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This is the book's answer. Seems that it MUST be incorrect. This is WITH replacement.

I calculate that the prob of drawing and ace first is 1/13 and a 10 second is also 1/13. So, it would seem that the prob of drawing an Ace first or a 10 second would be 1/13 + 1/13 - 1/169, correct?

Their math doesn't even seem to make sense with the subtrace the Pace 1st and ten 2nd) they use MY numbers not theirs!

Is the book incorrect?

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    The book is indeed incorrect: someone apparently intended to write $\frac4{52}$ and then reduce it to $\frac1{13}$ and instead used the numerator from the first form and the denominator from the second. – 2017-01-09
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    Thank you. That's what I thought. – 2017-01-09
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    I think you are also correct that even if we use the fractions $4/52$ or $1/13$, the calculation was set up for the "with replacement" case and not for probabilities "without replacement." – 2017-01-09
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    It's also true that, once the erroneous reduction of the fractions is accounted for, the probability is computed for the case of replacement. – 2017-01-09
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    And even a little common sense suggests that the probability of drawing either an ace first or a ten second can't be more than 0.6. It's easier to get either an ace or a ten in a single draw, and that only has probability $2/13$. – 2017-01-09
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    Someone forgot to make a quick sanity check. About 61% feels like way too much for such an event. Sanity check is our friend, if it does't feel right, either we are wrong, or we need to study it some more, or we are facing one of those rare mind bending situations. – 2017-01-10

2 Answers 2

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There are $52\cdot 51$ ways to draw the first two cards (without replacement).

Of these, $4\cdot 51$ have an Ace first (four possible aces, $51$ choices for the other card.

Of the hands without an Ace first, there are $44\cdot 4$ ways to choose a non-ten first and a ten second, and another $12$ ways to choose a ten first and a ten second.

The total is $204+176+12 = 392$ and the probability you want is $$ \frac{292}{2652} = \frac{98}{663} $$ which indeed is a tad less than $\frac{2}{13}$.

EDIT Corrections have been made, pointed out by Fabio Somenzi.

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    It seems to me that there are $44\cdot 4$ ways to draw neither ace nor ten as first card and a ten as second card. In addition, there are $4 \cdot 3$ ways to draw a ten both times. – 2017-01-09
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    I get $(204 + 204 - 16)/2652 = 392/2652 \approx 0.1478,$ which agrees to three digits with my simulation. – 2017-01-10
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    @BruceET That's correct. – 2017-01-10
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    @Fabio Somenzi You are aright on both points. I have now edited the answer. – 2017-01-10
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We have several answers. This drives me to try a simulation in R statistical software, which should be accurate to about 3 places.

Denote 'ace on the first' as $A$ (f.a in the simulation program), and 'ten on second' as $B$ (s.10 in program). Then $P(A) = P(B) = \frac{204}{2652} = 0.076923$ and $P(AB) = \frac{16}{2652}.$ Hence $$P(A \cup B) = 392/2652 = 0.147813.$$

m = 10^6;  f.a = s.10 = logical(m)
deck = rep(1:13, each=4)
for (i in 1:m) {
  draw = sample(deck,2)
  f.a[i] = (draw[1]==1)    # TRUE if first is Ace
  s.10[i] = (draw[2]==10)  # TRUE if 2nd is Ten
  }
mean(f.a);  mean(s.10);  mean(f.a|s.10)
## 0.076932    # aprx P(A)
## 0.076945    # aprx P(B)
## 0.147853    # aprx P(A or B)