Crooked coin


Two people have asked you out for a date and you’ve decided to decide who to go with by flipping a coin. The problem is that the only coin you have is biased: you know that heads comes up with a probability of $0.75,$ rather than $0.5$ as is the case with a fair coin. However, you do really want each of the two people to have a fair chance of being picked. Using a combination of two coin flips instead of one, can you find a way of making the random decision fair?

(Hint: the chance doesn't have to be 50:50.)


Two consecutive heads to go with person 1 (56.25%) else go with person 2

You could ignore both the Heads, Heads case (prob=0.5625) and Tails,Tails case (Prob=0.0625). The other two cases have equal probability (0.1875).

Assign an order to each person (eg person A = HT; person B = TH) then each person has an equal 3/16 chance on any pair of flips. If HH or TT shows up simply try again.

I don't know if this is legit...

1) assign 10 flips of coin (arbitrary) to each person
2) in theory each person shall end up with 7.5 heads and 2.5 tails. but with a small number of flips the uncertainty will be larger and each person will likely end up with a different number of heads and tails
3) date with the guy with more heads :) or repeat until one guy has more heads

there should be a better way of exploiting the probability of getting certain number of heads and tails, but I'm not a stat person so I can't tell. maybe use poisson distribution?

flip 2 times as (C1, C2) in {++, --, +-, -+},
flip again if (C1,C2) = ++ or --, we get +- or -+ in infinite step in practice, means "face up" and "face down" respectively,
So the probabilities for up and down are all 5/16.
This is a theorical solution.
Be patient, and don't do this forever by yourself or God.

First Coin: person 1 has heads,
person 2 has tails,
2nd Coin: person 1 has tails,
person 2 has heads
That way both people have 1.0 chance of getting it.
That is of corse if the 2nd coin is still biased!
If not just use the 2nd coin not the first coin!

Nice puzzle, thanks. But perhaps you could reword it? I understood we had to find a solution using EXACTLY two flips. The problem spec, as implied by the answer, is a little bit easier than what I was trying(?).

Anyway, would anyone reading this care to try and solve this version of the problem? Two and only two flips...