On 28/2/23 09:09, Jeff Zeitlin - editor at freelancetraveller.com (via
tml list) wrote:
> Assume that you have a Dx, and you need a Dy, where y < x.
>
> I've seen assertions, and experimentation _seems_ to bear it out, that if
> you roll Dx, and the result is greater than y, you ignore and reroll until
> you get y or less, this will give the same probabilities as rolling a Dy.
>
> Can someone point me to a proof/explanation of this?
>
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How about an explanation by way of example?

For sake of argument, assume FreeTrav has a d10, and needs to simulate a
d8.  Thus, x = 10, y = 8.

I'll return the favour he paid me a couple of days ago, and assume each
roll is independent and
identically distributed (i.i.d.), uniformly over all 10 faces.

I'll call 1 thru 8 collectively the "retained faces", and 9 and 10 the
"rerolled faces".

First roll:

Retained faces have their original probabilities, 0.1 each.
Rerolled faces also have probabilities of 0.1 each.
However, rolling a 9 or 10 trips a reroll.  Rerolling divvies up the
probability of
what was rolled and distributes it over all faces, in this case evenly
(by i.i.d. assumption)

Thus, 9 and 10 have a combined probability of 0.2, which divvies up to
0.02 for each face.

Second roll:
Retained faces have probabilities of 0.12 each.
Rerolled faces have probabilities of 0.02 each.

Thus, 9 and 10 have a combined probability of 0.04, which divvies up to
0.004 for each face.

Third roll:
Retained faces have probabilities of 0.124 each.
Rerolled faces have probabilities of 0.004 each.

Et cetera.

On the kth roll, the probability of a given rerolled face is 0.1 * 0.2 ^
( k - 1 ) - converging to 0 as k tends to infinity.

The retained face probabilities are a geometric sum, a = 0.1 (their
initial value) and r = 0.2.
Infinite sum of a geometric series with abs ( r) < 1 =  a / ( 1 - r) =
0.1 / (1 - 0.2) = 0.1 / 0.8 = 0.125.

The probabilities of each face on a uniform d8, whose rolls are likewise
i.i.d., are 0.125.

Not quite a proof, but that should be easily able to be generalised to one.

Alex