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Topic: Entertaining probability question
Replies: 10   Last Post: Jun 29, 2014 4:09 PM

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Mike Terry

Posts: 664
Registered: 12/6/04
Re: Entertaining probability question
Posted: Jun 28, 2014 10:21 AM
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"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...

Mike.






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