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(I do not know if I should post this in math.stackexchange.com, so I try to phrase it in a way that it makes sense for non-roleplayers. Just in case it gets moved)

In the latest two editions of Shadowrun, you roll several 6 sided dice, and every die showing a 5 or 6 is considered a hit.

How do I calculate the chance of rolling 3 dice giving more hits than rolling 4 dice?

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Your question is not entirely clear, could you provide some examples as to what you mean? a sample perhaps of the rolling itself, and what you are trying to use this for? As it will help in giving you an answer that is meaningful. "How do I calculate" is not what you want, you probably want a method, or a table for understanding the odds in your game. Either way - more description please. –  Inbar Rose Feb 24 at 10:07
    
@InbarRose maybe it's not clearly worded but I think the aim is really clear. He wants to roll 3d6 and 4d6, in sequence, and get more successes on the first roll than on the second. What's the chance of that happening? Please Andràs tell us if I'm wrong. –  Zachiel Feb 24 at 10:18
    
I deliberately did not use the DnD notation (3d6), because it is calculated entirely differently. 3d6 has an average of 10.5, rolling 3 dice in SR5 has an average of 1. –  András Feb 24 at 10:20
    
I am open to suggestions how to phrase the question better, but I think writing 3d6 is will make it only worse, not better. –  András Feb 24 at 10:21
    
I'd add a word about why you need to know that. I suppose Shadowrun has opposed checks and you want to know your winning chances when facing an opponent with an higher number of dice? –  Zachiel Feb 24 at 16:07
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I'm sure it's possible and not too hard to treat your dice rolls as if they were 3d3 and 4d3 giving a success on a roll of 3 (for easier math), calculate the probabilities of getting 1, 2 or 3 successes on the first set and then the probability to get 1, 2, 3 or 4 successes on the second, then seeing which combination of those gives the intended result.

I'm fairly sure someone can write a mathematical formula for this, and I think I could if I spent some time on it, luckily there's Anydice that can do the math for us.

output [count {5, 6} in 3d6] - [count {5, 6} in 4d6]

As you can see, the probability to get at least one more success on 3d6 than on 4d6 are 25.24%

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