Derek Nalls wrote on Sat, May 24, 2008 01:39 PM UTC:
'Actually the chance for twice the same flip in a row is 1/2.'
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Really?
You obviously need a lesson on probability.
Let us start with elementary stuff.
Mathematical Ideas
fifth edition
Miller & Heeren
1986
It is an old college textbook from a class I took in the mid-90's.
[Yes, I passed the class.]
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It says interesting things such as-
'The relative frequency with which an outcome happens
represents its probability.'
'In probability, each repetition of an experiment is a trial.
The possible results of each trial are outcomes.'
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An example of a probability experiment is 'tossing a coin'.
Each 'toss' (trial of the experiment) has only two equally-possible
outcomes, 'heads' or 'tails' ... assuming the condition that the
coin is fair (i.e., not loaded).
probability = p
heads = h
tails = t
number of tosses = x
addition = +
involution = ^
[This is a substitute upon a single line for superscript representation
of an exponent to the upper right of a base.]
probability of heads = p(h)
probability of tails = p(t)
p(h) is a base.
p(t) is a base.
x is an exponent.
p(h) = 0.5
p(t) = 0.5
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What follows are examples of the chances of getting the same result
upon EVERY consecutive toss.
1 time
x = 1
p(h) ^ x = 0.5 ^ 1 = 0.5
p(t) ^ x = 0.5 ^ 1 = 0.5
Note: In this case only ...
p(h) + p(t) = 1.0
2 times
x = 2
p(h) ^ x = 0.5 ^ 2 = 0.25
p(t) ^ x = 0.5 ^ 2 = 0.25
3 times
x = 3
p(h) ^ x = 0.5 ^ 3 = 0.125
p(t) ^ x = 0.5 ^ 3 = 0.125
Etc ...
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By a function that is the inverse of successive exponents of base 2,
the chance for consecutive tosses to yield the same result rapidly
becomes extremely small.
When this occurs, there are only two possibilities- 'random good-bad
luck' or an unfair advantage-disadvantage exists (i.e., 'the coin is loaded'). The sum of these two possibilities always equals 1.
random luck (good or bad) = l
unfair (advantage or disadvantage) = u
luck (heads) = l(h)
luck (tails) = l(t)
unfair (heads) = u(h)
unfair (tails) = u(t)
p(h) ^ x = l(h)
p(t) ^ x = l(t)
l(h) + u(h) = 1
l(t) + u(t) = 1
Therefore, as the chances of 'random good-bad luck' become extremely low in the example, the chances of an advantage-disadvantage existing for 'one side of the coin' or (if you follow the analogy) 'one side of the gameboard' or 'one player' or 'one set of piece values' become likewise extremely high.
Only if it can be proven that an advantage-disadvantage does not exist for one player, then can it be accepted that the extremely unlikely event by
'random good-bad luck' is indeed the case.
It is essential to understand that random good luck or random bad luck
cannot be consistently relied upon. From this fact alone, firm
conclusions can be responsibly drawn with a strong probability of
correctness.
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1 time
x = 1
p(h) ^ x = 0.5
u(h) = 0.5
p(t) ^ x = 0.5
u(t) = 0.5
2 times
x = 2
p(h) ^ x = 0.25
u(h) = 0.75
p(t) ^ x = 0.25
u(t) = 0.75
3 times
x = 3
p(h) ^ x = 0.125
u(h) = 0.875
p(t) ^ x = 0.125
u(t) = 0.875
Etc ...
