## Unfair coin

Posted by Chris on November 19, 2011 – 7:05 pm

An unfair coin has the property that when flipped four times, it has the same probability of turning up 2 heads and 2 tails (in any order) as 3 heads and 1 tail (in any order). What is the probability of getting a head in any one flip?

November 19th, 2011 at 10:30 pm

If 4 tosses yield 2h2t or 3h1t with equal probability then 8 tosses should yield 5h3t, so the probability of getting a head in one toss should be 5/8 or 62.5%.

November 20th, 2011 at 2:29 am

Hi Wiz. I again don’t know if you did it on purpose or not, but the argument is completely wrong and so is the answer Basically, you are introducing a “new fair coin” which single flip consists of 4 flips of the original coin and has sides 2H2T and 3H1T. Then you claim that if you flip the new fair coin twice it will always land on both of its sides exactly once. The first part is wrong, because your coin has 3 more sides you are not taking care of – 0H4T, 1H3T, and 4H0T! There is no point in explaining the mistake in the second assumption.

I don’t want to give the real answer for the moment being

November 20th, 2011 at 6:24 am

Wiz excels at making these seemingly plausible arguments

If I haven’t goofed, the probability of getting 5 heads and 3 tails is 108864/390625 = 0.27869184. It is the most probable combination, i.e. 4 heads and 4 tails or 6 heads and 2 tails is less likely.

November 20th, 2011 at 7:04 am

0.6

November 20th, 2011 at 7:28 am

Hi Mark53. That’s right. How did you do it?

November 20th, 2011 at 8:11 am

Let p=P(H), then 1-p=P(T)

Let X = number of heads out of 4

p=probability of a head

X~B(4,p)

Then you just need to work out P(X=2) = P(X=3)

4C2p^2(1-p)^2=4C3p^3(1-p)

Solving this gives the quadratic 5p^2-8p+3=0

And solving quadratic gives p=0.6 or 1.

p can’t equal 1 as no chance of a head so p=0.6

November 20th, 2011 at 8:53 am

Hi Mark53. Thank you. That looks good. Your notation may be hard for others to understand, so I’ll clarify.

B(n,p) is the binomial distribution. However, I’ll not bother with that. Also 4C2 is the combinations “function”, which I write as C(4,2) or more generally as

C(n,r) = n!/((n-r)! r!).

The number of ways of getting 2 heads (and 2 tails) in 4 flips is C(4,2) = 6. The number of ways of getting 3 heads (and 1 tail) in 4 flips is C(4,1) = C(4,3) = 4.

The probability of flipping any particular sequence of 2 heads and 2 tails is

p*p*(1-p)*(1-p) = p

^{2}(1-p)^{2}. The probability of flipping any particular sequence of 3 heads and 1 tail is p^{3}(1-p). Now accounting for the possible ways of flipping and noting the equality we get 6 p^{2}(1-p)^{2}= 4 p^{3}(1-p). Clearly p can’t be 0 or 1, so we may divide throughout by p^{2}(1-p) to get 6 (1-p) = 4 p => p = 3/5.Not that it matters, but the ways of flipping 2 heads and 2 tails are:

HHTT, TTHH, HTTH, THHT, HTHT, THTH and for 3 heads and 1 tail we have

THHH, HTHH, HHTH, HHHT

November 20th, 2011 at 12:33 pm

I live in a parallel universe where the laws of probability are what I choose them to be – and much simpler than yours.

That is why my answer is 0.625. Not that far away, really.

April 6th, 2012 at 2:39 pm

how about the two trivial cases of a double-headed or double-tailed coin?

the probability of 2H2T and 3H1T is in both cases the same (0)

this gives P(H) as 1 or 0