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You wish to use a fair coin to simulate occurrence or not of an event A that happens with probability 1/3. One method is to start by tossing the coin twice. If you see HH say that A occurred, if you see HT or TH say that A has not occurred, and if you see TT then repeat the process. Show that this enables you to simulate the event using an expected number of tosses equal to 8/3.

I would calculate the probability of having $${\frac{n(HH)+n(BB)}{N-n(HT)-n(TH)}}={\frac{1}{3}}$$ at the 2Nth toss, but I must also eliminate the probability, given this event, that it happens for other M<N: how to calculate such probability?

Edit: so I misunderstood the problem, since I thought the event to be simulated meant that we end tossing when ${\frac{n(HH)}{n(HH)+n(TH)+n(HT)}}={\frac{1}{3}}$, while it simply meant that one outcome representing the event (anything different from TT) happens.

StubbornAtom
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Pier Francesco Peperoni
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  • This is a nice, but well known method with which we can simulate every rational probability with a fair coin. But I wonder whether we need actually $\frac{8}{3}$ tosses to see a success. – Peter Aug 20 '20 at 11:25
  • @Peter, See my answer at https://math.stackexchange.com/questions/1356545/can-we-simulate-a-biased-coin-with-fair-die – Barry Cipra Aug 20 '20 at 11:56
  • A rather similar question here – Jean Marie Aug 20 '20 at 12:30
  • I would rather say that we need three times (in average) something not $TT$ and every time we need $\frac{4}{3}$ tosses in average. Since we have to repeat this in average three times to observse $HH$ , my result would be $4$ tosses in average. – Peter Aug 20 '20 at 12:36
  • Or even argumented easier : We need $4$ tosses in average to get $HH$ – Peter Aug 20 '20 at 12:42

1 Answers1

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The probability of having anything but TT in a single (double) toss is $\tfrac{3}{4}$. The number of tosses needed until the simulation is over is $Geom(\tfrac{3}{4})$ and the expected number of rounds is $\tfrac{4}{3}$. Since there are two tosses in each round, the expected number of tosses is $\tfrac{8}{3}$.

YJT
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