Conference rusure::math

276.0. "3-coin monte" by AURORA::HALLYB () Thu May 02 1985 12:35

An old paradox that has great potential as a bar bet.  Step right up, friend...

We're going to flip a fair coin repeatedly, thereby obtaining some sequence
of heads and tails, e.g.,  H H T H T T H H H . . . but before we start tossing, 
you and I will pick some outcomes ahead of time.  E.g., you may "pick" 
HHH, I might pick THH.  Let's confine ourselves to 3-coin sequences for now.
Then we start flipping and whoever's sequence comes up first wins.
(We are betting, aren't we?)  Fair enough?  Sure, it must be!  :-)

Strategy:  If you select HHH, I can almost guarantee a win by selecting THH.
Even though both events are equally likely on the first 3 flips, the fact that
we're dealing with sequences rather than sets of 3 tosses gives THH a huge
advantage over HHH.  If the first 3 flips are anything other than HHH, I'll win.
(On any start other than HHH, an HHH sequence would have to be preceded by a T).

One might represent this by 	P(THH) > P(HHH)

So, you say, you select THH next time?  OK, I'll select TTH.  By the same logic
TTH is a more likely outcome than THH (an exercise for the reader).  In other 
words, P(TTH) > P(THH).  Continuing this process, we eventually discover that

		P(THH) > P(HHT) > P(HTT) > P(TTH) > P(THH)
		   \\				      //
		    \\===============================//

By a coincidence of notation we seem to have proved that P(THH) > P(THH).

The general strategy is to choose your last two flips to be the same as your
opponent's first two flips, and choose your first flip to be whatever makes
your 3-flip sequence have exactly one pair of identical consecutive outcomes.
He chooses HTT, you choose HHT.  She chooses HTH, you choose HHT.  Your odds
are at least 2:1, sometimes 3:1, sometimes 7:1.  Hard to believe, but true!

  John

The use of the phrase "... is a more likely outcome than ..." is incorrect;
what can be said is that one sequence PRECEDES another in the course of
events. Like any sequence of things in a circular queue, each in some sense
precedes the others.

A similar paradoxical situation occurs if one goes to a subway station and
takes the first train in either direction. Over time, suppose one finds that
the northbound train is taken 5 times as frequently as the southbound train.
Does this imply that 5 times as many trains go north as go south? No, but
one may be able to infer that the distance between scheduled arrival times
differ by 5:1 intervals, e.g. that the northbound train arrives at 10 minutes
before the hour and the southbound train arrives on the hour. Again, each
train precedes the other; the interarrival times are what give the sequence
its peculiar statistics. - Lynn Yarbrough