\binom{4}{2}=6$ ways to choose them. (If we wanted order to matter, that could have been phrased "pick the first four, then pick the second: $4\times 3=12$ options, or pick the two simultaneously and choose the order: $\binom{4}{2}\times 2 = 6\times 2= 12$)

Taking all of these into account, we have then:

$$16\times 3 + 6 = 54$$

outcomes if we don't care about order and:

$$16\times 6 + 4\times 3 = 108$$

if we do care about order.

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Both calculations if used for probability yield the same result: $\dfrac{54}{\binom{52}{2}} = \dfrac{108}{52\times 51} \approx 0.040724$

Answer 2

For the $\{1,7\}$ pair, there are $4$ possible choices for an ace, and $4$ possible choices for a $7$-card, i.e. $16$ total combinations.

The same applies for the $\{2,6\}$ and $\{3,5\}$ pairs.

For the $\{4,4\}$ pair, we have to calculate the number of ways we can choose $2$ cards out of $4$ cards. This is given by the binomial coefficient $\binom{4}{2}=6$. (See here for an explanation.)

In total, there are $16+16+16+6=54$ possible outcomes.

(In case that you keep track of the order of the two cards, just double that number.)