3 people dealt 3 cards, probability that nobody recieves 3 of a kind

Question

If $3$ people are dealt $3$ cards from a standard deck, determine the probability that none of them is dealt three of a kind?

Here is my attempt:

The total number of hands is $${_{52}\mathsf C}_3\times{_{49}\mathsf C}_3\times{_{46}\mathsf C}_3=7.75262759\times 10^{12}.$$ The number of ways we can deal $3$ cards to all players so that player $1$ receives $3$ of a kind is:

$$13\times {_4\mathsf C}_3\times {_{49}\mathsf C}_3\times {_{46}\mathsf C}_3=1.454316864\times10^{10}.$$

The number is the same for player $1$ or player $2$.
The number of ways $2$ players receive three of a kind is: $$13\times{_4\mathsf C}_3\times 12\times{_4\mathsf C}_3\times{_{46}\mathsf C}_3 =37889280.$$ this number applies for p1&p2, p1&p3, p2&p3.

Finally the number of ways all $3$ players receive $3$ of a kind is:

$$13\times {_4\mathsf C}_3\times 12\times {_4\mathsf C}_3\times 11\times {_4\mathsf C}_3=109824.$$

Using the inclusion exclusion principle the number of ways no player receives a $3$ of a kind is: $$(7.75262759\times 10^{12})-(3\times 1.454316864\times 10^{10})+(3\times 37889280)+(109824) = 7.709111642\times 10^{12}.$$

Dividing this by the total number of hands gives a probability of $0.994386$.

This probability seems relatively high to me, I was just wondering if anybody agrees with my working, I not can you see where I have gone wrong. Cheers.

Answer 1

You have incorrectly written $+ 109824$ instead if $-109824$. But evidently you actually did subtract, because your answer is correct.

Answer 2

I haven't checked every detail, but it looks good, and the probability isn't too high. Three of a kind in three cards is quite unlikely – two players getting one is so unlikely that you can get a very good estimate just by calculating the probability that one player gets one and multiplying it by $3$:

$$ 1-3\cdot\frac{13\cdot\binom43}{\binom{52}3}=1-\frac3{425}\approx0.992941\;. $$