This will be the sum of the probability that exactly 10 have the same suit plus the probability that 9 of 10 have the same suit.
With replacement first. Let's focus on a single suit, say hearts.
The probability of drawing on a heart on the first draw is 1/4. Since we're replacing, that's the probability of a heart for cards two through ten. So the probability of all hearts is [tex](\frac 1 4)^{10}[/tex].
Since there are four suits, the probability of drawing all the same suit is four times the above, [tex](\frac 1 4)^9[/tex].
The probability of drawing the first 9 hearts then a different suit for the 10th is [tex](\frac 1 4)^9 \cdot \frac 3 4[/tex].
The different suit may be any of the cards so we multiply this by 10 to get the probability of 9 of 10 hearts: [tex]10 (\frac 1 4)^9 \cdot \frac 3 4[/tex].
We can have any suit, so we multiply this by a factor of 4, giving [tex]30 (\frac 1 4)^9[/tex].
The probability we seek is the sum P(10 the same) + P(9 the same)
[tex]p = (\frac 1 4)^9 + 30 (\frac 1 4)^9 = 31 (\frac 1 4)^9 = \dfrac{31}{262144}\approx 0.00012 = 0.012\% [/tex]
Answer: P(with replacement) = 0.012%
Now let's do it again without replacement.
P(Exactly 10 hearts). We have a 13/52 chance of picking a heart on the first card. Given we did, we have a 12/51 chance of picking a heart on the second card. So for all 10, we get
[tex] \frac{13}{52} \cdot \frac{12}{51} \cdot \frac{11}{50} \cdots \frac{4}{43}[/tex]
Let's write that as
[tex] \frac{13!}{3!} \cdot \frac{42!}{52!}[/tex]
Again the four suits gives us
[tex]P(\textrm{All 10 the same suit}) = 4 \cdot \frac{13!}{3!} \cdot \frac{42!}{52!}[/tex]
For 9 of 10, the last fraction 4/43 becomes 39/43, then we multiply by 10 because any of the cards might not match and 4 to cover the suits.
[tex]P(\textrm{9 of 10 the same suit}) = 4 \cdot 10 \cdot \frac{13}{52} \cdot \frac{12}{51} \cdot \frac{11}{50} \cdots \frac{5}{44} \cdot \frac{39}{43} [/tex]
[tex] = 40 \cdot 39 \cdot \frac{13!}{4!} \cdot \frac{42!}{52!} [/tex]
So a total probability of
[tex]p = 4 \cdot \frac{13!}{3!} \cdot \frac{42!}{52!} + 40 \cdot 39 \cdot \frac{13!}{4!} \cdot\frac{42!}{52!} = \dfrac{197}{27657385 } \approx .00071\%[/tex]
Answer: P(without replacement) = 0.00071%
A lot smaller.
I'd have to check this pretty hard to be sure it was right, which I haven't done.
