[Math] Probability of two cards with the same face value

Question

In (one-deck) blackjack, what's the probability that you are dealt two cards with the same face value?
I think it's 29/221…
Here's why: (Would love some feedback!)
P(2 aces)=(4/52)(3/51)=1/221
P(2 twos)=(4/52)(3/51)=1/221
P(2 threes)=(4/52)(3/51)=1/221
P(2 fours)=(4/52)(3/51)=1/221
P(2 fives)=(4/52)(3/51)=1/221
P(2 sixes)=(4/52)(3/51)=1/221
P(2 sevens)=(4/52)(3/51)=1/221
P(2 eights)=(4/52)(3/51)=1/221
P(2 nines)=(4/52)(3/51)=1/221
P(2 tens)=(16/52)(15/51)=20/221
I get this last line from 4 10's, 4 jacks, 4 queens, and 4 kings which make the 16/52…
So, 9(1/221)+20/221=29/221 which is approximately: 0.0905
Am I doing this right?

Answer 1

Yes, but that's the hard way.

The probability that you are dealt one card of any face and then another card of the same face value is: $$\begin{align}\frac{36}{52}\frac{3}{51}+\frac{16}{52}\frac{15}{51} ~ = & ~ \frac{29}{221} \\[1ex] \approx & ~ 0.1312...\end{align}$$

Since 36 of 52 cards are not 10-valued, and 16 of the 52 are all 10-valued.

Complication: an Ace is either 1-valued or 11-valued.   Should the event of drawing two aces count as having two of the same valued cards?   If not, the answer has to be adjusted.