Statistics help, d10s, variable targets.

[Voadam]

I am playing in a game with a system that uses d10s and I could use help on figuring out how to maximize success chances.

In the system to use one of my powers I have 7 d10 dice to roll. The base target is success on 4 or higher and I need three successes for the power to activate and bump up a stat by one. For each raise of the target number (5 or higher for successes, 6 or higher, etc.) the stat bump is one higher. The power is generally one shot per fight so I want to get the biggest boost chance with a reasonable chance of success.

I am not great with calculating statistics with complex dice mechanics. How do I calculate the chance of getting 3 successes on 7 d10 dice rolls at various probabilities for success? What is the right choice of difficulty to choose when using the power to maximize success and highest impact?

I've been going with bumping a stat to +4 which requires target 7 or higher on 7d10. In two fights this has worked once and not worked once.

Anybody with better statistics analysis would be appreciated.
Thanks

 

[Reynard]

Is that a 90s game? It feels like a 90s game.

To try and be helpful, I ran your question through Claude and this was the response:

Let's break this down systematically. I'll calculate the probabilities for different success thresholds.

Key insights from the probability calculation:

1. Success Thresholds:
- Target 4+: ~83.7% chance of 3+ successes
- Target 5+: ~47.7% chance of 3+ successes
- Target 6+: ~14.5% chance of 3+ successes

Recommendation:

• If you want a reliable activation with a stat bump, aim for the 4+ threshold
• For higher stat bumps, the 5+ or 6+ thresholds dramatically reduce your success chance
• 4+ gives you the most consistent power activation with ~84% success rate

Strategic advice:

• Use 4+ when you really need the power to work
• Consider 5+ or 6+ only if the potential stat bump is significantly more valuable than the activation risk

NOTE: I have no idea how accurate this math is. I have found Claude to be pretty good overall, but sometimes AI be AI

 

[fuindordm]

What is the RPG?

Legend of 5 Rings and 7th Sea run off d10 dice pools but they add up the highest k dice instead of counting successes vs a target number. I mention them because you can find detailed, precalculated probability tables online. You might find the same for your game.

Calculating the probabilities on paper is possible but elaborate. It becomes even more difficult if the dice "fumble" on a 1 or "crit" on a 10. Depending on your proclivities it may be easier to use an Excel sheet or a small program to generate a thousand random rolls and calculate the success rate for different target numbers. For 7 dice that's what I would do.

 

[Voadam]

It is a homebrew system called Seven Cities. Not a published system so I don't think more than two groups have used it and I don't expect to find online resources for it.

The system has a few complications as well. 3 successes is activation but 2 successes means partial activation, half the stat bonus. 1 or 0 success means no stat bonus. 4 or more successes means activation and longer duration which I don't care about.

Statistics are not my thing and I don't know the math for figuring out stuff much more complicated than target 4 on a d10 is a 70% chance of success. Chance of getting 3 successes out of 7 is beyond me but seems straightforward for someone who can figure out the math.

 

[Staffan]

This gets into some gnarly math. It all flows from fairly simple premises, but there's a lot going on.

Let's first start with the probability on a single die. If the probability of a success is P, then the probability of failure Q = 1-P. So on a target number of 4, P is 0.7 and Q is 0.3.

Expanding to two dice, we have three possible results: 0, 1, or 2 successes. These probabilities are as follows:
0: Q^2 = 0.3^2 = 0.09
1: This is a success on either one of the two dice, so it's P*Q + Q*P = 0.7*0.3 + 0.3*0.7 = 0.21 + 0.21 = 0.42.
2: P^2 = 0.7^2 = 0.49.

Going to three dice, things are getting complicated, but we'll soldier on:
0: Q^3 = 0.027. We'll hold off on 1 and 2 because that starts to get complicated, and go straight to
3: P^3 = 0.343
1: We want one failure and two successes. So this can be either PQQ, QPQ, or QQP. These are all the same result, so we get 3*(0.3*0.3*0.7) = 3*(0.063) = 0.189.
2: Same reasoning as above, but with two P and one Q. 3*(0.3*0.7*0.7) = 3*(0.147) = 0.441

With four dice, the options for 1-3 successes are:
1: PQQQ, QPQQ, QQPQ, QQQP.
2: PPQQ, PQPQ, PQQP, QPPQ, QPQP, QQPP
3: PPPQ, PPQP, PQPP, QPPP

This is becoming ridiculous. We need a more generic formula. What we need to know is: if you have n dice with k successes and n-k failures, how many ways are there to arrange those?

Well, if you have n of anything, and want to put those in any order, you have n options for what to put in the first slot, n-1 for the second, and so on. The mathematical term for this is a factorial, and is usually written n!

But in this case, there are a number of options that are functionally the same. So what you need to do is to divide the result by the number of ways to order each of the sub-divisions. In other words, if you want to know how many ways you can pick 3 out of 5, you take 5! (120) and divide it by 3! (6) and 2! (2), for 120/(6*2) = 120/12 = 10. This is usually written as C(n, k) or:

So for 7 dice, you get:
0: 1 combination of Q^7
1: 7 combinations of P*Q^6
2: 21 combinations of P^2*Q^5
3: 35 combinations of P^3*Q^4
4: 35 combinations of P^4*Q^3
5: 21 combinations of P^5*Q^2
6: 7 combinations of P^6*Q
7: 1 combination of P^7

At this point, it all becomes busywork which you probably want to put in a spreadsheet, like this one (feel free to copy and mod it to adapt it to your own needs). What you're looking for is in the "N or more" column, which tells you that there's a 0.97 probability of 4 or more successes, or 97%. If you modify P, you will see how that affects the outcome.

 

[GMMichael]

I am not great with calculating statistics with complex dice mechanics. How do I calculate the chance of getting 3 successes on 7 d10 dice rolls at various probabilities for success? What is the right choice of difficulty to choose when using the power to maximize success and highest impact?

It sounds like this might be past the interesting complexity threshold where you can no longer calculate the problem, and must instead feel the problem. When you choose 3 difficulty, does it feel better than 4 difficulty?

I'd love an RPG to run with feels, but I'd still have trouble doing it when I have 7d10 to count up.

 

[Len]

I just finished doing the math and making a spreadsheet, but @Staffan beat me to it! My spreadsheet gets the same results.

 

[Staffan]

If you want to do it the easy way, you could also go to anydice.com and use the formula
output 7d(d10>=4)
With the 4 being the target number. You want to use the "At least" data type.

 

[fuindordm]

LOL thanks for saving me the work @Staffan . I felt bad for leaving the OP hanging and was just about to show how to use the binomial distribution in this case.
It is really hard to have an intuition for success when the mechanic includes two redundant ways to improve their odds (lower target number or more dice).