4
\$\begingroup\$

The Wushu rulebook:

Mono-A-Mono

Speaking of diverted dice, it’s not fair for groups of players to gang up on a single Nemesis. Rolling more dice than the other guy is a serious advantage, so your dice pool limits will need some adjustment when fights aren’t one-on-one.

There are two ways to go...

If you’re feeling ambitious, let the Nemesis rack up a separate dice pool for each hero aligned against them. If two players get 6 dice each, you get a total of 12. Only 6 can be directed against a single hero, though. Fair is fair.

If you’re feeling exhausted, make your players split a dice pool between them. If the Nemesis gets 6 dice, two players get 3 dice each. Three players would get 2 dice each. If that’s not enough, raise the limit a little (you get 8 dice, two players get 4 dice each).

One on One

I know it means “monkey-to-monkey” in Spanish. The wordplay is Latin. Stuff it.

Let's call these two methods "Big Pool" and "Small Pool".

I suspect that the two distributions are comparable, but small pool is swingier. Therefore, my hypothesis is that the choice between big and small pool is merely a matter of style. However, before committing to one style or another, I would like to know whether style has consequences - is there a probabilistically significant difference between the two methods?

\$\endgroup\$

2 Answers 2

1
\$\begingroup\$

Big Pool makes defense more effective and favors higher skill

As an example, let's say we have four players, and our choices for the contest between the Nemesis and an individual player could either be 2v2 dice (Small Pool) or 8v8 dice (Big Pool). The player has a skill of 5, and the Nemesis has a skill of 3. We'll assume each side splits their pools evenly between offense and defense.

The damage done between the Nemesis and an individual player over a total of 8v8 dice (4 rounds for the Small Pool and 1 round for the Big Pool) can be expressed as follows:

output 4d( [highest of 1d(d6<=3) - 1d(d6<=5) and 0] ) named "Small Pool damage to player"
output 1d( [highest of 4d(d6<=3) - 4d(d6<=5) and 0] ) named "Big Pool damage to player"

output 4d( [highest of 1d(d6<=5) - 1d(d6<=3) and 0] ) named "Small Pool damage to Nemesis"
output 1d( [highest of 4d(d6<=5) - 4d(d6<=3) and 0] ) named "Big Pool damage to Nemesis"

We can see the following:

  • The Big Pool decreases the damage dealt to both sides.
  • The Big Pool especially decreases the damage dealt to the higher-skill side. In this case, the damage dealt to the Nemesis fell from an average of 1.67 to 1.42, which is proportionally only a small drop, but the damage dealt to the player fell from 0.333 to 0.0855, nearly a four-fold reduction. In other words, despite there being four times the dice being rolled, the player is barely taking any more damage per round!

Why are they different? By merging four rounds into one, the Big Pool effectively allows extra defensive hits that would otherwise be wasted to be floated into other rounds where they can be used. It's the higher-skill side that can better leverage this towards a shutout of the lower-skill side.

\$\endgroup\$
1
  • \$\begingroup\$ AMAZING answer. The round merge insight is revelatory. \$\endgroup\$
    – order
    Jul 6 at 3:18
5
\$\begingroup\$

There is no probabilistic difference

That is, over 2 rounds, the Small Pool gives exactly the same number of hits as the Big Pool does over 1 round as can be seen here.

However, that doesn't mean there aren't consequences, from the rules:

Smaller limits encourage shorter, punchier rounds of narration and give players more chances to alter their dice rolling strategies. Higher limits allow for longer, more elaborate bouts, but fortunes can change drastically from one round to the next.

In addition, a smaller pool increases the value of Chi as adding a guaranteed hit in rounds with fewer hits is more advantageous.

\$\endgroup\$
3
  • \$\begingroup\$ I was hoping for an AnyDice answer! I tried and failed to write up a function - overcomplicted by my attempt to implement skill level and yang/yin hits. Thanks! \$\endgroup\$
    – order
    Jul 4 at 2:03
  • \$\begingroup\$ I would also add that, as Director, this means you are bound to lose in the long run, all else being equal, since players win ties in Nemesis fights. Not that that's a bad thing of course! But it does remind me of a new question... \$\endgroup\$
    – order
    Jul 4 at 2:04
  • 1
    \$\begingroup\$ @order It can hardly remind you of a question that doesn't exist. I think you mean it suggested a new question. \$\endgroup\$
    – Dale M
    Jul 4 at 11:05

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .