Author: fca
Date: 12:49:14 08/16/98
Go up one level in this thread
On August 15, 1998 at 22:14:04, Bruce Moreland wrote:
>On August 14, 1998 at 21:48:40, fca wrote:
>>(a) In 1048576 trials (each of 20 throws) the probability of one or more "20 in
>>a row"s occurring is ________ .
>About 63%
1 - ( (2^20 - 1) / 2^20 )^1000000 == 61% == 63%
>>(b) In order for there to be at least a 99.9% probability of getting at least
>>one "20 in a row", the minimum number of such trials that can be a priori
>>expected to be needed is _______________ .
>Whatever value of N you need in order to generate whatever probability you want,
>in the above formula. I don't want to take the time to turn the formula around,
>but it's trivial.
( (2^20 - 1) / 2^20 )^x = 1 - 0.999
=> x = log(0.001) / (log(1023) - log(1024)) == c bruce's answer.
>>(c) In 1048576 such trials, where exactly 4 "20 in a row"s were observed, there
>>is a _______ % chance that this result does not contradict the "fair coin"
>>hypothesis.
>Good question, and maybe even on topic. I don't know how to do this one without
>spending more time with the statistics book. Do you?
Yup.
But Dann has emailed me that he is going away on holiday, so I wait till he
posts.
He raised questions about edge landings (I had said in the original post "let us
not count those edge terrace landings" - I guess he missed that, or distrusted
"terrace" - and also about ***whether the throws were continuous or in batches
of 20***.
No, he is not asking whether one or more coins was used (I had posted "let's
hypothesise the coin is fair", no mention of coins anywhere so it is obviously
only one coin).
Dann is a subtle man and used to my subtleties. I point out the word "batch" or
similar does not occur in my post, nor do I say or imply that the results of
each trial is independent of each other trial, or anything of that sort. So if
throws were a(i), trial k would be {a(k), a(k+1)... a(k+19)}; trial k+1 would be
{a(k+1), a(k+2) ... a(k+20)} per interpretation "X". ;-))
So, do you wish to change *any* of your answers, Bruce, assuming that
interpretation "X" is correct?
>>(d) "20 in a row" has been just been thrown for the first time in the
>>experiment. The best guess of the trial number on which this happened is
>>_____________ .
>
>One. The odds of getting 20 in a row on the first try are 1/2^20. The odds that
>you'll miss and get it on the next trial are (1 - 1/2^20) * 1/2^20, which is
>fractionally less, so by induction, blah blah blah.
There is more to Bruce than meets the eye!
But try Interpretation "X" here please...
>If two guys who are equally good marskmen take turns firing at each other, bet
>on the one that gets the first shot.
Doubtless, unless p(hitting) = 0, in which case you have to wait forever to get
your stake back ;-)
More interesting queries involve three marksmen (you, me and Dann say). We
shoot in that sequence, rotating. Our hitting probs are (0.1, 1 and 0.9), and
once one is hit, one is out. You can rely on Dann and I making perfect
decisions, and all of us (fools!) assume all the others will do so too... In all
other ways, assume this is RL.
What is your best strategy for your first shot, Bruce? ;-)
Get this right and I'll give you a hard one (designed by me: the above is
unoriginal) on the same theme.
>Why did you ask these questions?
To see what level of math sophistication was being used in drawing inferences.
> If you didn't have a reason, I'll ask you one.
Sorry, I missed the material below on my first read as I was putting everything
on ice till Dann posted. Now I see it!
Though I had a reason, I'll answer it of course!
>Describe a method wherein with nothing more than a single ordinary coin you can
>generate a probability of exactly 1/3.
By ordinary I deduce no cheapos (fair coin, no edge landings). And I'll try to
give the simplest way too.
Well, look at Pascal's Triangle. Each horizontal totals 2^n, so a third of it
ain't going to be integral. So forget using the whole line. Look at the 1 2 1
line i.e. the one that represents the frequencies of the HH, HT, TT
combinations. Need go no lower.
So, here is your answer (the "simplest" I see from an infinity of answers).
1. As a trial, take the said ordinary coin and flip it twice. Look at the
result.
2. If it is TT, disregard the trial altogether. It does not count as a trial
for any purpose. Goto 1.
3. If the result of the trial is HH, score 1 for the trial, else score 0.
Increment the trial counter by 1.
4. Goto 1 unless the trial counter has reached a suitably large number of your
choosing.
5. Divide the total score by the trial counter, and compare with 1/3.
6. And if the answer is not very close to 1/3, your large-enough number wasn't
large-enough!
;-)
Next!
btw I am a mathematician.
>OK. Back on topic for a second. I think it might be useful for the general
>computer chess population to be able to take a match result and figure out that
>odds that the weaker program won the match. I bet a lot of the time it is
>pretty near 50%, and it would be very interesting to know for sure. Since you
>asked such interesting questions, perhaps you know how we could generate an
>answer to this one?
Easy.
It is a matter of a posteriori analysis - Bayes Theorem territory. The answer
is near 50% for short matches... but "near" is subjective.
But Dann, Dan, Thorsten etc. may want to punt here too.
>bruce
In the vein of your 1/3 case, let me provide you with one. Do please be
careful. This has been used by me before, so pl do not search.
Given: A child may be only a boy (B) or a girl (G). Twins and higher do not
occur. Bs and Gs are equiprobable. The sex of any existing child does in no
way affect the probability of the sex of its later-born sibling or half-sibling
being B or G. "AttributeZ" is distributed identically among Bs and among Gs
(i.e. the two probability distributions are identical) - whether the
possession/degree of AttributeZ by/in one sibling affects the likelihood of its
possession/degree by/in another sibling is irrelevant to the question, but if
you like you may assume it does not have any effect.
Seriously, no funnies.
A man has exactly two children (whether they share a mother or not is irrelevant
as per above).
The man truthfully tells you:
"One of them is a B"
(1) What is the probability that both are Bs?
(2) And, what your answer be if *instead* the man had truthfully said "The one
with more AttributeZ is a B"?
Kind regards
fca
PS re chess: sensitivity analysis to variation in p(draw) is required to
accompany many of the conclusions being drawn.... In RL, it is crucial,
variable between 2 players as in "circumstantial" (like Anand's draw 2 vs Fritz
a couple of months back).
"Only trying to help"
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