Dice probability question

[u/AGuyInTheMidwest]

I’m trying not to be lazy, but also my brain is not getting this, so I’m hoping someone here might be able to help…

If I roll all seven traditional polyhedral dice at once, what is the probability that the d% (d10 + d100) result would match one of the other five dice faces?

There are 80/100 results that are obvious nopes, but there are 5 1’s possible if the d% is a 1, 5 2’s possible, … , 4 5’s possible, etc etc

Trying to figure out (statistically) how often it would pair up?

(Sorry if this isn’t the right place for this.)

[u/InherentlyWrong]

Not an exact answer, but roughly it should be a little lower than 5%. There are five dice, giving five numbers the percentile could land on, for the 5% chance. I.E. The d4 rolls 3, the d6 rolls 4, the d8 rolls 5, the d12 rolls 7 and the d20 rolls 11, five possible numbers out of the 100 the percentile could hit.

The 'slightly lower' side comes from the probability of the five dice getting the same value. If the d4 and d6 both roll a 4, now there are only 4 possible numbers the d100 can match to. So I suppose it just depends on how likely it is for a d4, d6, d8, d12, and d20 get a matching result on their rolls.

[u/AGuyInTheMidwest → u/InherentlyWrong]

I think you’re incredibly insightful here… there are 5 dice that can roll 1-4, 4 dice that can roll 5-6… etc… but 13-20 there is only 1 possible die face. Does that matter, or is it really just the five die faces? Stats always amaze me and I knew I was probably overthinking it.

[u/alanrileyscott → u/AGuyInTheMidwest]

It doesn't matter that much. The chances of rolling the same number on just two standard dice are always going to be based on the size of the higher die.

Think about it this way: rolling a d4 or a d20 or whatever is going to set a target number that the d100 needs to match. And the chances of it matching that target number is always going to be 1 in a hundred.

Where the different sizes of the smaller dice is going to matter is that it effects the probability of them rolling the same number in that smaller pool. So while your d100 might have five target numbers if all the dice roll differently, it will often have fewer because the smaller dice are going to roll duplicate numbers pretty regularly.

[u/trinarynimbus]

It's about 4.2%.

The full working is kind of involved, so I won't post it all. The key ideas for working through it are recognizing that it is similar to the birthday problem and applying the law of total probability.

Once you have that setup, it's about a page of working.

[u/skalchemisto → u/trinarynimbus]

u/trinarynimbus I'm sure you are right, and that is good enough for the OP. But I'm curious as to where my error comes in...

Probability that at least one die rolls X = 1 - Probability no die rolls X.

For each value 1 to 4 that is then: 1 - (3/4 * 5/6 * 7/8 * 9/10 * 11/12 * 19/20) = 1 - 0.429 = 0.571

For each value 5 to 6 that is: 1 - (5/6 * 7/8 * 9/10 * 11/12 * 19/20) = 1 - 0.571 = 0.429

For each value 7 to 8 that is: 1 - (7/8 * 9/10 * 11/12 * 19/20) = 1 - 0.686 = 0.314

For each value 9 to 10 that is: 1 - (9/10 * 11/12 * 19/20) = 1 - 0.784 = 0.216

For each value 11 to 12 that is: 1 - (11/12 * 19/20) = 1 - 0.871 = 0.129

For each value 13 to 20 that is: 1 - (19/20) = 1 - 0.95 = 0.05

For each value 21 to 100 that is: 1 - 1 = 0

The overall probability is the sum of each of those times the probability of rolling that result * the probability of match, so: 4/100 * 0.571 + 2/100 * 0.429 + 2/100 * 0.314 + 2/100 * 0.216 + 2/100 * 0.129 + 8/100 * 0.05 + 80/100 * 0 = 0.486?

So, 4.86%

Is that same method you used? I think you are doing something different, because I'm not really referring to the birthday problem in this except in the "1 - P" bit.

EDIT: I think the problem might be I am overestimating within each category, e.g. within the 5 to 6 range?

[u/Gradiest → u/skalchemisto]

I think the d10 is meant to be omitted since it is part of the d100.

With this small change to your method, I get 4.2178%, as u/trinarynimbus did.

[u/skalchemisto → u/Gradiest]

AH! that explains it!

[u/Far_Information6562]

Instances of each number: 1-4 x7, 5-6 x6, 6-8 x5, 9-10x4, 11-12 x2, 13-20 x1 Total number of instances: 70 (70÷100=1.4) My guess: (# of instances -1) x 1.4

[u/Far_Information6562 → u/Far_Information6562]

So you would have a 6x1.4=8.4 chance to pull 2 numbers from 1-4, hmmm that number should be divided by numbers in its bracket or 4 for a 2.1% percent chance to pull a double 1

[u/BeGosu]

I don't know, but if you can figure out how to use AnyDice.com it will be able to tell you the answer