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Re: Entertaining probability question
Posted:
Jun 29, 2014 2:35 PM


"Port563" <reader80@eternalseptember.org> wrote in message news:lopjgm$ne7$1@dontemail.me... > "Mike Terry" <news.dead.person.stones@darjeeling.plus.com> wrote... > > "Michael Press" <rubrum@pacbell.net> wrote in message > > news:rubrumF6BA62.21323027062014@news.albasani.net... > >> In article <lojgns$bto$1@dontemail.me>, > >> "Port563" <reader80@eternalseptember.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 nonace between > them. >
Thanks, I wasn't thinking "holes"! :)



