Open polo

anna_gg · 2016-03-02 02:26:24

In the final open polo tournament, the games are held alternatively in each team's home court and away court and the home advantage is 75%. The champion team will be the one to achieve exactly 2 wins more than its opponent. What is the expected number of games in the tournament?

Last edited by anna_gg (2016-03-03 04:54:44)

bobbym · 2016-03-02 11:33:05

I assume that a final is between two teams?

Nehushtan · 2016-03-02 15:39:26

NB: Notice that

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so I'm very confident my formula is correct.

Last edited by Nehushtan (2016-03-02 21:28:12)

anna_gg · 2016-03-02 20:17:42

Yes, two teams.

bobbym wrote:

I assume that a final is between two teams?

Nehushtan · 2016-03-02 21:20:41

Okay, I think I've worked it out. Is it...

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Last edited by Nehushtan (2016-03-02 21:33:56)

anna_gg · 2016-03-03 01:45:36

How do you take into account the requirement for the winning team to achieve a winning streak of 2 games (i.e. to have exactly 2 more winning games from its opponent)?

Nehushtan wrote:

Okay, I think I've worked it out. Is it...
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Nehushtan · 2016-03-03 04:03:48

I took it that the winner was the first to win two games in a row. Is that correct?

anna_gg · 2016-03-03 04:54:02

So maybe it's my fault; I did not express it correctly: the winning team is the one who achieves exactly 2 wins more than its opponent (not necessarily 2 consecutive wins. For example, it can be WLWW or WW or LWWW).

Nehushtan wrote:

I took it that the winner was the first to win two games in a row. Is that correct?

Last edited by anna_gg (2016-03-03 21:46:09)

bobbym · 2016-03-03 14:07:28

I am confused over the wording of the problem. How is WLW 2 more? Also which team gets the home court first?

anna_gg · 2016-03-03 21:47:43

My fault again. I corrected my previous comment. WLW is not correct, it must be WLWW. It doesn't matter which team gets the home court first. We must examine both cases and calculate the expected number of games.

bobbym wrote:

I am confused over the wording of the problem. How is WLW 2 more? Also which team gets the home court first?

Nehushtan · 2016-03-04 00:37:28

anna_gg wrote:

So maybe it's my fault; I did not express it correctly: the winning team is the one who achieves exactly 2 wins more than its opponent (not necessarily 2 consecutive wins. For example, it can be WLWW or WW or LWWW).

Right. This means that in order to have a winner we must have one team winning two consecutive games from a position of parity. Hence an even number of games is required, so our sample space is {2, 4, 6, …}.

This problem is harder than I thought.

bobbym · 2016-03-04 07:13:02

Can we differentiate between the two teams? If so, which one gets the first game at home?

anna_gg · 2016-03-04 08:22:35

I don't think this is important, because we are looking for the "expected" number of games (the problem is symmetric).

bobbym wrote:

Can we differentiate between the two teams? If so, which one gets the first game at home?

bobbym · 2016-03-04 15:30:36

If I am to put this into a Markov chain then I must assign probabilities to who goes first. I can assign .5 to each one. Before I start, do you have the answer to this problem so we can tell if I get there?

anna_gg · 2016-03-04 21:32:42

Good morning to all of you
I believe the answer is 5,3333 and I think that Nehushtan's post #3 is very close to the solution. Indeed we have an even number of games, and we can examine two different cases, one is WW (2 wins, or LL, symmetrically) and all others that start with an even number of games where nobody wins and then we have a winning streak of 2 more games (for example, WHWHWH plus HH at the end). The first case is with probability 2 . 3/4 . 1/4 = 3/8. Thus the other cases (those with more than 2 games in total) occur with probability 5/8. But then what?

bobbym wrote:

If I am to put this into a Markov chain then I must assign probabilities to who goes first. I can assign .5 to each one. Before I start, do you have the answer to this problem so we can tell if I get there?

Last edited by anna_gg (2016-03-04 21:35:05)

bobbym · 2016-03-05 04:22:31

Hi;

The Markov Chain of this process is,

We can use an initial state vector of

or

to compute the chance that the team starts at home versus on the road. Getting the mean of the first passage time to the appropriate absorbing state (one of the last 4 rows) we get a expected number of games of 16 / 3

anna_gg · 2016-03-08 07:50:41

That's very hard for me to understand, but the result is the same as the one I got with my simplistic approach

Many thanks!!

bobbym wrote:

Hi;
The Markov Chain of this process is,
We can use an initial state vector of

or

to compute the chance that the team starts at home versus on the road. Getting the mean of the first passage time to the appropriate absorbing state (one of the last 4 rows) we get a expected number of games of 16 / 3

bobbym · 2016-03-08 07:55:17

Absorbing Markov Chains are a good way to describe random walks like this. They only look tough and they are a breeze to compute.

anna_gg · 2016-03-08 10:14:02

So now you can focus on my rotten apples

bobbym wrote:

Absorbing Markov Chains are a good way to describe random walks like this. They only look tough and they are a breeze to compute.

bobbym · 2016-03-08 10:20:20

I am afraid I did a lot of research on that problem and there was no known solution that I could find. That is not to say there is none but just that the general census on it was that it might be possible in 3 weighings but 4 was necessary in most cases. Mind you, this was on a less tough problem.

bobbym · 2016-03-14 12:39:51

Hi;

By using a tree there is another way to get the expectation:

So, we have two ways to get the same answer, I am happy.

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#1 2016-03-02 02:26:24

Open polo

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