El conjunto $\Omega(d,n)$ de distinta identificable resultados en $n$ independiente rollos de morir con $d=6$ caras ha $d^n$ elementos. Cuando el morir es justo, que significa cada uno de los resultados de un rollo ha probabilidad de $1/d$ e independencia significa cada uno de estos resultados por lo tanto, tienen probabilidad de $(1/d)^n:$ es decir, tienen una distribución uniforme $\mathbb{P}_{d,n}.$
Supongamos que se han ideado algunos de procedimiento $t$ determina que cualquiera de las $m$ resultados de una $c (=150)$colindado mueren, es decir, un elemento de $\Omega(c,m)$- o más informes de fallo (lo que significa que usted tendrá que repetir con el fin de obtener un resultado). Es decir,
$$t:\Omega(d,n)\to\Omega(c,m)\cup\{\text{Failure}\}.$$
Let $F$ be the probability $t$ results in failure and note that $F$ is some integral multiple of $d^{-n},$ say
$$F = \Pr(t(\omega)=\text{Failure}) = N_F\, d^{-n}.$$
(For future reference, note that the expected number of times $t$ must be invoked before not failing is $1/(1-F).$)
The requirement that these outcomes in $\Omega(c,m)$ be uniform and independent conditional on $t$ not reporting failure means that $t$ preserves probability in the sense that for every event $\mathcal{A}\subset\Omega(c,m),$
$$\frac{\mathbb{P}_{d,n}\left(t^{*}\mathcal{A}\right)}{1-F}= \mathbb{P}_{c,m}\left(\mathcal{A}\right) \tag{1}$$
where
$$t^{*}\left(\mathcal A\right) = \{\omega\in\Omega\mid t(\omega)\in\mathcal{A}\}$$
is the set of die rolls that the procedure $t$ assigns to the event $\mathcal A.$
Consider an atomic event $\mathcal A = \{\eta\}\subconjunto\Omega(c,m)$, which must have probability $c^{m}.$ Let $t^{*}\left(\mathcal A\right)$ (the dice rolls associated with $\eta$) have $N_\eta$ elements. $(1)$ becomes
$$\frac{N_\eta d^{-n}}{1 - N_F d^{-n}} = \frac{\mathbb{P}_{d,n}\left(t^{*}\mathcal{A}\right)}{1-F}= \mathbb{P}_{c,m}\left(\mathcal{A}\right) = c^{-m}.\tag{2}$$
It is immediate that the $N_\eta$ are all equal to some integer $N.$ It remains only to find the most efficient procedures $t.$ The expected number of non-failures per roll of the $c$ sided die is
$$\frac{1}{m}\left(1 - F\right).$$
There are two immediate and obvious implications. One is that if we can keep $F$ small as $m$ grows large, then the effect of reporting a failure is asymptotically zero. The other is that for any given $m$ (the number of rolls of the $c$-sided die to simulate), we want to make $F$ as small as possible.
Let's take a closer look at $(2)$ by clearing the denominators:
$$N c^m = d^n - N_F \gt 0.$$
This makes it obvious that in a given context (determined by $c,d,n,m$), $F$ is made as small as possible by making $d^n-N_F$ equal the largest multiple of $c^m$ that is less than or equal to $d^n.$ We may write this in terms of the greatest integer function (or "floor") $\lfloor*\rfloor$ as
$$N = \lfloor \frac{d^n}{c^m} \rfloor.$$
Finally, it is clear that $N$ ought to be as small as possible for highest efficiency, because it measures redundancy in $t$. Specifically, the expected number of rolls of the $d$-sided die needed to produce one roll of the $c$-sided die is
$$N \times \frac{n}{m} \times \frac{1}{1-F}.$$
Thus, our search for high-efficiency procedures ought to focus on the cases where $d^n$ is equal to, or just barely greater than, some power $c^m.$
The analysis ends by showing that for given $d$ and $c,$ there is a sequence of multiples $(n,m)$ for which this approach approximates perfect efficiency. This amounts to finding $(n,m)$ for which $d^n/c^m \ge 1$ approaches $N=1$ in the limit (automatically guaranteeing $F\a 0$). One such sequence is obtained by taking $n=1,2,3,\ldots$ and determining
$$m = \lfloor \frac{n\log d}{\log c} \rfloor.\tag{3}$$
The proof is straightforward.
This all means that when we are willing to roll the original $d$-sided die a sufficiently large number of times $n,$ we can expect to simulate nearly $\log d / \log c = \log_c d$ outcomes of a $c$colindado mueren por rollo. Equivalentemente,
Es posible simular un gran número de $m$ independiente de rollos de una $c$colindado muere el uso de una feria de $d$colindado mueren utilizando un promedio de $\log(c)/\log(d) + \epsilon = \log_d(c) + \epsilon$ rollos por resultado que $\epsilon$ puede hacerse arbitrariamente pequeña eligiendo $m$ lo suficientemente grande.
Ejemplos de algoritmos y
En la pregunta, $d=6$ e $c=150,$ dónde
$$\log_d(c) = \frac{\log(c)}{\log(d)} \approx 2.796489.$$
Thus, the best possible procedure will require, on average, at least $2.796489$ rolls of a d6 to simulate each d150 outcome.
The analysis shows how to do this. We don't need to resort to number theory to carry it out: we can just tabulate the powers $d^n=6^n$ and the powers $c^m=150^m$ and compare them to find where $c^m \le d^n$ are close. This brute force calculation gives $(n,m)$ pairs
$$(n,m) \in \{(3,1), (14,5), \ldots\}$$
for instance, corresponding to the numbers
$$(6^n, 150^m) \in \{(216,150), (78364164096,75937500000), \ldots\}.$$
In the first case $t$ would associate $216-150=66$ of the outcomes of three rolls of a d6 to Failure and the other $150$ outcomes would each be associated with a single outcome of a d150.
In the second case $t$ would associate $78364164096-75937500000$ of the outcomes of 14 rolls of a d6 to Failure -- about 3.1% of them all -- and otherwise would output a sequence of 5 outcomes of a d150.
A simple algorithm to implement $t$ labels the faces of the $d$-sided die with the numerals $0,1,\ldots d, d-1$ and the faces of the $c$-sided die with the numerals $0,1,\ldots, c-1.$ The $n$ rolls of the first die are interpreted as an $n$-digit number in base $d.$ This is converted to a number in base $c.$ If it has at most $m$ digits, the sequence of the last $m$ digits is the output. Otherwise, $t$ returns Failure by invoking itself recursively.
For much longer sequences, you can find suitable pairs $(n,m)$ by considering every other convergent $n/m$ of the continued fraction expansion of $x=\log(c)/\log(d).$ The theory of continued fractions shows that these convergents alternate between being less than $x$ and greater than it (assuming $x$ is not already rational). Choose those that are less than $x.$
In the question, the first few such convergents are
$$3, 14/5, 165/59, 797/285, 4301/1538, 89043/31841, 279235/99852, 29036139/10383070 \ldots.$$
In the last case, a sequence of 29,036,139 rolls of a d6 will produce a sequence of 10,383,070 rolls of a d150 with a failure rate less than $2\veces 10^{-8},$ for an efficiency of $2.79649$--indistinguibles desde el límite asintótico.
