How do I design the most unfair dice for this game? I was thinking about a simple dice game where the goal is to roll all the face values of a six-sided die consecutively in order (1,2,3,4,5,6). If I'm not mistaken, the probability of doing this with a fair die is $\left(\frac{1}{6}\right)^6$. And I think I'm correct in saying that there is no way to bias the die to increase my chances of winning. In other words, a fair (uniform) die maximizes the probability of winning this game.
So, I added a new rule: if the rolled value is smaller than the "target" value, then the player simply rolls again. For example, {1,2,3,4,5,2,6} counts as a win because rolling a 2 when the target is 6 can be ignored because $2 < 6$.
By removing the ignored values from the equation, I think the probability of winning (with a fair die) has increased from $\left(\frac{1}{6}\times\frac{1}{6}\times\dots\times\frac{1}{6}\right)$ to $\left(\frac{1}{6}\times\frac{1}{5}\times\dots\times\frac{1}{1}\right)$. In other words, it's now $\frac{1}{6!}$.
My question is: how should I (unfairly) weight my die in order to maximize the probability of winning? I feel like I should favor small values because they carry less risk (1 never loses, 2 only loses if the target is exactly 1, etc). However, I'm not sure how to formulate this problem.
Also, I feel intimidated by the fact that 1 never loses. If I design a die that always rolls 1, then I can never lose... but I can never win either. So I'm not sure if maybe I need to constrain my question in some way.
 A: Nice question! You can make the probability arbitrarily close to $1$ with a die where the probability of landing on $i$ decreases rapidly.  For example, $99\%$ chance of landing on 1, $.99\%$ chance landing on 2, $.0099\%$ chance landing on 3, etc.  Then you will have $99\%$ chance of getting a $1$ on the first throw; and on the second throw, you throw away $99\%$ of the rolls because they are $1$s, and of the remaining $1$% of rolls you keep, $99\%$ of them will be 2s, etc.  Of course, this means you will waste a LOT of time rolling dice and ignoring the results, so it might not be the most fun game...
Now, if you want to model this formally, we let the result from the die roll be distributed as a variable $X$, letting $X_i$ be the $i$-th roll (not counting the rolls we ignored and re-rolled). Then we have $P(X_i = i) = P(X = i | X\geq i)$; ignoring the too-low rolls is equivalent to conditioning on the event $X \geq i$.
Let $p_i = P(X=i)$. We want a high chance $p$, close to $1$, of having $X_i = i$. So, set $p_1 = p$; then of the remaining probability $1-p$, we want a proportion of $p$ of it on $2$, so we set $p_2 = p(1-p_1)$; and in general $p_i = p(1-(p_1 + p_2 + \cdots + p_{i-1}))$, except for $p_6$, where we don't have any choice but $p_6 = 1 - p_1 - \cdots - p_5$.
Then we have
\begin{align*}P(\text{win}) &= P(X_1 = 1) \cdot P(X_2 = 2) \cdots P(X_6 = 6)\\
&=  P(X = 1) \cdot P(X = 2 | X \geq 2) \cdots P(X = 6 | X \geq 6).\end{align*}
So, by design
\begin{align*}
P(X_i = i | X \geq i) &= P(X = i) / P(X \geq i)\\ 
&= \frac{p_i}{1 - p_1 - \cdots - p_{i-1}}\\ 
&= \frac{p(1 - p_1 - \cdots - p_{i-1})}{1 - p_1 - \cdots - p_{i-1}} = p.\end{align*}
and then $P(\text{win}) = p^5$. (note $P(X_6 = 6) = P(X = 6 | X \geq 6) = 1.$)
So, for $p = .99$ we have a $.99^5 \approx 95$% chance of winning.
A: The probability of rolling an element of a finite set $A$ before an element of a finite set $B$ is $\mathbb{P}(A)/(\mathbb{P}(A)+\mathbb{P}(B))$.
The original problem is to maximize
$$\prod_{i=1}^n \Bigl(p_i\Big/\sum_{j=1}^n p_j\Bigr)$$ subject to
\begin{align}
\sum_{i=1}^n p_i &= 1 \tag1 \\
p_i &\ge 0 &&\text{for all $i$} \tag2
\end{align}
An equivalent objective function is $\sum_{i=1}^n \log p_i$, and $p_i=1/n$ is optimal.
Your modified problem is to maximize $$\prod_{i=1}^n \Bigl(p_i\Big/\sum_{j=i}^n p_j\Bigr)$$ subject to $(1)$ and $(2)$.  Uniform probabilities indeed yield $1/n!$.  Numerically, it appears that the maximum approaches $1$.
