Those damn coins...

Coin tossing (or coin flipping) is one of the popular settings for probability questions. It might be of interest to contrast coin tossing and dice rolling:

- Coin tossing follows a binomial distribution (which converges to Gaussian); die rolling follows a uniform distribution.

- Tossing m coins gives an expectation value of m/2; rolling an m-sided die gives an expectation value of (m+1)/2.

Because of the binomial distribution, coin tossing problems usually are more involved than dice rolling in terms of arithmetics. The most straight-forward kind of coin problems would simply ask you about the mean and variance of the number of Head or Tail:

CT01
What is the expected gain if you are paid $1 for each head in tossing 4 fair coins? What is the standard deviation?

CT01 - Answer:
Remember that in n binary trials with single success probability p, the mean is np and the variance is np(1-p). So the expected gain is $2 and the standard deviation is $1.

Sometimes the question is framed as a game:

CT02
What is the probability that the number of Heads is greater than or equals to 2 when tossing 4 coins?

CT02 - Answer:
The Pascal numbers for 4 trials are {1,4,6,4,1}. Hence the required probability is (6+4+1)/16 = 11/16.

Once the number of tossing becomes large, the Law of Large Number is handy.

CT03
What is the probability that there are more than 60 Heads out of tossing 100 coins?
CT03 - Answer:
The last thing you want to do is to calculate the binomial coefficients explicitly. Instead, observe that the probability distribution converges to a Gaussian as number of trials goes to infinity, with mean equals fifty and variance equals 25 (hence the standard deviation is 5). 60 is two standard deviations away from the mean, and the 68-95-997 rule gives (1 - 95%)/2 = 2.5%.

Another way to ask coin-tossing brainteeasers is to play with Bayesian statistics.

CT04
There are three coins, one with a H side and a T side (coin A); one with both sides being T (coin B); one with both sides being H (coin C).If you draw one among them at random and toss it to observe a Head, what is the conditional probability that coin C is picked?
CT04 - Answer:
This is a typical Bayesian inference question. Let X = [coin C is picked] and Y = [the toss gives an H], then
P(X|Y) = P(Y|X)*P(X)/P(Y) = 1 * (1/3) / (1/3*0 + 1/3*1/2 + 1/3*1) = 2/3

Hope you find this helpful. More to come!

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