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VtM5 has many quirks that make their calculation of probabilities really hard. But ignoring hunger dice, how can I calculate the distribution of success of a given pool considering that each pair of 10's counts as two successes (a regular success is just d10>5).

So 10,1 is just one success, but 10,10 is four.

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Turns out it is simpler than I thought, but I must use a function:

function: roll ROLL:s  {
 result: (ROLL > 5) + ((ROLL = 10)/2)*2
}

I just sum every success and for every pair of 10's I add 2 successes.

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