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 Post subject: Re: math puzzle
Post #21 Posted: Fri Jun 17, 2011 6:08 am 
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jts wrote:
Isn't it fairly obvious? If on average a coin comes up heads 50% of the time, then after it comes up heads on one flip, it needs to come up heads less than fifty percent of time on subsequent flips in order for the asserted probability to hold true. If you doubt me, flip a coin five or six times to test this.



that's NOT true :scratch: ... but That was a joke right ? Beware you might plant some weird idea on ppl mind, probas are tricky enough as they are

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 Post subject: Re: math puzzle
Post #22 Posted: Fri Jun 17, 2011 7:17 am 
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@ percival
I think we both read the question wrong. I (and probably you too) interpreted the question as; Flip a coin 3 times, which sequence is more likely: HHT or HTH. They are obviously the same with each being 1/8. This is because each coin flip is a separate event and the result is not influenced by previous flips. You can easily draw a decision tree and then you just have to walk along one of the branches and multiply the probabilities.

But the question was: I play a large number of games (10000+) which sequence is likely to appear first and how often will it appear. This kind of math is more advanced than the first one because it depends on the previous results. If the first two flips are HH then you have 50% to hit HHT but 0% to hit HTH :mrgreen:

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 Post subject: Re: math puzzle
Post #23 Posted: Fri Jun 17, 2011 8:05 am 
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Mnemonic wrote:
@ percival
I think we both read the question wrong. I (and probably you too) interpreted the question as; Flip a coin 3 times, which sequence is more likely: HHT or HTH. They are obviously the same with each being 1/8. This is because each coin flip is a separate event and the result is not influenced by previous flips. You can easily draw a decision tree and then you just have to walk along one of the branches and multiply the probabilities.

But the question was: I play a large number of games (10000+) which sequence is likely to appear first and how often will it appear. This kind of math is more advanced than the first one because it depends on the previous results. If the first two flips are HH then you have 50% to hit HHT but 0% to hit HTH :mrgreen:


i agree with that but my "that's NOT true" was to jts first comment: saying that if you got 1 H first you have less than 1/2 prob to have H next to "maintain the balance" :shock: but i think that was a joke . Or i didnt understand the statement
Hyperpape explanation is the right one (IMHO) even if it still hurts my common sense so much that i want to find a flaw.

i'll come up with my own Proba anecdote:

There was a probability teacher who would ask his class, onthe first day, to generate a random string of H and T in two ways:
Th
The one whose mother last name started with an L or before should flip a coin 200 times, the other should just make out a sequence from the top of their heads.
the "mother last name" thing is choosen so that the professor doesnt know which student used which way (but still some students will have to).

Then the professor would guess which one was simulated and which wasnt at a glance wit ha very high sucess rate. How did he do that ?

.. i don t remember where i heard that story hope it wasnt on that board....
He would look for a string of 5 H in a row, which is highly likely to happen in a 200 H or T random seq, but a human would never produce it while trying to be pseudo random
Edit: proba is much closer to 1 with 5 H in a row than 6

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 Post subject: Re: math puzzle
Post #24 Posted: Fri Jun 17, 2011 12:21 pm 
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Back to Araban Question, here is a more complete version of hyperpape answer; (with the inital 3 valued event)

let P(TZT ;n)= prob that pattern TZT appear for the first time at match n AND TTZ did not happened before
P(TTZ;n) = prob that pattern TTZ appear for the first time at match n AND TZT did not happened before

Q(n): neither pattern arise at rank n AND nth match was vs a T;

we want to know if sum P(TTZ;n) > sum P(TZT;n)
P(TTZ;n) P(TZT;n) and Q(n) are hard to compute exactly and the sum are worse
BUT it is enough to prove that P1(n) >= P2(n) for all n.
Hyperpape proves it for n=3 and 4;
more generally
we have P(TZT;n+2)=[Q(n) - p(Match n-1 was T knowing Q(n))]/9
( we need to substract p(Match n-1 was T ...) becasue in that case first occurrence is TTZ at rank n+1)
P(TTZ;n+2)=Q(n)/9
so P1(n+2)<=P2(n+2) for each n we do not know pp(Match n-1 was T) but it does not matter
so pattern TTZ has a greater chance of appearing first but computing it exactly is not easy (it probably converge quickly though)

with that i am more or less convinced

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