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




Re: Entertaining probability question
Posted:
Jun 28, 2014 10:21 AM


"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...
Mike.



