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In a baseball series, the first team to win 4 games wins the series. No more than 7 games are played.

  • a) If Team A wins the first 3 games, how many ways can the series be completed?
  • b) Suppose team A wins the first 2 games, how many ways can the series be completed?
  • c) How many ways can the world series be played if team A wins 4 games in a row?
  • d) how many ways an a world series be played if no team wins 2 games in a row?

My attempt:

  • a) 4 ways
  • b) I was thinking of using C(5,2) which was = 10
  • c) 3 ways
  • d) Having hard time on this one.

Any help is appreciated.

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    $\begingroup$ Is $A$ supposed to win the series? $\endgroup$ – drhab Sep 17 '15 at 9:36
  • $\begingroup$ How many different teams are playing? $\endgroup$ – TravisJ Sep 17 '15 at 13:13
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  • a) There are $4$ games left to play. If $A$ wins the series then $A$ wins $1$ of those games: $\binom41$ possibilities. If $B$ wins the series then $B$ wins $4$ of those games: $\binom44$ possibilities. So in total $\binom41+\binom44=4+1=5$ possibilities.
  • b) There are $5$ games left to play. If $A$ wins the series then $A$ wins $2$ of those games: $\binom52$ possibilities. If $B$ wins the series then $B$ wins $4$ of those games: $\binom54$ possibilities. So in total $\binom52+\binom54=10+5=15$ possibilities.
  • c) $AAAA$, $BAAAA$, $BBAAAA$ or $BBBAAAA$ so $4$ possibilities.
  • d) $ABABABA$ or $BABABAB$ so $2$ possibilities.
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(a) Ok.

(b) Ok. $A$ needs to win any 2 in the next 5, so C(5,2) = 10

(c) Ok.

(d) There are only two possible alternating sequences, $ABABABA$ or $BABABAB$, so only $1$ way for A to win.

Edit

I took it that $A$ is to win the series.

If either can win, I'll give a general formula for $(a)$ and $(b)$, add ${R\choose 4-W}$ for each player

where $R$ = games remaining, $W$ = games won

$(a)\;\; {4\choose 4-3} + {4\choose 4} = 5$

$(b)\;\; {5\choose 4-2} + {5\choose 4} = 15$

$(c)\;\;$ A wins starting $1,2,3,4 = 4$

$(d)\;\;$ Only ways for wins to alternate are $ABABABA$ ot $BABABAB = 2$

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  • $\begingroup$ a), b) and c) are not correct (unless $A$ is supposed to win the series, but I see no reason for that). $\endgroup$ – drhab Sep 17 '15 at 9:32
  • $\begingroup$ You are right, I took it that A is supposed to win. Waiting for OP's response. $\endgroup$ – true blue anil Sep 17 '15 at 9:38
  • $\begingroup$ OP hasn't responded so far, so ans added for alternate interpretation in edit. $\endgroup$ – true blue anil Sep 17 '15 at 10:04
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a) 5: AAAA, AAABA, AAABBA, AAABBBA, AAABBBB

b) 15:

AAAA, AAABA, AAABBA, AAABBBA

AABAA, AABABA, AABABBA

AABBAA, AABBABA

AABBBAA

AABBBB, AABBBAB, AABBABB, AABABBB, AAABBBB

c) 4: AAAA, BAAAA, BBAAAA, BBBAAAA

D) 2: ABABABA, BABABABA

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Assume team A wins series. Case 1: win in 4:1 possibility Case 2: win in 5: C(4,3)=4 possibilities Case 3: win in 6: C(5,3)=10 possibilities Case 4: win in 7: C(6,3)=20 possibilities

Total= 35 possibilities if team A wins. Therefore 35 possibilities if team B wins.

Total = 70 possibilities.

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