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I have a probability question that I've solved the "long" way using a probability table, and I'm wondering if there is a faster way to solve this?

"For a 3-player game, the first person to win 4 rounds wins. They are equally matched, so the probability any one of them wins a round is 1/3. The players are forced to quit after 6 rounds, with player A having won 3 rounds, player B having won 2 rounds, and player C 1 round. At this point, the probability of player A winning should be...?"

I figured out that they need to do at most 4 more rounds (rounds 7, 8, 9, and 10), and that there are 81 total rows in my table. They're filled out and I have an answer (player 1 wins 57/81 possible rounds). BUT --- it just seems like so much work! Is there a more efficient way to solve a problem like this?

Edit to add: I'm going to look more into Pascal's work where I think I may find an answer (since this is based on some of his early work) --- but other ideas are always appreciated!

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    I may be misreading the question, but it seems to me that the players have already gone for $5$ rounds, and the game ends after $6$ rounds (at most), so why are you saying "they need to do at most $4$ more rounds"? Isn't there just one more? – 2017-01-04
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    I think OP meant *atleast*. – 2017-01-04
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    If they did just one more round and could guarantee that player A wins, then that would be all they need. But - since they have equal probability of winning each round, player C could win 2 rounds, then player B could win 2 rounds (leaving player B as the winner in this case). If they only played once more it doesn't guarantee a win....if that makes sense. – 2017-01-04
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    No, definitely meant at most (although this could be wrong). They need at least one more round (if player A were to win again), but will have a definite winner in at most 4 more rounds. – 2017-01-04
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    It's based on Pascal and Fermat's "Problem of Points"...if that helps? – 2017-01-04
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    I think I just realized - I need to look more into Pascal's work (my last post gave me that realization). – 2017-01-04
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    This is very simple, if all players are forced to quit at this point, then nobody reaches $4$ wins, hence the probability of player A winning is $0$. – 2017-01-04
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    @barak: That, however, is clearly not what is intended. – 2017-01-04

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You don’t actually have to make the whole chart; you can count the relevant outcomes fairly easily without doing so.

As you say, the game must finish in at most $4$ more rounds; it takes $4$ more rounds if $B$ wins one and $C$ wins two of the next $3$ rounds, at which point each player has won $3$ rounds.

Imagine that they go ahead and play $4$ more rounds, even if one of them wins the game before the $4$ rounds are complete. There are $3^4=81$ possible sequences of winners of these $4$ rounds. We want to count the sequences in which $A$ gets a win before $B$ gets $2$ or $C$ gets $3$ and divide this number by $81$. We can count them separately for each possible position of $A$’s first win.

  • There are $3^3=27$ sequences that start with $A$, since there are $3$ choices for the winner of each of the last $3$ rounds; all of them are wins for $A$.

  • There are $2\cdot3^2=18$ sequences in which $B$ or $C$ wins the first round and $A$ wins the second; these are also all wins for $A$.

  • There are $2\cdot3=6$ sequences that start $BCA$ or $CBA$, and $3$ more that start $CCA$; these too are wins for $A$. (That is, if $A$’s first win is the third round, $A$ wins the game unless $B$ won the first two rounds.)

  • Finally, $A$ wins the game in the fourth round if and only if $B$ wins one of the first three rounds and $C$ wins two, so $A$ wins the game with the $3$ sequences $CCBA,CBCA$, and $BCCA$.

$A$ therefore wins the game in $27+18+9+3=57$ of the $81$ possible outcomes, or with probability $\frac{57}{81}=\frac{19}{27}$.