[seqfan] Re: game series
Jonathan Post
jvospost3 at gmail.com
Fri Jan 28 17:22:08 CET 2011
http://riot.ieor.berkeley.edu/~baseball/detail_calc.html
As the end of the season approaches, baseball fans like to engage in
speculation about which teams will advance to the post season and
whether their favorite team is still in contention for a playoff
berth. Newspapers frequently run headlines in their sports sections
declaring that a particular team has been eliminated from contention
or that the first place team has clinched a playoff berth. These
determinations can often be made by looking at the standings and
making a few simple calculations. The RIOT baseball standings can
actually tell you when a team has locked up a playoff spot or fallen
out of contention days before it is reported anywhere else!
The key is taking into account the actual match-ups that remain in the
season and enumerating all possible outcomes. Until the final weeks of
the season, however, the number of possible outcomes can be
astronomically large which makes such an enumeration impractical even
on a very fast computer. By using a powerful and pervasive technique
of industrial engineering, called Network Optimization, it is possible
to accurately perform the necessary calculations without explicitly
enumerating all possibilities. Consequently, it is possible for RIOT
to determine the playoff prospects of a team a few days, or even
weeks, before it is reported by the popular media. The following
example, using National League standings as of 9 am, EST, Sunday
September 8 1996, shows how this works....
On Fri, Jan 28, 2011 at 8:06 AM, Robert Israel <israel at math.ubc.ca> wrote:
>
>
> On Fri, 28 Jan 2011, Peter Pein wrote:
>
>> On 28.01.2011 10:10, Dmitry Kamenetsky wrote:
>>>
>>> Hello Sequence Fans,
>>>
>>> Recently I have been looking at the following problem. There is a series
>>> of
>>> n games played between two teams. The outcome of each game is either a
>>> win
>>> or a loss (no draws).
>>> A team wins the whole series if it wins k=floor(n/2)+1 games or more. Now
>>> if
>>> a team reaches the magic number of k wins then the games that follow (if
>>> there are any) are
>>> dead games, because their outcome cannot affect the outcome of the
>>> series.
>>> So a natural question arises: out of all the possible 2^n series how many
>>> of
>>> them will have
>>> at least one dead game? This forms the sequence
>>> 0,0,4,4,20,24,88,116,372,...
>>> This sequence is not in the OEIS and neither is its version for all odd
>>> n.
>>
>> Hi Dimitry,
>>
>> what did I do wrong while trying to reconstruct the sequence
>> {0,0,4,4,20,24,88..} ?
>>
>> I tried to find those sequences of wins/losses which contain a sequence of
>> wins/losses of length >= Floor[n/2]+1 followed by at least one "dead" game:
>>
>> In[1]:=
>> f[n_]:=Count[Tuples[{0,1},{n}],({___,0,s:1..,0,__}|{___,1,s:0..,1,__})/;Length[{s}]>=Floor[n/2]+1]
>> In[3]:= f/@Range[20]
>> Out[3]= {0,0,0,0,0,0,4,4,20,20,68,68,196,196,516,516,1284,1284,3076,3076}
>>
>> Did I misunderstand your definition? Please enlighten me!
>>
>> Cheers,
>> Peter
>
> The last game is "alive" if and only if the result of the first n-1 games
> is either
> (if n is odd) (n-1)/2 wins for both teams, or
> (if n is even) n/2 wins for one and n/2-1 for the other.
> So if n is odd the number of series with at least one dead game is
> 2^n - 2 (n-1 choose (n-1)/2)
> and if n is even it is 2^n - 4 (n-1 choose n/2).
> This gives me 0, 0, 4, 4, 20, 24, 88, 116, 372, 520, 1544, 2248, 6344, 9520,
> 25904, ...
> agreeing with Dmitry.
>
> Robert Israel israel at math.ubc.ca
> Department of Mathematics http://www.math.ubc.ca/~israel University
> of British Columbia Vancouver, BC, Canada
>
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