Boston Celtics VS. LA Lakers- Expectation of series of games? 
Boston celtics & LA Lakers play a series of games. the first team who win 4 games, win the whole series.

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*The probability of win or lose game is equal (1/2)

a. what is the expectation of the number of games in the series?
    


So i defined an indicator: $x_i=1$ if the game $i$ was played.
It is clear that $E[x_1]=E[x_2]=E[x_3]=E[x_4]=1$.
for the 5th game, for each team we have 5 different scenarios (W=win, L=lose) and the probability to win is:


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*W W W L   $((\frac12)^3=\frac18)$

*W W L L   $((\frac12)^2=\frac14)$

*W L L L   $((\frac12)^1=\frac12)$


$\frac12+\frac14+\frac18=\frac78$
We can use the complementary event: if only one game is L in a series of 4 games (so the 5th game will be played): $1-\frac18=\frac78$
The same calculation is for the 6th and 7th game:
6th:  W W W L L or W W L L L    => $(\frac18+\frac14=\frac38)$
7th:  W W W L L L =>  $(\frac18)$
And the expectation is $E[x]=1+1+1+1+\frac78+\frac38+\frac18=\frac{43}8$
What am i missing here, and how can I fix it?

If we have the family of all length-7 sequences composed of W and L, we see that each of these sequences represent one of $2^7$ outcomes to our task at hand with equal probability. Then, we see that the number of games played is pre-decided for each such given sequence (e.x.: WWWLWLL and WWWLWWW both result in five games played, while WLWLWLW result in seven games). So, we can find the probability of each event (number of games played) by counting how many sequences fall into each category.
Note: O indicates that this can be either W or L; this occurs when the outcome of the series is already decided and additional games become irrelevant to the total number of games played. Also, let us assume that the W team wins the seven-game series without loss of generality.
Four games: WWWWOOO
$2^3=8$ sequences
Five games: (combination of WWWL)WOO
$2^2{{4}\choose{3}} = 16$ sequences
Six games: (combination of WWWLL)WO
$2{{5}\choose{3}} = 20$ sequences
Seven games: (combination of WWWLLL)W
${{6}\choose{3}} = 20$ sequences
So, the expected value for number of games played is:
$4(\frac{8}{64})+5(\frac{16}{64})+6(\frac{20}{64})+7(\frac{20}{64})=\frac{93}{16}=5.8125$ games
