Duane Bozarth <dpbozarth@[EMAIL PROTECTED]
> wrote:
: BlankMondragon wrote:
: >
: > > But OP wanted a run of 6 to be only one "event" when I asked what
: > > comprised a "run of five"...
: >
: > In the simulation I showed a run of 6 as only one event.
: >
: Yes, I was commenting back to Dave Chamberlain who noted a run of 7 as
: "3 runs of 5" when doing his calculations of P(5 H|T) independently that
: that didn't follow the rules as outlined by OP.
: I had noted a similar approach some time much earlier in looking at the
: 100 tosses as a series of 96 independent sets of 5 but when OP responded
: that a run of N >= 5 is to be counted as on run, that eliminates that
: simplification.
I thought I had replied to this but I don't see my reply.
You're right, the OP (in a later followup I believe) did say that a
run of 6 wouldn't count as two runs of 5, which is fine. I like to
see if the math works for it though. So consider this:
A run of 5 that must be EXACTLY 5 is really a run of 6 with the last one
being opposite. Meaning:
H H H H H T or
T T T T T H
Thus a run of 6 has 64 combinations, 2 of which are acceptable (shown just
above).
2 out of 64 (2/64) is 1 out of 32, so roughly 3%. In the sample run there
were 2 runs of exactly 5 tails and 1 run of exactly 5 heads. So the math
does work, 3 out of the 100 tosses resulted in a run of exactly 5.
--
-davec
-----------------------------------------------------------------
--
submissions: post to k12.ed.math or e-mail to k12math@[EMAIL PROTECTED]
e-mail to the k12.ed.math moderator: kem-moderator@[EMAIL PROTECTED]
website: http://www.thinkspot.net/k12math/
newsgroup charter: http://www.thinkspot.net/k12math/charter.html
|
