# the first player to win three games in a row or a total of four games wins.

In a competition between players X and Y, the first player to win three games in a row or a total
of four games wins.
a. How many ways can the competition be played in total?
b. How many ways can the competition be played if X wins the first game and Y wins the second and third games?

• Did you try solving it using a probability tree – Cloverr Dec 8 '15 at 2:45
• Without any other condition, minimal number of games needed is 3 and maximal is 8. I believe it wouldn't be hard just to do case by case study from this. One need to do minor change for (b), since the minimal number of games needed is 4 now. – Xuqiang QIN Dec 8 '15 at 3:24
• The maximum number of games is 7. After 7 games there one player have already won at least 4 games. – Sobi Dec 8 '15 at 3:38
• What have you tried ? – true blue anil Dec 8 '15 at 4:07

I don't see a really elegant way to do this.

One simplification is to calculate the ways in which X wins and multiply by 2 because when you interchange all the Xs and Ys in all the "X wins" scenarios you get all the possible "Y wins" scenarios.

X wins in 3 - XXX - 1 way
X wins in 4 - YXXX - 1 way
X wins in 5 - YYXXX - XYXXX - XXYXX - 3 ways

for 6 and 7 organize yourself by considering the six possible results for the first 3 games (it must have been 2-1 after 3 games if no one wins in 3)

e.g. for X to win in 6 there are 2 scenarios starting with "XXY" but no scenarios starting with "YYX" .

For X to win in 7 all six 3 game starts are possible leading to either 2 or 3 scenarios.

I count 7 ways for X to win in 6 and 14 ways for X to win in 7

total ways for X to win $= 14 + 7 + 3+ 1+ 1 = 26$

So there must be 52 possible ways to play out the series, of which 7 start with XYY ( using the "X wins" list count sequences starting with either XYY or YXX )

An interesting result is that given you win in exactly 7 games the conditional probability that you were 2-1 down after 3 games is exactly 50% !

We shall count $A's$ wins, and multiply by $2$.

Firstly, let us list the "special" cases :
Wins due to $3$ games in a row: $WWW, LWWW,\;$ and $\;LLWWW$
Losses due to $3$ games in a row:$LLLWWWW, WLLLWWW, WWLLLWW Now we will count the general cases, winning$4$games, and subtract the special cases: To win$4$games in$5$,$A$must win the$5th$game, and$3$of the previous$4$in$\binom43 = 4$minus$2$special wins in$4$games or less$=2$ways. To win$4$games in$6, A$must win the$6th$game, and$3$of the previous$5$minus$3$special wins in$5$games or less$= 7$ways, Only when$7$games are played is there also chance of losing due to special cases. To win$4$games in$7, A$must win the$7th$game, and$3$of the previous$6$, thus$\binom63 - 3$special wins -$3$special losses$= 14$Thus total # of ways =$2(3+2+7+14) = 52$ways ps: You should now try the easier second part. • I think the correct answer is 56 (counting exhaustively on a binary tree). – Sobi Dec 8 '15 at 14:58 • @trueblueanil think there are two more losing cases that should be subtracted WLLLWWW and WWLLLWW - that would bring us to agreement at 52 – WW1 Dec 8 '15 at 21:22 • @WW1: I agree with the "missing link", and have adjusted, thanks. It is very late here, I will try and see if a more elegant solution is possible later. – true blue anil Dec 8 '15 at 23:06 Suppose four games are played and one player wins them all. WWWW. There are 2 ways that can happen. (Either one player wins all four or the other player does. Suppose five games are played and one player winning 4 and losing one. LWWW or WLWW or WWLW. The winner can't lose the last game (because then the winner would have already won four games and they'd never play the fifth game). So there are three choices for the winner to lose. That's${3 \choose 1}$. And then there are 2 possible winners so$2{3 \choose 1}$Suppose six games are played, one player wins 4 and loses two. There are${4 \choose 2}$to do this and two players.$2{4 \choose 2}$. Suppose seven games... winner wins 4 loses 3.$2{5 \choose 3}$. Total;$2 + 2{3 \choose 1}+ 2{4 \choose 2} + 2{5 \choose 3}\$.