Question related to Ellenberg Transylvanian lottery On page 255 of "How Not to Be Wrong: The Power of Mathematical Thinking" Ellenberg, there is mention of a lottery game called the Transylvanian Lottery. Also shown in this video: https://youtu.be/kZTKuMBJP7Y?t=1661
From the set 1, 2, 3, 4, 5, 6, 7 - choose 3 numbers. Which gives 35 (7 choose 3) possible lottery tickets with numbers like 147, 245... etc.
And if a player of this lottery gets 2 numbers out of the three on their ticket to match - it is referred to as a 'deuce'. Now there is a table showing the odds of getting 0 thru 7 deuces if the player buys 7 random tickets.
Here are the numbers from that table:
5.3% chance of no deuces
19.3% chance of exactly one deuce
30.3% chance of two deuces
26.3% chance of three deuces
13.7% chance of four deuces
4.3% chance of five deuces
0.7% chance of six deuces
0.1% chance of all seven tickets being deuces.

My question is how are these percentages from this table calculated?
To me, it seems like if the called numbers in the lottery are 1, 4 and 7 - then there are 12 out of the 35 tickets that would be considered deuces. 
So I was assuming that there are a total of 35!/28! ways of picking 7 tickets from 35 tickets. And to get the number of 'no deuces' it would be 23!/16!. Which I think would mean the percentage comes to be (23!/16!)/(35!/28!) or 3.65%. So I don't seem to get the 5.3% shown in the table. What am I getting wrong? Thanks.
 A: There are $C(7,3) = 35$ different tickets, and for a given drawing, $C(3,2) \cdot C(4,1) = 12$ of them will be deuces, so the probability of a given ticket being a deuce is $\frac{12}{35} \approx 0.3429$.
Then, for 7 randomly chosen tickets and assuming numbers may be repeated, the probabilities follow a binomial distribution.
$$P(X = 0) = C(7,0)(\frac{12}{35})^0(1 - \frac{12}{35})^{7-0} \approx 0.0529$$
$$P(X = 1) = C(7,1)(\frac{12}{35})^1(1 - \frac{12}{35})^{7-1} \approx 0.1933$$
$$P(X = 2) = C(7,2)(\frac{12}{35})^2(1 - \frac{12}{35})^{7-2} \approx 0.3025$$
$$P(X = 3) = C(7,3)(\frac{12}{35})^3(1 - \frac{12}{35})^{7-3} \approx 0.2631$$
$$P(X = 4) = C(7,4)(\frac{12}{35})^4(1 - \frac{12}{35})^{7-4} \approx 0.1372$$
$$P(X = 5) = C(7,5)(\frac{12}{35})^5(1 - \frac{12}{35})^{7-5} \approx 0.0430$$
$$P(X = 6) = C(7,6)(\frac{12}{35})^6(1 - \frac{12}{35})^{7-6} \approx 0.0075$$
$$P(X = 7) = C(7,7)(\frac{12}{35})^7(1 - \frac{12}{35})^{7-7} \approx 0.0006$$
Your 3.65% assumes that no repeats are allowed, which means the number of deuces follows a hypergeometric distribution.
