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### Finishing a Best-of-Seven Series

```Date: 06/19/2009 at 03:05:40
From: Jakob
Subject: 2-1 chance of getting to 4?

Adam and Britta are playing Backgammon.  They are playing to 4 and the
stop and want to find out who had the biggest chance of winning.
Each of them has 50% chance of winning a game.  How big is Adam's
chance of winning?

The combinatorics in this question is a little hard, the tree we use
to sort it gets hugely big and in some cases Adam wins after two
games, but they might have to play to 3-3 in that case it gets too
confusing.

The three games they already played have 3 possibilities.  AAB, BAA,
ABA.  Now they play again, then it might look like this: AABA, AABB,
BAAA, BAAB, ABAB, ABAA.  But adding the next game to the chain makes
it very large (might have missed some): AABAA, AABAB, AABBA, AABBB,
BAAAAx, BAAAB, BAABA, BAABB, ABABA, ABABB, ABAAAx, ABAAB.  Some of
those combinations will end the game, Adam winning with 4, those are
marked with an x.  But they can also play on.  From now on it gets too
confusing for me.

```

```
Date: 06/19/2009 at 11:33:14
From: Doctor Ian
Subject: Re: 2-1 chance of getting to 4?

Hi Jakob,

Let's use a simpler representation, like the score.  For your problem
it doesn't matter how they got to 2:1, we just know they start there.
equally likely to go to either of them:

+-- 3:1
|
2:1 --+
|
+-- 2:2

We can keep building that tree out with two outcomes for the next game
in each case.  In cases where Adam's score reaches 4, I'll mark it
with an A, and where Britta reaches 4 I'll use a B:

+-- 4:1 A
|
+-- 3:1 --+
|         |
|         +-- 3:2
|
2:1 --+
|         +-- 3:2
|         |
+-- 2:2 --+
|
+-- 2:3

At this level, we have a win for Adam.  What is the probability of
that win?  It's (1/2)^2, or 1/4.  So after two games, the probability
is 1/4 that Adam wins, and 3/4 that play continues.

Let's go another level, by playing the third game:

+-- 4:1 A
|
+-- 3:1 --+
|         |         +-- 4:2 A
|         |         |
|         +-- 3:2 --+
|                   |
|                   +-- 3:3
|
2:1 --+
|                   +-- 4:2 A
|                   |
|         +-- 3:2 --+
|         |         |
|         |         +-- 3:3
|         |
+-- 2:2 --+
|
|         +-- 3:3
|         |
+-- 2:3 --+
|
+-- 2:4 B

What are the probabilities associated with these nodes?

Let's look at the second 4:2, for instance.  How can we get here?
Half the time, Britta will win the first game; after that, half the
time Adam will win the second game; and half the time, Adam will win
the third game.  So the probability of ending up here is (1/2)^3 = 1/8.

Similar reasoning for the other cases gives us

+-- 4:1 A (1/2)^2
|
+-- 3:1 --+
|         |         +-- 4:2 A (1/2)^3
|         |         |
|         +-- 3:2 --+
|                   |
|                   +-- 3:3 (1/2)^3
|
2:1 --+
|                   +-- 4:2 A (1/2)^3
|                   |
|         +-- 3:2 --+
|         |         |
|         |         +-- 3:3 (1/2)^3
|         |
+-- 2:2 --+
|
|         +-- 3:3 (1/2)^3
|         |
+-- 2:3 --+
|
+-- 2:4 B (1/2)^3

As a quick check the combined probability of a win (for either player)
at this point is

1/4 + 3/8 = 5/8

And the combined probability of having reached 3:3 without a winner is

1/8 + 1/8 + 1/8 = 3/8

Note that the probabilities of having or not having a winner add to 1,
as they should.

And if there's a 3:3 tie, that means the next game decides the
contest.  Let's play that out:

+-- 4:1 A (1/2)^2
|
+-- 3:1 --+
|         |         +-- 4:2 A (1/2)^3
|         |         |
|         +-- 3:2 --+
|                   |         +-- 4:3 A (1/2)^4
|                   |         |
|                   +-- 3:3 --+
|                             |
|                             +-- 3:4 B (1/2)^4
|
2:1 --+
|                   +-- 4:2 A (1/2)^3
|                   |
|                   |
|         +-- 3:2 --+         +-- 4:3 A (1/2)^4
|         |         |         |
|         |         +-- 3:3 --+
|         |                   |
|         |                   +-- 3:4 B (1/2)^4
|         |
+-- 2:2 --+
|                   +-- 4:3 A (1/2)^4
|                   |
|         +-- 3:3 --+
|         |         |
|         |         +-- 3:4 B (1/2)^4
+-- 2:3 --+
|
+-- 2:4 B (1/2)^3

1/4 + 1/8 + 1/8 + 1/16 + 1/16 + 1/16

= 4/16 + 2/16 + 2/16 + 3/16

= 11/16

and the probability of Bitta's wins,

1/8 + 1/16 + 1/16 + 1/16

= 2/16 + 3/16

= 5/16

and they add up to 1, which they should--since one of them HAS to win
4 games eventually.

Does this make sense?

- Doctor Ian, The Math Forum
http://mathforum.org/dr.math/

```

```
Date: 06/19/2009 at 12:25:14
From: Jakob
Subject: Thank you (2-1 chance of getting to 4?)

Thanks for helping, that one bugged me all day.  Very simple when you
put it like that!
```
Associated Topics:
College Probability
High School Permutations and Combinations
High School Probability

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