Hi, I am working on a simulation game where attacking is done with a certain number of dice, and the player selects the higher value for their attack.

So, if your attack is 1, you roll 1d12, and take the value as your attack, but if your attack is 7, you roll 7d12 and take the highest number as your attack.

I would love if someone could explain to me the math involved in calculating average scores for different number of die, from 1 to 15. I know a bit about probability, but I have tried and tried and have not found any reliable way to come with the answer.

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Advance task:

What if rolling a 12 meant that the player could roll a new die and add to the result? How would that affect the average value? The idea is that for every 12 rolled on the original roll, the player rolls a die on a second roll and then add it to the original highest number (12, in this case).

Example: a player rolls 9 dice: 8, 12, 1, 4, 5, 12, 4, 12, 10.

The highest value is 12, and he rolled 3 of them, so he would roll again 3 dice, and add the highest value to his original roll, which was 12. So he rolls: 3, 12, 7.

Another 12, which means he has again the chance to increase his roll. He adds the new highest value to the original one (12+12) and rolls another die. If he rolls a 12, he would repeat the process. But he rolls a 1. So the final value of the roll would be 12+12+1=23.

Is there any way to calculate the average score for this kind of rolls with 1-15 dice?

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Absurd level task:

I would like to introduce an automatic fail feature, and so I thought about turning 1s into fails in the following manner: each 1 will cancel the highest number rolled. For example, rolling 7 dice: 10, 12, 1, 4, 1, 10, 5. One of the 1s would cancel the 12, and the other 1 would cancel one of the 10s, so, the highest valid number would be 10.

I know (this math was approachable for me) that increasing the number of dice will decrease the possibility of a complete fail (more than half 1s) which is perfect to my intentions: higher attack values have lower risk of complete failure.

But I find it out of my reach how to calculate average values of rolls.

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Collateral absurd level task:

When applying the 1=fail system, rolling an even number of dice becomes a problem. For example, two dice: 7, 1. The 1 cancels the 7, and there is no number left. This could be taken as a fail, but then rolling 2 dice would be worse than rolling only 1, which is completely stupid in a system where higher attack values should be better than lower attack values. The same would happen for 3 and 4 dice, and 5 and 6, etc.

So, does anybody have any idea on how to balance even number of dice with a system of 1=fail like I explained above?

PS: I already thought about the obvious solution: having only odd values for attack. That will remain as a last-resort solution, but I would like a more universal solution that would not mess with the possibility of attack bonus (+1,+2, -3, etc.)

Thanks a lot.

mmm... i do not know if i am following you (heck, more than probably i am not). I do not want to know the probability of rolling a 34, I would like to know the expected value of a roll of n dice with the rules above so that i can track the improvement in rising the attack of a unit from 4 to 7 and know more or less what it means. Like, is the average roll going to increase from 9.3 to 10.7, or is going to rise up to 13.5?

I hope i explained myself better this time. It is very difficult to talk about unkown stuff in a second language.

But thanks a lot for your help.