Re: Sporkhack's illiterate oracle

AngleWyrm wrote:
"Philip Potter" <pgp@xxxxxxxxxxxx> wrote in message news:fi7o8v$k32$2@xxxxxxxxxxx
There are up to a maximum of 676 experiments possible, which is not the same thing as 676 experiments. Overlaying the numbers like that produces a dynamic number of experiments. And it still doesn't answer how the odds were achieved.

To make this clear, here's a short version: coin flip to get three heads in a row, with ten flips in the logfile. Question: How many experiments were conducted?

logfile:
1234567890 <- flip number
hthhhhthhh <- (h)eads or (t)ails

There are of course {1~3, 2~4,3~5,4~6,5~7,6~8,7~9,8~10} eight possible locations (three of which are all heads), but that does not answer the question of how many times we tried. Nor does it seem reasonable to me to say that we succeeded three times.

I don't follow your argument. I don't know what you mean by "how many times we tried" - how does "trying" affect probability?

To go *really* simple, lets look for a run of two heads in three coin flips. By my analysis, any given run of two has a 1/4 chance of being two heads. In three coin flips, there are two runs of two. The chances of both /not/ being runs is (3/4 * 3/4 = 9/16); the chance of there being one run is 1/4*3/4*2 = 6/16; the chance of there being two runs (three heads) is 1/4*1/4 = 1/16. Evaluating the possibilities:

hhh (two runs)
hht (one)
hth
htt
thh (one)
tht
tth
ttt

....so the real probablility is 1/8 two, 1/4 one, and 5/8 none. The problem is that the runs are not independent; if there has just been a run of two, it is much easier to get another run of two. Therefore, having two runs (three heads) is more likely than I predicted, and having zero runs is also more likely than I predicted.


Let's try to refine the analysis to take the dependence into account:

Probability of first run = 1/4
Probability of second run given that the first run happened = 1/2
Probabliity of second run given that the first run didn't happen = 1/6
[because of the three non-run possibilities for the first run {ht,tt,th}, only one will allow a run]

Probability of two runs: 1/4 * 1/2 = 1/8
Probability of one run: 1/4 * 1/2 + 3/4 * 1/6 = 1/8 + 1/8 = 1/4
Probability of zero runs: 3/4 * 5/6 = 5/8

A similar analysis for runs of n events in a series of N, each of probability p:

Probability of first run = p^n
Probability of each subsequent run, given previous run happened = p
Probability of each subsequent run, given previous run didn't happen
= (p^(n-1))(1-p)/(1-p^n) * p
This is the probablility that the previous run was thhhh given that we know it wasn't hhhhh, multiplied by the probability of another head.
It simplifies to this:
= p^n(1-p)/(1-p^n)

Probability of zero runs (which is the one we're interested in):
(1-p^n) * [1-(p^n)(1-p)/(1-p^n)]^(N-n)
This is the probability that the first run doesn't happen, times the probability that the N-n subesquent runs don't happen.

Taking N=680, n=5, and p = 0.26 (taken from misc-343.txt):

(1-0.26^5) * [1 - 0.26^5*0.74/(1-0.26^5)]^675

= 0.55121

or 55%. So in 680 games, there's a 45% chance that you'll have at least one run of five games in a row with zero Oracle spellbooks.

*gets anorak*
.

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