Given a biased coin, with probability of Heads equal to x. How to do unbiased coin tossing?
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Lets define an event, E as tossing the biased coin twice. The possible outcomes with probabilities is as follows
P(h,h) = x^2
P(h,t) = x(1-x)
P(t,h) = x(1-x)
P(t,t) = (1-x)^2
The event h,t or t,h are equi-likely, without any bias we can call that if Event h,t occurs it means head, t,h means tails but if h,h or t,t occurs we repeat the experiment.
Try to find the expected number of coin toss that would be required to call heads or tails?
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Let expected coin toss be e.
Probability that we get outcome in 1st event is 2x(1-x)
Total number of coin toss would be 2
Probability that we get no outcome in 1st event is [1 - 2x(1-x)]
Total number of coin toss would be 2 + e (we wasted 2 coin toss and still we expect e)
==> e = 2 * 2x(1-x) + (2+e) * [1 - 2x(1-x)]
==> e = 4x(1-x) + 2 - 4x(1-x) + e - 2ex(1-x)
==> e = 1/ [x(1-x)]
so for x = 1/2, e = 4
for x=2/3, e = 4.5
for x=1, e = INF (as expected, because all we get is chain of h h h h h)
What is the probability that there wont be any outcome in e coin tosses (expected outcomes)?
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p = prob of e heads + prob of e tail
==> p = x^e + (1-x)^e
To find a bound on this p in polynomial terms can be done by using binomial expansion and
Newtonian series is out of the scope of this blog. But some number crunching is.
For x = 1/2, e = 4, p = 0.125
For x = 2/3, e = 4.5, p = 0.168
For x = 3/4, e = 5.33, p = 0.216
For x = 0.99, e = 101, p = 0.362
This tells us that probability that outcome comes is quite good even for high coin biases.
How can we improve on the expected number of coin tosses?
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In above method, we say that HT or TH terminates experiment. and continue the experiment a fresh when the outcome is HH or TT.
We can further combine outcome of two such events to increase the probability of outcome e.g. say HH TT => heads and TT HH means tails.
How much expected coin tosses are we doing in this case?
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