If you fix only the number of dice, but let the number of faces be arbitrary (or if you simply find a way to make arbitrary probabitities for different faces), then the answer for $n$ faces is $\displaystyle 1-\frac1{4\cos^2(\pi/(n+2))}$. This is proved (that's quite immodest, I know...) in my paper "intransitive roulettes" in Matematicheskoe prosveschenie, III series, 2010, Vol, 14, pp.~240--2556 but only in Russian, sorry. Similar results (lacking, perhaps, only the explicit constants) you may find in the following papers:

S. Trybula, On the paradox of $n$ random variables, Zastos. Mat. (Appl. Math.) 8 (1965), 143--154.

Z. Usiskin, Max--min probabilities in the voting paradox, Ann. Math. Statist. 35 (1964), 857--862.

The optimal example is the following. Let $q=\frac1{4\cos^2(\pi/(n+2))}$. Define $r_n=0$, $r_{i}=q/(1-r_{i+1})$. Then one can show that $1-q=r_1>r_2>\dots>r_n=0$. Now let us make the following "dice": $i$th die ($1\leq i\leq n$) makes $i$ with probability $1-r_i$ and makes $i+n$ with probability $r_i$. Then each die wins the (cyclically) previous one with probability $1-q$.

Consequently, $2/3$ is the optimal number for $n=4$ for any number of faces. Pitifully, the answers for $n>4$ are irrational, hence they are not achievable on regular dice.

It seems that the optimal configuration for $n>4$ on regular dice can be made by the corresponding modification of the general optimal example. E.g., for $n=5$ we have $r_4=q=0.30797\dots$, $r_3=0.445\dots$, $r_2=1-r_3$, $r_1=1-r_4$, so we cannot achieve 70%. On the other hand, these values can be approximated to make the following 5 icosahedral dice: $$(6\times 1, 14\times 6), \quad (9\times 2, 11\times 7), \quad (11\times 3, 9\times 8), \quad (14\times 4, 6\times 9), \quad (20\times 5), $$ where each wins the (cyclically) next one with probability at most $\frac{9\times 14}{20^2}=0.315$.

Next, there is a bound for the answer when the number of faces is bounded (or fixed, as in our case). If the number of faces is $2k$ for each die, then consider the $k$th maximal numbers on each die. Consider the die which contains the maximal number among them; then it wins the next one with the probability at least $\frac{k+1}{4k}$. Hence for the icosahedral dice the result cannot exceed $\frac{29}{40}=0.725$. THis can be achieved on the following set: $$ (5\times 1,15\times 11), \quad (7\times 2,13\times 12), \quad (8\times 3,12\times 13), \quad (9\times 4,11\times 14), \quad (10\times 5,10\times 15), $$ $$ (11\times 6,9\times 16), \quad (12\times 7,8\times 17), \quad (13\times 8,7\times 18), \quad (15\times 9,5\times 19), \quad (20\times 10), $$ but not on a smaller one. Some less optimal answers with smaller number of dice are $$ (5\times 1,15\times 9), \quad (7\times 2,13\times 10), \quad (9\times 3,11\times 11), \quad (10\times 4,10\times 12), $$ $$ (11\times 5,9\times 13), \quad (13\times 6,7\times 14), \quad (15\times 7,5\times 15), \quad (20\times 8) $$ with losing probability at most $\frac{13\times 9}{20^2}=0.2925$, and $$ (5\times 1,15\times 7), \quad (8\times 2,12\times 8), \quad (10\times 3,10\times 9), (12\times 4,8\times 10), \quad (15\times 5,5\times 11), \quad (20\times 6) $$ with losing probability at most $\frac{15\times 8}{20^2}=\frac{12\times 10}{20^2}=0.3$ --- exactly 30% on 6 dice.

Finally, for dodecahedral dice the bound is $\frac{17}{24}=0.7083\dots$, hence it is also possible to make it more than 70% (but this is impossible for octahedral dice...). The example is as follows: $$ (3\times 1,9\times 9), \quad (4\times 2,8\times 10), \quad (5\times 3,8\times 11), \quad (6\times 4,8\times 12), $$ $$ (7\times 5,8\times 13), \quad (8\times 6,8\times 14), \quad (9\times 7,8\times 15), \quad (12\times 8). $$

I cannot claim that the numbers of dice presented above are optimal for these probabilities, but it seems so.

If you fix only the number of dice, but let the number of faces be arbitrary (or if you simply find a way to make arbitrary probabitities for different faces), then the answer for $n$ faces is $\displaystyle 1-\frac1{4\cos^2(\pi/(n+2))}$. This is proved (that's quite immodest, I know...) in my paper "intransitive roulettes" in Matematicheskoe prosveschenie, III series, 2010, Vol, 14, pp.~240--255, but only in Russian, sorry. Similar results (lacking, perhaps, only the explicit constants) can be found in the following papers:

S. Trybula, On the paradox of $n$ random variables, Zastos. Mat. (Appl. Math.) 8 (1965), 143--154.

Z. Usiskin, Max--min probabilities in the voting paradox, Ann. Math. Statist. 35 (1964), 857--862.

The optimal example is the following. Let $q=\frac1{4\cos^2(\pi/(n+2))}$. Define $r_n=0$, $r_{i}=q/(1-r_{i+1})$. Then one can show that $1-q=r_1>r_2>\dots>r_n=0$. Now let us make the following "dice": $i$th die ($1\leq i\leq n$) makes $i$ with probability $1-r_i$ and makes $i+n$ with probability $r_i$. Then each die wins the (cyclically) previous one with probability $1-q$.

Consequently, $2/3$ is the optimal number for $n=4$ for any number of faces. Pitifully, the answers for $n>4$ are irrational, hence they are not achievable on regular dice.

It seems that the optimal configuration for $n>4$ on regular dice can be made by the corresponding modification of the general optimal example. E.g., for $n=5$ we have $r_4=q=0.30797\dots$, $r_3=0.445\dots$, $r_2=1-r_3$, $r_1=1-r_4$, so we cannot achieve 70%. On the other hand, these values can be approximated to make the following 5 icosahedral dice: $$(6\times 1, 14\times 6), \quad (9\times 2, 11\times 7), \quad (11\times 3, 9\times 8), \quad (14\times 4, 6\times 9), \quad (20\times 5), $$ where each wins the (cyclically) next one with probability at most $\frac{9\times 14}{20^2}=0.315$.

Next, there is a bound for the answer when the number of faces is bounded (or fixed, as in our case). If the number of faces is $2k$ for each die, then consider the $k$th maximal numbers on each die. Consider the die which contains the maximal number among them; then it wins the next one with the probability at least $\frac{k+1}{4k}$. Hence for the icosahedral dice the result cannot exceed $\frac{29}{40}=0.725$. THis can be achieved on the following set: $$ (5\times 1,15\times 11), \quad (7\times 2,13\times 12), \quad (8\times 3,12\times 13), \quad (9\times 4,11\times 14), \quad (10\times 5,10\times 15), $$ $$ (11\times 6,9\times 16), \quad (12\times 7,8\times 17), \quad (13\times 8,7\times 18), \quad (15\times 9,5\times 19), \quad (20\times 10), $$ but not on a smaller one. Some less optimal answers with smaller number of dice are $$ (5\times 1,15\times 9), \quad (7\times 2,13\times 10), \quad (9\times 3,11\times 11), \quad (10\times 4,10\times 12), $$ $$ (11\times 5,9\times 13), \quad (13\times 6,7\times 14), \quad (15\times 7,5\times 15), \quad (20\times 8) $$ with losing probability at most $\frac{13\times 9}{20^2}=0.2925$, and $$ (5\times 1,15\times 7), \quad (8\times 2,12\times 8), \quad (10\times 3,10\times 9), (12\times 4,8\times 10), \quad (15\times 5,5\times 11), \quad (20\times 6) $$ with losing probability at most $\frac{15\times 8}{20^2}=\frac{12\times 10}{20^2}=0.3$ --- exactly 30% on 6 dice.

Finally, for dodecahedral dice the bound is $\frac{17}{24}=0.7083\dots$, hence it is also possible to make it more than 70% (but this is impossible for octahedral dice...). The example is as follows: $$ (3\times 1,9\times 9), \quad (4\times 2,8\times 10), \quad (5\times 3,8\times 11), \quad (6\times 4,8\times 12), $$ $$ (7\times 5,8\times 13), \quad (8\times 6,8\times 14), \quad (9\times 7,8\times 15), \quad (12\times 8). $$

I cannot claim that the numbers of dice presented above are optimal for these probabilities, but it seems so.

Hm. If you fix only the number of dice, but let the number of faces be arbitrary (or if you simply find a way to make arbitrary probabitities for different faces), then the answer for $n$ faces is $\displaystyle 1-\frac1{4\cos^2(\pi/(n+2))}$. This is prived (that's quite immodest, I know...) in my paper "intransitive roulettes" in Matematicheskoe prosveschenie, III series, 2010, Vol, 14, pp.~240--2556 but only in Russian, sorry.

The optimal example is the following. Let $q=\frac1{4\cos^2(\pi/(n+2))}$. Define $r_n=0$, $r_{i}=q/(1-r_{i+1})$. Then one can show that $1-q=r_1>r_2>\dots>r_n=0$. Now let us make the following "dice": $i$th die ($1\leq i\leq n$) makes $i$ with probability $1-r_i$ and makes $i+n$ with probability $r_i$. Then each die wins the (cyclically) previous one with probability $1-q$.

Consequently, $2/3$ is the optimal number for $n=4$ for any number of faces. Pitifully, the answers for $n>4$ are irrational, hence they are not achievable on regular dice.

It seems that the optimal configuration for $n>4$ on regular dice can be made by the corresponding modification of the general optimal example. E.g., for $n=5$ we have $r_4=q=0.30797\dots$, $r_3=0.445\dots$, $r_2=1-r_3$, $r_1=1-r_4$, so we cannot achieve 70%. On the other hand, these values can be approximated to make the following 5 icosahedral dice: $$(6\times 1, 14\times 6), \quad (9\times 2, 11\times 7), \quad (11\times 3, 9\times 8), \quad (14\times 4, 6\times 9), \quad (20\times 5), $$ where each wins the (cyclically) next one with probability at most $\frac{9\times 14}{20^2}=0.315$.

Next, there is a bound for the answer when the number of faces is bounded (or fixed, as in our case). If the number of faces is $2k$ for each die, then consider the $k$th maximal numbers on each die. Consider the die which contains the maximal number among them; then it wins the next one with the probability at least $\frac{k+1}{4k}$. Hence for the icosahedral dice the result cannot exceed $\frac{29}{40}=0.725$. THis can be achieved on the following set: $$ (5\times 1,15\times 11), \quad (7\times 2,13\times 12), \quad (8\times 3,12\times 13), \quad (9\times 4,11\times 14), \quad (10\times 5,10\times 15), $$ $$ (11\times 6,9\times 16), \quad (12\times 7,8\times 17), \quad (13\times 8,7\times 18), \quad (15\times 9,5\times 19), \quad (20\times 10), $$ but not on a smaller one. Some less optimal answers with smaller number of dice are $$ (5\times 1,15\times 9), \quad (7\times 2,13\times 10), \quad (9\times 3,11\times 11), \quad (10\times 4,10\times 12), $$ $$ (11\times 5,9\times 13), \quad (13\times 6,7\times 14), \quad (15\times 7,5\times 15), \quad (20\times 8) $$ with losing probability at most $\frac{13\times 9}{20^2}=0.2925$, and $$ (5\times 1,15\times 7), \quad (8\times 2,12\times 8), \quad (10\times 3,10\times 9), (12\times 4,8\times 10), \quad (15\times 5,5\times 11), \quad (20\times 6) $$ with losing probability at most $\frac{15\times 8}{20^2}=\frac{12\times 10}{20^2}=0.3$ --- exactly 30% on 6 dice.

Finally, for dodecahedral dice the bound is $\frac{17}{24}=0.7083\dots$, hence it is also possible to make it more than 70% (but this is impossible for octahedral dice...). The example is as follows: $$ (3\times 1,9\times 9), \quad (4\times 2,8\times 10), \quad (5\times 3,8\times 11), \quad (6\times 4,8\times 12), $$ $$ (7\times 5,8\times 13), \quad (8\times 6,8\times 14), \quad (9\times 7,8\times 15), \quad (12\times 8). $$

I cannot claim that the numbers of dice presented above are optimal for these probabilities, but it seems so.

If you fix only the number of dice, but let the number of faces be arbitrary (or if you simply find a way to make arbitrary probabitities for different faces), then the answer for $n$ faces is $\displaystyle 1-\frac1{4\cos^2(\pi/(n+2))}$. This is proved (that's quite immodest, I know...) in my paper "intransitive roulettes" in Matematicheskoe prosveschenie, III series, 2010, Vol, 14, pp.~240--2556 but only in Russian, sorry. Similar results (lacking, perhaps, only the explicit constants) you may find in the following papers:

S. Trybula, On the paradox of $n$ random variables, Zastos. Mat. (Appl. Math.) 8 (1965), 143--154.

Z. Usiskin, Max--min probabilities in the voting paradox, Ann. Math. Statist. 35 (1964), 857--862.

The optimal example is the following. Let $q=\frac1{4\cos^2(\pi/(n+2))}$. Define $r_n=0$, $r_{i}=q/(1-r_{i+1})$. Then one can show that $1-q=r_1>r_2>\dots>r_n=0$. Now let us make the following "dice": $i$th die ($1\leq i\leq n$) makes $i$ with probability $1-r_i$ and makes $i+n$ with probability $r_i$. Then each die wins the (cyclically) previous one with probability $1-q$.

Consequently, $2/3$ is the optimal number for $n=4$ for any number of faces. Pitifully, the answers for $n>4$ are irrational, hence they are not achievable on regular dice.

It seems that the optimal configuration for $n>4$ on regular dice can be made by the corresponding modification of the general optimal example. E.g., for $n=5$ we have $r_4=q=0.30797\dots$, $r_3=0.445\dots$, $r_2=1-r_3$, $r_1=1-r_4$, so we cannot achieve 70%. On the other hand, these values can be approximated to make the following 5 icosahedral dice: $$(6\times 1, 14\times 6), \quad (9\times 2, 11\times 7), \quad (11\times 3, 9\times 8), \quad (14\times 4, 6\times 9), \quad (20\times 5), $$ where each wins the (cyclically) next one with probability at most $\frac{9\times 14}{20^2}=0.315$.

Next, there is a bound for the answer when the number of faces is bounded (or fixed, as in our case). If the number of faces is $2k$ for each die, then consider the $k$th maximal numbers on each die. Consider the die which contains the maximal number among them; then it wins the next one with the probability at least $\frac{k+1}{4k}$. Hence for the icosahedral dice the result cannot exceed $\frac{29}{40}=0.725$. THis can be achieved on the following set: $$ (5\times 1,15\times 11), \quad (7\times 2,13\times 12), \quad (8\times 3,12\times 13), \quad (9\times 4,11\times 14), \quad (10\times 5,10\times 15), $$ $$ (11\times 6,9\times 16), \quad (12\times 7,8\times 17), \quad (13\times 8,7\times 18), \quad (15\times 9,5\times 19), \quad (20\times 10), $$ but not on a smaller one. Some less optimal answers with smaller number of dice are $$ (5\times 1,15\times 9), \quad (7\times 2,13\times 10), \quad (9\times 3,11\times 11), \quad (10\times 4,10\times 12), $$ $$ (11\times 5,9\times 13), \quad (13\times 6,7\times 14), \quad (15\times 7,5\times 15), \quad (20\times 8) $$ with losing probability at most $\frac{13\times 9}{20^2}=0.2925$, and $$ (5\times 1,15\times 7), \quad (8\times 2,12\times 8), \quad (10\times 3,10\times 9), (12\times 4,8\times 10), \quad (15\times 5,5\times 11), \quad (20\times 6) $$ with losing probability at most $\frac{15\times 8}{20^2}=\frac{12\times 10}{20^2}=0.3$ --- exactly 30% on 6 dice.

Finally, for dodecahedral dice the bound is $\frac{17}{24}=0.7083\dots$, hence it is also possible to make it more than 70% (but this is impossible for octahedral dice...). The example is as follows: $$ (3\times 1,9\times 9), \quad (4\times 2,8\times 10), \quad (5\times 3,8\times 11), \quad (6\times 4,8\times 12), $$ $$ (7\times 5,8\times 13), \quad (8\times 6,8\times 14), \quad (9\times 7,8\times 15), \quad (12\times 8). $$

I cannot claim that the numbers of dice presented above are optimal for these probabilities, but it seems so.