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Re: Entertaining probability question
Posted:
Jun 29, 2014 2:35 PM
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"Port563" <reader80@eternal-september.org> wrote in message
news:lopjgm$ne7$1@dont-email.me...
> "Mike Terry" <news.dead.person.stones@darjeeling.plus.com> wrote...
> > "Michael Press" <rubrum@pacbell.net> wrote in message
> > news:rubrum-F6BA62.21323027062014@news.albasani.net...
> >> In article <lojgns$bto$1@dont-email.me>,
> >> "Port563" <reader80@eternal-september.org> wrote:
> >>
> >> > A standard deck of 52 cards is thoroughly shuffled.
> >> >
> >> > What is the probability that exactly two aces** occupy adjacent
> > positions
> >> > within the deck?
> >> >
> >> >
> >> > "Hint": Some may wish to consider the adjacent aces as if they were
one
> >> > card.
> >> >
> >> > (-:
> >> >
> >> > While I think the above is unambiguous, for the avoidance of doubt it
> > means
> >> > that none of these orderings qualify:
> >> > ...xxxxxAxxxAxxxxxxAxxxxxAxxxx...
> >> > ...xxxxxxxxAAxxxxxxxxxxxAAxxxx...
> >> > ...xxxxxxxxAAAxxxxxxxxxxAxxxxx...
> >> > ...xxxxxxxxAAAAxxxxxxxxxxxxxxx...
> >>
> >> Number of ways of putting 4 distinguishable aces in 3 distinguishable
> > places
> >> times the number of ways of selecting 3 from 49 without replacement
> >> divided by 52!.
> >
> > I don't think I can be looking at this right - if you select 3 from 49,
> > some
> > of the 3 could be adjacent, which would upset the calculation it seems
to
> > me...
>
>
> Think "gaps" or "holes".
>
> Adjacency is not possible to represent (and therefore to occur) in this
> model, as apparently adjacent gaps would still have a non-ace between
> them.
>
Thanks, I wasn't thinking "holes"! :)
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