This is for the one player game and a initial hand of five cards. I will consider that you have a hand with $k$ cards if after throwing away all groups of four cards of the same value, you have $k$ cards in your hand. For instance, if your initial hand is $\{2,3,3,3,3\}$ I will consider it a one card hand, not a five card hand.
Let $P(k)$ be the probability that at some moment in the game you have a $k$ card hand. The following values are easy: $P(0)=1$ (at the end of the game you have no cards in your hand) and $P(40)=P(41)=\dots=P(52)=0$ (if you have $40$ cards, there must be a group of four cards of the same value). A little thought gives
$$
P(39)=\frac{4^{13}}{\dbinom{52}{13}}=0.000105681
$$
For the rest of the values I have run a simulation in Mathematica of $10^7$ games. These are the results:
k H(k)
0 483
1 25839
2 596131
3 10000000
4 230004
5 10000000
6 10000000
7 10000000
8 10000000
9 10000000
10 10000000
11 10000000
12 10000000
13 10000000
14 10000000
14 9999996
16 9999945
17 9999720
18 9998600
19 9994328
20 9980514
21 9942513
22 9854106
23 9675123
24 9348731
25 8822787
26 8068444
27 7080395
28 5919525
29 4675434
30 3458156
31 2381202
32 1514558
33 876208
34 458693
35 213203
36 84852
37 28160
38 7216
39 1067
For each $k$, $H(k)$ is the number of games in which a hand of $k$ cards has been held before reaching the end of the deck. Observe that the value of $H(39)$ is in accordance with the exact value of $P(39)$. A graph of the results:
It is surprising (at least to me) that for certain values of $k$, like $k=5$, a hand of $k$ cards was held in all $10^7$ games, even if a deck like
1,1,1,1,2,2,2,2,3,3,3,3,...
will give only hands of $1$, $2$, $3$ and $4$ cards.
The probability of getting a war on the first hand is 1/17 because whatever Player 1 card is there are 3 out of 51 cards that are the same rank.
From then on it's complicated conditional probabilities.
Call Player 1's first card R1, and Players 2's first card S1. If R1 $\ne$ S1 (16/17 probability) and Players 2's second card (R2) is neither R1 nor R2 (a 44/50 = 22/25 probability) then there 3/49 probability that player 2's second card is a war (16/17*22/25*3/49). But if R2 = (R1 or S1) (a 6/50 = 3/25 probability then there is a 2/49 that the second hand is a war. Or
So the probability of hand1 not a war hand2 being war is 16/17(22/25*3/49 + 3/25*2/49).
But if hand1 was a war the probability of the second hand being a war is: if R2 = R1 = S1, (a 2/50 = 1/25 probability) then the probability is 1/49 (total: 1/17*1/25*1/49) but if R2 $ne$ R1 (a 48/50 = 24/25 probability) then the probability of a war is 3/49 (total: 1/17*24/25*3/49)
So the probability of hand 2 being a war is
16/17(22/25*3/49 + 3/25*2/49) + 1/17(1/25*1/49 + 24/25*3/49) =
(16(22*3 + 3*2) + 1 + 24*3)/(17*25*49)=
(16(72)+72 + 1)/17*25*49 = (17*72 + 1)/17*25*49 = 1/17
So basically every hand the probability of a war is 1/17.
Best Answer
The winning combinations are ABC, AABC, ABBC, AABBC, where the unshown cards are among the other $25$. This gives ${2 \choose 1}{2 \choose 1}{25 \choose 8}+2{2 \choose 1}{25 \choose 7}+{25 \choose 6}=6426200$. Dividing by ${30 \choose 11}$ gives about $11.76\%$