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Card Orders

Consider a deck of 120 cards, 60 red and 60 black. From a shuffled
deck, which of the following is most likely when dealing the cards?
- Alternating red, black, red, black, ...
- Alternating 2 reds, 2 blacks, 2 reds, 2 blacks, ...
- Alternating 3 reds, 3 blacks, 3 reds, 3 blacks, ...
- Alternating 4 reds, 4 blacks, 4 reds, 4 blacks, ...
- Alternating 5 reds, 5 blacks, 5 reds, 5 blacks, ...
- Alternating 6 reds, 6 blacks, 6 reds, 6 blacks, ...

Assigning the least likeliest option a value of 1, how much more likely
is each other option?
Source: Original.

Solutions were received from Nick McGrath, Al Zimmermann,
Jayavel Sounderpandian, Lance Nathan, Ross Millikan,
Paul Botham, Rajendra Shah, Jozef Hanenberg.
Jozef's summary says it all:

There are 120! / ( 60! * 60! ) different ways to choose a group of 60
places from 120 available places to put your 60 sixty red
cards. The remaining 60 places can then be filled with the 60 black
cards. This means that the chance that you will get a fully specified
order of 60 red and 60 black cards is:

( 60! * 60! ) / 120! = 1.035..* 10 ^ -35.

Because in all the proposed cases the order of red and black is fully
specified, the chance is the same in all these cases.

Mail to Ken