Civil Liberties

[ppillai at sprint.ca]

I just noticed the thread and Im throwing in my two cents worth in the interest of
putting a quick end to this thread! If I am correct in figuring out everyones
position after my quick glance, its clear Carrol and Luke's mathematical
intuitions are correct. Max, Matt and Zak, go back and redo your homework.
Luke and Carrol were claiming the chance of an event appearing at least once in a
series of trials increases as the number of trials increase -- correct.
Max/Matt/Zak's response was that each trial was independant of each other -- also
correct , but irrelevant. What Max/Matt/Zak actually calculated was the
probability of consecutively throwing Heads in a row n times (or for that matter
any other specific sequence):   1/2^n where n = # of tosses;
so after 1 toss(n =1) Probality of Heads : P(H) = 1/2;
2 tosses P(H)*P(H) = 1/2^2 =1/4;
3 tosses P(H)*P(H)*P(H) = 1/2^3 = 1/8
etc. etc. etc

The sequence (A) of results for n=1 to n=5 looks like (1/2, 1/4, 1/8, 1/16, 1/32)
After 5 throws, the prob. of coming up with any *specific* sequence is 1/32, eg.
{HHHHH} or {HTHTH} or {TTTTT}. . . etc.. But after 5 throws we actually want the
total number of outcomes that have at least 1 H somewhere in them out of a total
probability space of 32 eg. {HTTTT} or {THTTH} . . . . and so on. So it is
actually easier to calculate what we dont want -- any outcome that has only tails,
which it turns out is a very specific outcome: {TTTTT} and as a specific outcome
is calculated by 1/2^5 = 1/32. Any other outcome will have at least 1 H somewhere
in the set. So the probabilityof coming up with at least 1 Heads in 5 tosses is
actually:
1 - (1/2)^5 = 1 - 1/32 = 32/33
Notice that the sequence (A) above for n tosses is decreasing:
(1/2, 1/4, 1/8, . . ., (1/2)^n  )
But if we take an extremely large number of trials, ie n --> infinity, the above
sequence converges at 0 (as do all sequences of the form (r^n) where -1< r < 1).
So the probability of us getting at least one Head approaches 1 as the number of
trials icrease and approaches infinity:
1 -  [limit n-->infinity (1/2)^n]   = 1-0 = 1

Similarly the probability of eventually getting *at least one* 6  popping up after
a series of throws of the dice increases as the sample space of dice throws
increases. *Notice* that each individual dice throw, as Max/Matt/Zak correctly
pointed out, is still 1/6 (So its easy to see how they came by their mistake). But
as the sample size increases the chances of 6 popping up also increases. Again its
eassier to measure prob. of 6 appearing by measuring the prob. of it not popping
up ie. (1-1/6)^n = (5/6)^n where n is the number of throws.
Similarly when n is sufficienly large the probability that a series of dice throws
will *not* have a 6 somewhere in there progressively decreases until it become
infinitesimally small (though never 0)
Therefore the probability that 6 appears at least once increases as n throws of
the dice increase.
( 1-(5/6)^n )

solidarity
-pradeep