Re: Coin tossing guessing strategy...

On Apr 21, 9:43 am, Chula Pittayapinun <pastelsa...@xxxxxxxxx> wrote:
In a coin tossing game consisting of a single player, a single fair
coin is tossed several times. After each toss, if the player can
correctly guess the outcome (H/T) of that toss then he wins and the
game ends. However, if after finite n tosses the player has missed all
of n tosses, he loses.

This game is simply the coin gambling with Martingale strategy
(doubling the bet every time if lose, so only a single win can recover
all the loses plus a reward equals to initial bet) with finite
resource (after finite tosses the resource ran out and the player
loses all).

A player's strategy can be represented by a string of H's and T's of
length n, e.g. HTHTTT for n = 6. And since each tosses are considered
independent, a strategy can be predetermined before the actual tosses
(?).

The question is : does the strategy HHH...H (or TTT...T) have more
probability of winning than any other specific arbitrary strategies,
says, HTHTTT for n= 6? What about strategies with equal H's and T's
(e.g. THTHTH...TH)?

===========================================

I have seen some arguments saying that HHH...H have high prob of
winning, since the outcome TTT...T (which make you lose) have lowest
prob of occuring-- with big n, the chance of tosses being all T is low
(according to common sense and some theory...).

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.

R.G. Vickson


But then again, in my
tinkling opinion, since for a specific arbitrary strategy, there is
also only one string that can make you lose (i.e. the 'opposite'
string), and that string should also has the prob of 1/(2^n), thus the
winning prob of 1-1/(2^n) for any strategy. Therfore such argument is
absurd. And we are back to the 50-50 chance of winning regardless of
tosses order.

As you can see, I have a very little knowledge in Probability and I
have poorly stated the problem. So please feel free to restate it if
necessary. Thank you for your time helping this confused soul.

.

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