1. ## Good problem

Each of three cards has an integer written on it. The three integers
p, q, r satisfy
the condition 0
p < q < r. Three players A, B, C mix the cards and pick one each.
The number on the card they select is added to their scores.
This process is repeated at most ten times, after which
A has 20 points, B has 10
points, and
C has 9 points. Also we know that B got the r card in the last round.

q card in the first round?

2. With these assumptions also:
There are only 3 cards, and their initial score each is zero (0).

The total value at the end of all the rounds is 39 (20+10+9).

Total value is (p+q+r)*rounds = 39.
Number of rounds: 1, or 3, or 39 (the factors of 39)

Since the number of rounds is "at most ten", can't have 39 rounds.

If there were only 1 round, then B could not have received the
highest valued card, since player A has a higher score.

There were three rounds.

Since all three cards total to 13, player B could not have received
all three cards, nor could he have received the same card 3 times to
get a total of 10.
Thus B received 1 of the cards twice.

No cards are equal. Player B scores a 10 count this possible way:
10 = 1+1+8
10 = 2+2+6
10 = 3+3+4
10 = 4+4+2
The remaining card is:
13 = 8+1+ 4
13 = 6+2+ 5
13 = 4+3+ 6
13 = 4+2+ 7

Since player B received the highest card in the last round, then
remaining card cannot be a 6 or 7.
Thus the cards are numbered:
:: p=1, q=4, r=8
or
:: p=2, q=5, r=6 <-- no combination of these gives a score of 20.
The max score possible in three rounds is 18 (3*6).

Therefore
player A: 8+8+4 = 20
Player B: 8+1+1 = 10
Player C: 4+4+1 = 9

Player B received the 8 in the last round, Player A the 4, and Player C the 1.

The order of rounds must be:
A..B..C
8..1..4 first round
8..1..4 second round
4..8..1 third round

Player C received card q in the first round.