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Help crunching the numbers

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Thu, 06/01/2006 - 14:36
Mosse
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Joined: 12/31/1969

I've been browsing this forum for a while and my impression is that there are people here who are good at crunching statistics, especially when it comes to dice. It's been a while since my school days, and I can't for the life of me come up with how to count this.

I've been working on a hack'n slash-game for a while, and I'm starting to get movement rules and such, but am still looking for a good, simple and fast way for the combat rolls.

This is how I see it right now:
Count the distance of the shot in spaces, and throw that amount of d6:s. Any 6:s and the shot is a miss. Now I know the risk of a 6 on 1d6 is 1/6. But how is it with more dice?

I like this system because it's fast in the game, and wouldn't slow it down. And it would allow the use of special abilities - for example a sniper could rethrow up to three dice, for example...

What do you think of a system like this? Is it already used in any game, or can you - who have more exprience in game design - see any flaws in this system?

My thanks in advance for any help you can offer!

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Thu, 06/01/2006 - 15:29
#1
Epigone
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Joined: 12/31/1969
Help crunching the numbers

Larienna's dice probabilities

Your chance of getting a 6 will be the last number in the first row for each of Xd6.

To analyze your sniper idea, follow this example. Say you're rolling 7d6 with a sniper and you get to reroll up to 3 dice.

First, you could roll 4 or more 6s and miss. The table shows this has a 1.76% chance of happening.

Or you could roll exactly 3 6s and reroll them. The chance to roll exactly 3 is the chance to roll 3 or more minus the chance to roll 4 or more: 9.58%-1.76%=7.82%. When rerolling, the chance miss is the chance to roll 1 or more 6s on 3d6, or 42.13%. So the total probability of missing here is 0.076*0.4213=0.0320, or 3.2%.

Similarly there is a (0.3302-0.0958)*(0.3056)=0.0716, or 7.16% chance to roll 2 6s initially and then roll at least one 6 on the reroll, and a (0.7209-0.3302)*(0.1667)=0.0651=6.51% chance to roll one 6 and then reroll it a 6.

So the chance of a sniper missing with 7 dice is 1.76% + 3.2% + 7.16% + 6.51% = 18.63%.

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Thu, 06/01/2006 - 15:31
#2
dsavillian
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Joined: 06/21/2010
Help crunching the numbers

I think this is the easiest way to figure it out:

The odds that you will not roll a 6 on a die is 5/6. So the odds that you will hit one square away is 5/6 (83%). The odds that you will not roll a 6 on 2 dice is 5/6 * 5/6 or 25/36 (69%). 3 dice is 5/6 * 5/6 * 5/6 = 125/216 (57%)... etc etc.

For every square, add another 5/6 multiplier to the mix. Those are the odds that you will *not* roll a 6.

I think the next steps down are 48%, 39%, 33%, 27%, 22%, 19% (give or take).

You will find the larger differences occur early on and as it gets further away you chances stay about the same.

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Thu, 06/01/2006 - 15:33
#3
dsavillian
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Joined: 06/21/2010
Help crunching the numbers

Epigone wrote:
Larienna's dice probabilities

Your chance of getting a 6 will be the last number in the first row for each of Xd6.

To analyze your sniper idea, follow this example. Say you're rolling 7d6 with a sniper and you get to reroll up to 3 dice.

First, you could roll 4 or more 6s and miss. The table shows this has a 1.76% chance of happening.

Or you could roll exactly 3 6s and reroll them. The chance to roll exactly 3 is the chance to roll 3 or more minus the chance to roll 4 or more: 9.58%-1.76%=7.82%. When rerolling, the chance miss is the chance to roll 1 or more 6s on 3d6, or 42.13%. So the total probability of missing here is 0.076*0.4213=0.0320, or 3.2%.

Similarly there is a (0.3302-0.0958)*(0.3056)=0.0716, or 7.16% chance to roll 2 6s initially and then roll at least one 6 on the reroll, and a (0.7209-0.3302)*(0.1667)=0.0651=6.51% chance to roll one 6 and then reroll it a 6.

So the chance of a sniper missing with 7 dice is 1.76% + 3.2% + 7.16% + 6.51% = 18.63%.

Instead of adding up the probabilities of rolling one 6, 2 sixes, 3 sixes... etc...

you can just figure out the probability of rolling no sixes and subtract that from 1 :)

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Thu, 06/01/2006 - 15:41
#4
Epigone
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Joined: 12/31/1969
Help crunching the numbers

dsavillian wrote:
Epigone wrote:
To analyze your sniper idea, follow this example. Say you're rolling 7d6 with a sniper and you get to reroll up to 3 dice.

Instead of adding up the probabilities of rolling one 6, 2 sixes, 3 sixes... etc...

you can just figure out the probability of rolling no sixes and subtract that from 1 :)
Not if you get to reroll, you can't. :)

But yes, if you're just rolling Xd6 and you want to know the probability of getting no 6s, it is 1-(5/6)^X, as you said.

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Thu, 06/01/2006 - 16:55
#5
dsavillian
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Joined: 06/21/2010
Help crunching the numbers

Ahh, yes. The dreaded re-roll. I failed to read that part :P

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