Suppose that we want to generate the outcome of the flip of a fair coin, but tha

Question

Suppose that we want to generate the outcome of the flip of a fair coin, but that all we have at our disposal is a biased coin which lands on heads with some unknown probability p that need not be equal to 1/2 . Consider the following procedure for accomplishing our task: 1. Flip the coin. 2. Flip the coin again. 3. If both flips land on heads or both land on tails, return to step 1. 4. Let the result of the last flip be the result of the experiment.

1. Show that the result is equally likely to be either heads or tails.

2. Could we use a simpler procedure that continues to flip the coin until the last two flips are different and then lets the result be the outcome of the final flip?

Explanation / Answer

Ans:

(a) The procedure can be reformulated in the following way: We flip the coin twice. If the outcomes agree, then our flips do not matter, we restart the experiment. If the two outcomes disagree, then we consider the second outcome. From this formulation one can see that the outcomes of our procedure follow a conditional distribution:

P{the procedure gives H} = P{(T, H)|(T, H) or (H, T)}

= P{(T, H)}/ P{(T, H) or (H, T)}

= (1 p)p /[(1 p)p + p(1 p)] = 1/2

. Similarly, the procedure gives tails with probability 1/2.

(b) For any 0 < p < 1, we will surely find both heads and tails eventually. Thus method (b) gives H if and only if the first flip comes out T, which happens with probability 1p. The method gives T if and only if the first flip is H, this happens with probability p. Hence method (b) does not give fair coin flip results. The difficulty in this problem is figuring out why our arguments in (a) do not work in the case of method (b). In the above formulation of (a) we consider a fixed pair of coin flips (the first two flips), while in method (b) we take a randomly selected pair of coin flips, those where we first see a change in the outcomes. This innocent-looking difference essentially modifies the probabilities of the outcomes (H, T) and (T, H). Our arguments in (a), applied on case (b), would look like this:

P{the procedure gives H} = P{(T, H)|(T, H) or (H, T)}

= P{(T, H)} /P{(T, H) or (H, T)}

= (1 p)/ [(1 p) + p] = 1 p,

and similarly the probability of outcome T is p.

Related Questions

Navigate

• Browse (All)
• Browse S
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.