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.