Poker hand - difference from any full house vs a specific full house

Question

I'm having trouble understanding the solution for the first part of the question below:

Problem: In a poker hand consisting of 5 cards, find the probability P(A) of holding 2 aces and 3 jacks.

My Solution: I started solving this using the same logic as the solution for any full house, but like this:

Part 1

for the 2 aces: 13C1 * 4C2 (because out of the 13 ranks, I am interested on only 1 - the Ace)
for the 3 jacks: 13C1 * 4C3 (same reason as above, I am interested only in the appearance of the Jacks, that's why I use 13).
This would give me 13*6*13*4 = 4,056

Part 2

Then: P(A) = 4,056/2,598,960 = 0.00156, which is wrong.

Solution from book: I split in 2 parts too, where part 2 is similar to what I did.

Part 1:

The number of ways of being dealt 2 aces from 4 cards is 4C2 = 6. And the number of ways of being dealt 3 jacks from 4 cards is 4C3 = 4. By the multiplication rule, there are n = (6)(4) = 24 hands with 2 aces and 3 jacks.

Part 2

Therefore, the probability of getting 2 aces and 3 jacks in a 5-card poker hand is P(A) = 24/2,598,960 = 0.9 x 10^-5.

Question: Why is the first part calculated like that, considering only 4 cards and not the whole rank of 13 cards like in calculating any full house?

Thanks.

Thanks Jose. I really tried but now I got what was missing: the $ before and after the code. — Priscila Freitas, Commented Jan 18, 2018 at 0:06

Nick Peterson · Accepted Answer · 2018-01-17 23:56:11Z

2

When you include the $\binom{13}{1}$ term, you are saying that you can make ANY one choice out of thirteen possibilities.

What you actually want is to say that there is only one choice that works -- hence you want to multiply by $1$, not by $13$.

answered Jan 17, 2018 at 23:56

Nick Peterson

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$\begingroup$ Hi Nick, I think I'm getting it now... So just to be sure, the solution would actually be 1C1 *4C2 * 1C1 * 4C3 for the first part. Because 1C1 is 1, it was omitted. $\endgroup$
– Priscila Freitas
Commented Jan 17, 2018 at 23:58
$\begingroup$ Yes. It is quite usual to omit $^1C_1$ because it is in fact $1$, and we rarely feel the need to point out how we count ways to "select one from that set of one thing". You only ever see it when the author wants to be extra obvious; such as hen showing why a term vanishes when deriving a special case from a generic expression . $\endgroup$
– Graham Kemp
Commented Jan 18, 2018 at 0:25

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