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New Wold of Darkness Success Probabilites
06/18/09
New Wold of Darkness Success Probabilites
I previously posted about the probabilities of successes for dice roll for the old White Wolf World of Darkness role-playing game system. A few years ago White Wolf release a new system for their games set in the World of Darkness.
The new system does not have varying difficulties. Instead the difficulty of all rolls are fixed at 8. The variable in the new system is the number of dice you throw for a given action: the size of your dice pool. Rolls of one no longer cancel out successes. Also, you only botch when you roll a one with a dice pool of one.
The chart below shows the success probabilities for the new system.
Everything should be pretty self explanatory. For a dice pool of 1, the probability for 0 successes also includes the probability to botch. If the dice pool is one, the botch probability is 10% and can be subtracted from the "0 successes" probability value (70%) if you're interested in the chance to get zero successes without botching (60%).
All stats are based on 1 million simulated rolls for each dice pool size. 10s are re-rolled as per the rules. An 8 or better is considered a successes. MATLAB was used for all calculations.
This is the link for a PDF that can be used for printing: new_WoD_prob.pdf
The next figure is the percent distribution of successes for each of the dice pool sizes. While the figure above shows the cumulative probability of meeting or surpassing a certain success threshold, this figure shows the actual distribution of successes rolled and gives you a better idea of what a roll of a certain dice pool size will produce.
3 comments
Additionally, just for clarification, the botch rule is if your dice pool is reduced to less than 1 die, you can still roll a "chance die", which succeeds on a 10 and "dramatically fails" on a 1. This is also subject to the 10-again rule, giving a 1.11111.../10 chance of success.
First, thanks for your comment.
The plots that I posted do take into account the 10-again rule. Since a 8, 9, or 10 produces a success, that means that there is a probability of 3/10 or 30% to gain 1 or more successes on a single die roll. You'll note that this is reflected on the successes probability chart. The probability of only 1 success for a single die is 0.2 + (0.1 * 0.7) or 27%. That requires a roll of 8, 9, or 10 but with the 10 followed by a re-roll of 7 or less. 2 or more successes require a roll of a 10 and then a subsequent roll of 8, 9, or 10. That equates to a probability of 1/10 * 3/10 or 3%. Again, that is reflected on the successes probability chart. Given the numbers that you give in your comment it appears that you have a fundamental misunderstanding of the probabilities for a 10-sided die, particularly with the added complexity of considering the potential for additional rolls.
I have not included the zero dice pool chance roll in my plots since it uses a different set of rules than a standard roll. Furthermore, since the probabilities surrounding botches can easily be subtracted from the zero successes probabilities, I haven't denoted them separately from the 0 success curve.
Again, thanks for taking a look at my work.
Cheers,
Matthew Miller
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