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You have 2 coins.

one (A) has probabilities 1/2 for H and 1/2 for T, The other (B) has probabilities (n-1)/n for H and 1/n for T.

You wish to use these 2 coins to simulate a balanced n-faces dice. with a finite and pre-known number of flips.

Is it possible to do so?

If the probabilities where (for example) A = (1/2,1/2), B=(4/5,1/5) and you wished to simulate a 5-faces dice you would roll B, if you got T you would return 4 otherwise you would flip A 2 times and return the binary number you got ( 0 , 1 , 2 or 3 )

But if you had only one coin A=(1/2,1/2) and you wished to simulate a 7-faces dice it's provable that it is not possible

Ofer Magen
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  • You said: "But if you had only one coin A=(1/2,1/2) and you wished to simulate a 7-faces dice it's provable that it is not possible". But it is, isn't it? Roll the die three times, and assign a value based on the sequence: HHH is 1, HHT is 2, ... THT is 6, TTH is 7, and for TTT you just reroll and try again. Is the "reroll and try again" why this doesn't "count" as a simulation? (It would obviously work for any practical purpose....) – Aaron Montgomery Nov 05 '17 at 18:26
  • @AaronMontgomery "with a finite and pre-known number of flips" – orlp Nov 05 '17 at 20:09
  • @orlp Ah, yes - of course. Silly miss on my part. – Aaron Montgomery Nov 05 '17 at 20:11

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An algorithm is given for simulating a fair die with $n$ sides using two biased coins, one with $p = 1/2$ and one $p = 1/n$, for arbitrary integer $n$ in On Dice and Coins: Models of Computation for Random Generation by Feldman et. al in section 2.1.

The algorithm reproduced here is (with two corrections by me, $\leq$ replaced with $<$):

  1. Flip the $1/n$ coin once and the fair coin $k = \lceil 2\log n\rceil$ times. Let $v$ be the number that represents the value of the sequence of the $k$ fair coin flips when viewed as a binary number.

  2. If the $1/n$ coin is tails then if $v \color{red} < d(n-1)$ then output $(v \bmod n-1) + 1$ else output $n$.

  3. If the $1/n$ coin is heads then if $v \color{red}< r(n-1)$ output $(v \bmod n-1) + 1$ else output $n$.

Where a $1/n$ coin is heads $1/n$ of the time and $d = \lfloor 2^k/(n-1)\rfloor$ and $r = 2^k \bmod n - 1$.

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orlp
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  • for n = 7 then k = 6, d = 10, r = 4 the odds of getting a 7 are: 1 - ((6/7)(61/64) + (1/7)(25/64)) which is 57 / 448 rather than 1/7 – Ofer Magen Nov 05 '17 at 19:31
  • the algorithm is incorrect – Ofer Magen Nov 05 '17 at 19:31
  • @OferMagen I transcribed the algorithm exactly as it is in the reference, as far as I can see. But it does appear that there is an off-by-one error in the reference, and that the $\leq$ checks in the algorithm should be $<$ checks. In that case replace $61$ with $60$ and $25$ with $24$ giving the correct $1/7$. – orlp Nov 05 '17 at 19:59
  • yeah that's also pretty easy to prove, thank you – Ofer Magen Nov 05 '17 at 20:58