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View Full Version : Help me calculate the odds, please!

2012-05-07, 09:53 AM
So, once again (http://www.giantitp.com/forums/showthread.php?t=227383) I come to you fine fellow playgrounders for help with that daunting beast known as... Calculus! :-D

So, what are the odds of:
- rolling 12 or higher on 4d6?
- rolling 15 or higher on 4d6?
- rolling 15 or higher on 5d6?
- rolling 20 or higher on 5d6?
- rolling 15 or higher on 6d6?
- rolling 20 or higher on 6d6?

(Oh, and if anyone's curious about the results from the previous thread, I ran some "Dragonlance 5th Age" games with dice instead of cards and they ran quite well! Thanks again to those who helped me the first time. :-))

Yora
2012-05-07, 10:04 AM
The answer to all these questions is here (http://anydice.com/).

Ask again, if you need help using the tool.

Andrewmoreton
2012-05-07, 10:15 AM
No calculus involved, just probabilities. Entirely different branch of mathematics.

2012-05-07, 11:49 AM
No calculus involved, just probabilities. Entirely different branch of mathematics.

...Which just goes to show how my knowledge of mathematics stopped pretty much at Arithmetic. See why I asked for help? :-P

The answer to all these questions is here (http://anydice.com/).

Ask again, if you need help using the tool.

Thanks a lot! :-D
So, if I'm reading things correctly, rolling 5d6 means there's a higher than 70% chance of rolling 15 or higher? Is that correct?

prufock
2012-05-07, 12:20 PM
So, if I'm reading things correctly, rolling 5d6 means there's a higher than 70% chance of rolling 15 or higher? Is that correct?

Yep. If you hit the "at least" button under "Data," it will show the probability of that number or higher. At 5d6 there is 77.85% of getting 15 or higher.

2012-05-07, 01:12 PM
Yep. If you hit the "at least" button under "Data," it will show the probability of that number or higher. At 5d6 there is 77.85% of getting 15 or higher.

Oh, thanks! Didn't realize that. :-)
But to be honest, I was expecting the odds to be a lot closer to 50%. There's a big leap from 4d6 to 5d6, huh?

Eldest
2012-05-07, 01:40 PM
5d6 means the average rolled number is a 17.5. That's around where a 50% would be. 4d6 would have the 50% around 14. It's really just where the bell curve is centered.

NecroRebel
2012-05-07, 02:11 PM
When you roll multiple dice, you're much more likely to get a result near the middle of the possible range because there are so many more combinations that reach those numbers than the outliers. For instance, let's say you're rolling 2 6-sided dice. You get a 2 only if you roll a 1 on both dice, so you only have a 1/36 chance of getting a 2. You get a 7, though, if you get a 1 and 6, a 2 and 5, a 3 and 4, or the reverse of those rolls. 6 possible rolls give a 7, in other words, so you have a 6/36 chance of rolling a 7.

The up shot of this is that you have a greater than 50% chance of rolling the average on any roll of multiple dice, since you have an even chance of rolling below the average but not the average itself and above the average but not the average itself. If your target is below the average, your chance of hitting it goes up, quite rapidly at first since the numbers directly above and below the mode (meaning most common result) have the second-greatest probability of occurring.

In the examples from your questions, 14 is the mode of 4d6, so targets below 14 will have a higher-than-50% chance while those above it will have lower. 17 and 18 are both the modes of 5d6, while 21 is the mode of 6d6.

Yora
2012-05-07, 02:16 PM
With every event added to the total, the final result will be more likely to be closer to average.
So if you roll 5 dice instead of 4, an average result will be more likely.

At the same time, you add more dice so the average itself will also be greater.

These two things combined make a quite significant difference.

Probability is one of the fields of math, for which you don't need to be actually good at math. All you need is addition, substraction, multiplication, and decimal numbers. After that, you can imagine everything taking place in the pysical world with actual objects.
I am a complete failure at math, but there's two fields I truly understand: Probability and Geometry. :smallbiggrin:

Tyndmyr
2012-05-07, 02:19 PM
The answer to all these questions is here (http://anydice.com/).

Ask again, if you need help using the tool.

Is there a good way to calculate the probability of say, 5d10b3, when a 10 results in an additional die being rolled and kept?

NecroRebel
2012-05-07, 02:30 PM
Is there a good way to calculate the probability of say, 5d10b3, when a 10 results in an additional die being rolled and kept?

You can do output [highest 3 of 5d10] to get the odds of what would be rolled with 5d10b3 on the forum roller... Aha! I think I've got it:
set "explode depth" to 1
output [highest 3 of 5d[explode d10]]

Edit: ...Actually, no, that's not it... It's only keeping 3 dice. Nuts, and here I thought I was on to something.

EditEdit: I think you'd have to do something with a custom variant function, but I don't understand the tool well enough to manage it. On the left column of anydice, check the "function library" page.

nedz
2012-05-07, 02:34 PM
Yes, but you are aware that the maximum of exploding dice is infinite ?

Yora
2012-05-07, 03:44 PM
There are some dice rollers that can do that. Finding one is probably the most difficult part.

Tyndmyr
2012-05-08, 09:54 AM
Yes, but you are aware that the maximum of exploding dice is infinite ?

Yeah, that's an issue.

I'm more interested in getting good average breakdowns and stuff for practical percentages...so it need not actually be infinite. I don't need perfect precision.

Jay R
2012-05-08, 11:14 AM
output [highest 3 of 5d[explode d10]]

This won't iterate forever, but gets you well past the 99.99 percentile. That ought to do for all practical purposes. You have every probability to the .0001 place.

Yora
2012-05-24, 05:24 PM
How do you calculate the average result for "role two dice, ignore the lower" and situations like that?

reddir
2012-05-24, 07:01 PM
How do you calculate the average result for "role two dice, ignore the lower" and situations like that?

output [highest 1 of 2d20]

would do it for a d20. Can substitute any d#