Imagine we have three people and we want to select one of them at random.

If we had a fair coin we could have tossed the coin twice to get one of the 4 outcomes and could have related the occurrence of any three of them with the first, second and third person..

How can we do this if we have a biased coin instead?

  • $\begingroup$ It isn't at all clear what you're actually asking, and, as I suspect this is homework or self-study, you should add the self-study tag. $\endgroup$
    – jbowman
    Commented Mar 31, 2018 at 18:31
  • $\begingroup$ stats.stackexchange.com/questions/50295/… $\endgroup$
    – whuber
    Commented Mar 31, 2018 at 20:43

1 Answer 1


One way to do this, is by what is known as the von-Neumann trick.

Imagine you have a biased coin where you get heads($H$) with probability $p$ and tails($T$) with probability $1-p$. Notice that if you throw the coin twice, the outcomes $HT$ and $TH$ are equiprobable with probability $p(1-p)$. You can simulate a fair coin by assigning the outcome $HT$ to $H$ and the outcome $TH$ to $T$. If you get one of the other two cases($HH$ or $TT$) you ignore them and repeat the experiment.

Now that you know how to simulate a fair coin, you can easily extend this to simulate throwing a fair coin multiple times by repeating the above experiment.

Note that this method can be very inefficient, especially when $p$ approaches $1$ e.g $p=0.99$. Since you are waiting for the first event $HT$ or $TH$, you are wasting a lot of throws since those two events only occur with probability $p(1-p)=0.0099$ each.

  • $\begingroup$ Mathematically correct...but practicality is unclear. ¿What if $p=99.99999\%$? That seems like a long time to wait for a TH or HT $\endgroup$
    – Gregg H
    Commented Mar 31, 2018 at 18:40
  • $\begingroup$ Yes indeed I should add that the method is inefficient. $\endgroup$
    – Andreas G.
    Commented Mar 31, 2018 at 18:41
  • 2
    $\begingroup$ I hope the moderators would agree a smiley emoji would be appropriate here. $\endgroup$
    – Gregg H
    Commented Mar 31, 2018 at 18:44
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    $\begingroup$ @Andreas G. wrote "Yes indeed I should add that the method is inefficient." On the contrary. The OP's method for getting someone else to do her/his homework or take-home test problem was very efficient. $\endgroup$ Commented Mar 31, 2018 at 19:10
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    $\begingroup$ @Mark L. Stone If it's homework, they are only hurting themself. Nevertheless, it's not clear(at least to me) that this is a homework exercise. $\endgroup$
    – Andreas G.
    Commented Mar 31, 2018 at 19:16