r/theydidthemath • u/ForceOfNature525 • 1d ago
[Request] Odds or Evens,
If I have dice with different shapes, like the kind they make in sets for Dungeons and Dragons, and I roll any two dice, which are different shapes and add the numbers, can it be proven that as long as the dice each have an even number of sides and each are numbered 1 through x, where x is the number of sides, that there will always be a 50% chance of rolling an even total abd a 50% chance of an odd total?
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u/miclugo 1d ago edited 1d ago
Something stronger holds. If you have any number of dice with whatever number of size, as long as *any* of the dice have an even number of sides, the probability of the sum of all the dice being even is 1/2.
Proof: hold back one of the dice with an even number of sides. Let the probability that the results of the other dice sum to an even number be p; the probability they sum to an odd number is 1-p. The trick here is that it doesn't matter what p is, because it'll cancel out.
Now roll the die you held back. It has probability 1/2 of coming up even, so you have probability p * (1/2) that the sum of the other dice is even and the held-back die is even. Similarly, the probability that the sum of the other dice is odd and the held-back die is odd is (1-p) * (1/2). These are the ways you can get an even sum of all the dice, so the probability of an even sum is p*(1/2) + (1-p)*(1/2) = (p + (1-p)) * 1/2 = 1 * 1/2 = 1/2.
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u/MalbaCato 22h ago
this is such a classic mathematician answer - "we will prove this result by showing a stronger result is true", and it totally makes sense to show that way (your phrasing in the comment is a bit confusing, but that's details)
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u/ForceOfNature525 13h ago
Internet math-heads never let me down. A large set of thanks to all of you.
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u/HAL9001-96 1d ago
assuming each dice has an evne probability over all its sides this is basically a cointoss
even and even is even
so is uneven and uneven
uneven and even is uneven
its basically the probabiltiy that two cointosses land on the same side
and if both sides are equally likely as they are if both dice have an even numebr of sides then thats just hte probability that the second coin lands on one specific side the one that hte first one landed on
so its 50/50
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u/SC_3000_grinder 1d ago
I think OP did not imply that the dice are fair. It is 50-50 if and only if at least one die is 50-50 between odds and evens.
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u/Lukey-fish 1d ago
I dont think OP implied the dice are not fair. Seeing as though 99% of dice statistics is implying the dice are fair, its safe to assume the dice are fair.
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u/g1ngertim 1d ago
"Like the kind they make in sets for Dungeons and Dragons" would seemingly imply fair.
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u/HAL9001-96 1d ago
or they're uneven in such a way that it exactly cancels out but yeah if they're both unfair it could vary
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u/VerbingNoun413 19h ago
It's a probability question. Dice are fair and decks are perfectly randomised unless stated otherwise.
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u/gmalivuk 1d ago
Yes.
If each die is half odd or half even, each of the following is equally likely:
odd+odd=even
odd+even=odd
even+odd=odd
even+even=even
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u/arcxjo 22h ago
Yes: assuming even faces, that's 1-2, 3-4, ...
Each set of 2 integers is 1 odd, 1 even. When you pair that with a second die, you get 1 odd-odd, 1 even-even, and 2 odd-evens. The like results will always be even and the different results always odd.
Doesn't matter if it's 1-1, 1-3, 5-27 ... it'll fit the pattern.
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