My computer simulation program is quite simple. It just draws random cards (without replacement) and starts checking for a winner after $3$ cards are drawn. I use a counter array of the $13$ possible ranks do determine how many of each rank I have at any time (after $3$ cards are drawn) then I check for triples or a quad. It is interesting to me that even with $1$ million simulated hands, the maximum win threshold of A never seems to exceed $10$ (triples). So that would mean there is a winner no later than the $36$ drawn card since we would have to get all the ranks twice (that is $26$ cards) then get $10$ more ranks making $10$ triples. Of course it is MUCH more likely to get a single quad way before then.
Here is some interesting info I observed by placing checks/counters in my program and using at least $10,000$ decisions (wins) and at most $10$ million:
$4.32$ : Average # of triples required for player A to win.
$22.4$ : Average # of cards drawn to determine a winner.
$~~~~~3$ : Minimum number of cards drawn to determine a winner (obviously a single triple).
$~~~36$ : Maximum number of cards drawn to determine a winner.
$21:19$ : ratio of A wins vs. B wins at exactly $35$ cards drawn ($5$M wins total).
about$~1/2$ : Average difference from half wins for player A ($500$K out of $1$M for example).
If anyone would like other checks/counts like this just ask an I will try to add them into my program and report the results here.
This game seems interesting since it is not some fixed probability game but rather one that adjusts the make the game almost dead even as far as who is likely to win on average.
$UPDATE~1$ - I have some data from $10,000$ decisions (wins) as requested by Greg Martin. I will express them here as 4 tuples with the following format: # of triples during the win, total # of wins (A or B) with that # of triples, # of times A wins with that many triples, # of times B wins with that many triples. Here goes:
$9,10,11,12$, and $13$ triples did not show up in this table cuz I only had $10,000$ wins (as requested by Greg Martin). I can run it overnight and get $100$ million ($100$M) and create a 2nd (new) table. I would like someone else to simulate this on computer as well to make sure my numbers are correct.
I got exactly $5000$ wins for A and exactly $5000$ wins for B.
$UPDATE~2$ - Same format as above but $100$ million decisions (I ran it overnight)
$11,12$, and $13$ triples did not show up in the simulation.
I got exactly $50,000,001$ wins for A and exactly $49,999,999$ wins for B.
$UPDATE~3$ - Same format as above but now I am reporting the # of wins based on A's # of triples win threshold instead. $1$ million decisions.
An interesting pattern is shown here. Let's take $2$ and $3$ for example. The # of wins for A when the win threshold for A is $2$ is $54791$, but that is exactly the number of wins for B when the win threshold is $3$. This pattern is also true for $1$ and $2$, $3$ and $4$....
Using tt as an abbreviation for A's triple win threshold (tt = triple threshold), the actual # of wins for A is the same as actual # of wins for B when tt is one higher. For example, when tt=$4$, A will have some number of wins, call it x. When tt=$5$, that same exact number x appears as # of wins for B. Perhaps it is that whenever A wins and tt gets bumped up, on average, B will win the very next hand and thus there is even more equilibrium/balancing going on than I thought. I think that may be the answer. Note that I award the win before updating tt value in the simulation program. So for example, if A wins the very first hand when tt=$4$, then the counter for the # of wins when tt=$4$ will get incremented by $1$ and then tt will get bumped up to $5$, so at that point, the counter for the # of wins when tt=$5$ will still be at $0$.
$11,12,13$, and $14$ triple threshold did not show up in the simulation.
I got exactly $500,000$ wins for A and exactly $500,000$ wins for B.
$UPDATE~4$ - I tried $2$ manual hands using a deck of cards. The first hand I got $4$ triples (5,9,Q,7) and $27$ cards were dealt spanning all $13$ ranks. The win threshold for player A was $4$ (triples) for that hand. For the 2nd hand (since A won the first hand), A's win threshold was then bumped up to $5$. 2nd hand also got $4$ triples first but because the threshold was raised to $5$, the quad appeared before the $5$th triple. In that 2nd hand, only $22$ cards were dealt spanning only $10$ ranks. So look at how quickly the "law of averages" took place. For only $2$ games, I got $24.5$ average cards drawn and $50/50$ (split) on the wins. I did not practice any games. They were the first $2$ I tried with real cards. This gives me a better appreciation for how quickly a computer can simulate this type of game. Likely in the time it takes me to play $1$ manual game, a computer can play millions if not billions of games.
$UPDATE~5$ - I tried (for fun and to satisfy my curiosity), altering A's win threshold by $2$ instead of by $1$. The pattern above was retained here in that the # of wins for A when tt=$2$ is the same as the number of wins for B when tt=$4$. The only change I see is B is now a favorite to win at about $51.3$% to A's $48.7$% (if tt starts at $4$), and of course there are no wins where tt is odd. Interestingly, if I instead change tt (A's win threshold for # of triples) to start at $3$ instead of at $4$, then it goes back to exactly $50/50$. So apparently the initial value of tt makes a difference as far as the fairness of the game which seems somewhat surprising since I would think that with thousands or even millions of hands, it should not make a difference (but it does). We lose the "straddling" effect (where $50/50$ is achievable), when certain conditions are met. For example, when tt is an even number (such as $4$) and we change tt by an even number (such as by $2$). That makes it so there is no way for there to be a $50$% chance for each player to win. Not sure why this is true but it is what my simulation program is showing. Maybe someone else can explain why.