Re: Coin tossing guessing strategy...


"Chula Pittayapinun" <pastelsalad@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, a coin is blind.

How would a coin even know whether it came up H or T?

The simplest way to defeat these fools who think a fair coin is 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 10 coin sequence?

In fact, assuming a fair coin, all 10 coin sequences are equally
likely.

Don't confuse the equiprobability of n coin sequences 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 n coin sequence 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 next coin flipped 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 next coin flipped only has a 1 in 32 chance of coming
up heads."

This is the fallacious step in the argument. If the coin is 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 the coin is 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 the coin has 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 a coin toss ? 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.
Coin Toss has no memory. <<<
Got it ?



.

Relevant Pages

  • Re: Coin tossing guessing strategy...
    ... You need a precise string in exactly one order, ... with respect to sequences of n flips, is to note that if some sequence ... In fact, assuming a fair coin, all 10 coin sequences are equally ... heads and tails. ...
    (sci.math)
  • Re: Coin tossing guessing strategy...
    ... You need a precise string in exactly one order, ... with respect to sequences of n flips, is to note that if some sequence ... In fact, assuming a fair coin, all 10 coin sequences are equally ... heads and tails. ...
    (sci.math)
  • Re: Coin tossing guessing strategy...
    ... ANY *particular* string, such as HTTHHHTHTHTT has the same probability ... How would a coin even know whether it came up H or T? ... with respect to sequences of n flips, is to note that if some sequence ... it's not reasonable to claim the bias happens ...
    (sci.math)
  • Re: Coin tossing guessing strategy...
    ... ANY *particular* string, such as HTTHHHTHTHTT has the same probability ... How would a coin even know whether it came up H or T? ... with respect to sequences of n flips, is to note that if some sequence ... it's not reasonable to claim the bias happens ...
    (sci.math)
  • Re: Complexity; was: SQL
    ... then the sequence is not random. ... > This becomes a realization, the realized sequence is no longer random. ... > They both have the same probability of occuring. ... You gamble by tossing a coin and bet on heads. ...
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