Re: Coin tossing guessing strategy...

On Apr 22, 11:45 am, Chula Pittayapinun <pastelsa...@xxxxxxxxx> wrote:
On Apr 22, 5:31 am, "Krieg Suden" <dd...@xxxxxxxxx> wrote:



"Chula Pittayapinun" <pastelsa...@xxxxxxxxx> wrote in message

news:a96d423f-0466-4e9f-a355-ae54101671bc@xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx

On Apr 21, 7:55 pm, quasi <qu...@xxxxxxxx> wrote:
On Tue, 22 Apr 2008 00:39:43 -0700 (PDT), Chula Pittayapinun

<pastelsa...@xxxxxxxxx> wrote:

On Apr 21, 6:19 am, quasi <qu...@xxxxxxxx> wrote:

On Mon, 21 Apr 2008 10:57:54 -0700 (PDT), Chula Pittayapinun wrote:

On Apr 21, 5:27 am, Ray Vickson <RGVick...@xxxxxxx> wrote:

ANY *particular* string, such as HTTHHHTHTHTT has the same
probability
of occurrence as any other string of the same length, such as
HHHHHHHHHHHH or TTTTTTTTTTTT (strings of length 12 in this case).
It
is true that strings of length 12 that have 5 heads and 7 tails are
much more probable than all H or all T, but that is not what we are
talking about here. You need a precise string in exactly one order,
because according to your description of the game, you lose as soon
as
the element in the nth position (H or T) fails to match the actual
result of the nth toss, if you have not already lost before the nth
toss. Just guessing the right number of H's and T's is not good
enough.

That is what I also thought. But could you please elaborate on 'that
is not what we are talking about here'? I need a precise formal
argument, if possible, to explain to my friends. (One of them has
invoked the argument of random walk- that HHH...H has lower prob of
occurance than HTTHHHTHTT.)

Coins have no memory.

Moreover, acoinis blind.

How would acoineven know whether it came up H or T?

The simplest way to defeat these fools who think a faircoinis biased
with respect to sequences of n flips, is to note that if some sequence
of say 10 flips was more likely than some other sequence of 10 flips,
then a similar bias (possibly a little less) should hold for 9 flips,
right? In other words, it's not reasonable to claim the bias happens
only for n > 9. Once you convince your challenger that n = 9 would
also show a bias, then ask what about n = 8? When you reach n = 2, get
out some coins and experiment. Does your challenger really believe any
of the 4 sequences HH, HT, TH, TT is more likely than any other?

There's something dependent on the order of n in the argument of one-
dimensional random walk (please refer to wiki page of the topic
'random walk'), although I'm not sure how that argument is related to
this problem. Furthermore: 'for any random walk in one dimension,
every point in the domain will almost surely be crossed an infinite
number of times.' Does this mean that the longer the string (as n
approach /infty), it is more likely that the number of H's will equal
T's? If not, then why is it not applicable in this problem setting?

Read my prior explanation.

For 2 coins, do you think there a bias, even a slight bias with
respect to the 4 possible sequences HH, HT, TH, TT?

If not, then why would you think a bias suddenly develops for a longer
sequence, say a 10coinsequence?

In fact, assuming a faircoin, all 10coinsequences are equally
likely.

Don't confuse the equiprobability of ncoinsequences with the
probability distribution for the _difference_ between the number of
heads and tails. Thus, for an n-term sequence, it's much more likely
that the difference is 0 than n, but that has nothing to do with the
game in question. You don't care about the totals. Effectively, you
are asking if one particular ncoinsequence is more likely than
another. The answer is "no".

quasi

Hahaha what a way to settle an argument.

That's exactly what I am saying to them: each *particular* sequence
has equal probability of occurrence i.e. 1/(2^n). In fact, saying 1/
(2^n) is absurd since the past tosses are done and really don't
count.

I found this well-explained argument in wiki article 'gambler's
fallacy':

"Now suppose that we have just tossed four heads in a row. A believer
in the gambler's fallacy might say, "If the nextcoinflipped were to
come up heads, it would generate a run of five successive heads. The
probability of a run of five successive heads is (1 / 2)5 = 1 / 32;
therefore, the nextcoinflipped only has a 1 in 32 chance of coming
up heads."

This is the fallacious step in the argument. If thecoinis fair, then
by definition the probability of tails must always be 0.5, never more
or less, and the probability of heads must always be 0.5, never less
(or more). While a run of five heads is only 1 in 32 (0.03125), it is
1 in 32 before thecoinis first tossed. After the first four tosses
the results are no longer unknown, so they do not count. The
probability of five consecutive heads is the same as that of four
successive heads followed by one tails. Tails isn't more likely. In
fact, the calculation of the 1 in 32 probability relied on the
assumption that heads and tails are equally likely at every step. Each
of the two possible outcomes has equal probability no matter how many
times thecoinhas been flipped previously and no matter what the
result. Reasoning that it is more likely that the next toss will be a
tail than a head due to the past tosses is the fallacy."

However, the above paragraph ends with the following 2 sentences:

"The fallacy is the idea that a run of luck in the past somehow
influences the odds of a bet in the future. This kind of logic would
only work if we had to guess all the tosses' results before they are
carried out."

The last sentence is how the law of large numbers (or random walk
concept) comes into play. They (my freinds) say, with big n, "HHH...H"
is much less likely to occur than other *specific* strings with equal
number of H's and T's.

Wrong. The Weak Law of Large Numbers does not apply.

Which is quite sound, so now I guess the problem is at "if we had to
guess all the tosses' results before they are carried out.". So maybe
the first question to ask is: is guessing the entire string of H's and
T's before the tosses the same as guessing the tosses one by one after
each toss?

Why guess at all ?? Will guessing change the outcome of acointoss ? I
think not.

That's why I said my problem might be ill-defined :(

your problem is trivial.
If you cannot see this, do not gamble, for sure you would loose it all.>>> CoinToss has no memory. <<<

Got it ?

As I understand, LLN states that as n->/infty, the sample average
converges to the expected values. In this case of coin tossing, that
means with big n, HH...H is less likely to occur than a *specific*
(not a set of) string of equal H's and N's, no?

No. Every individual string of length n has probability 1/2^n. That
means that an individual string with 5H's and 5T's, such as the very
specific string HHTHTTTHTH, has probability 1/2^10, exactly the same
as HHHHHHHHHH or TTTTTTTTTT. The law of large numbers does not come
into it. What the LLN DOES say is that for large n, the proportion of
strings having nearly the same number of heads as tails is close to 1,
and becomes 1 in the limit of n going to infinity. Note: this says
that for any small number e > 0, the limiting proportion of strings
having |#(H)/n - 1/2| < e is one. It DOES NOT SAY that you can put e =
0 and conclude that the exactly-equal H-T strings are probable;
indeed, they have probabilities that go to zero like 1/sqrt(n). But,
anyway, this has nothing at all to do with your original question,
which was about INDIVIDUAL strings, not Head-Tail counts or
proportions.



So if would be nice if you could explain why this argument is not
applicable in this case.

And wiki has also stated some thing interesting that leads me to
question the setting of the problem, please reread my post with quotes
from wiki.

By the way, I believe that the chance is always 50-50 and every string
is equal in probability of occurrence.

I don't know what your 50-50 refers to, but your statement about every
string (of given length, of course) has the same probability is true,
at least for fair coins and independent tosses. In that case, your
very own statement contradicts your statement about all heads being
less probable than another specific string.

So no point getting so
aggressive on me;

Nobody is getting aggressive with you. They are saying you are
incorrect, and are giving you corrected arguments. If you refuse to
listen then that is up to you.

I just want a really good precise explanation.
Because when I want to debate about something to someone, I usually
also try to disprove his/her point, not only asserting my point.

You have been given such arguments several times already, but for some
reason you either refuse to acknowledge them or you dismiss them, or
something---it is not clear what the remaining problem is.

R.G. Vickson


Thank you.

.

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