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Odds of five of a kind dice3/2/2024 ![]() Chessex, especially, is guilty of making dice with the primary focus of LOOKING good. ![]() Chessex, and other major producers of polyhedral dice are certainly not. HOWEVER, most dice are not that accurate. As are some dice manufacturer's products (like GameScience dice). Most virtual dice apps are designed to do exactly this. you are using something scientifically proven to give you EXACTLY even probabilities of each side being rolled. I just want to point out that the numbers posted by BRC are absolutely correct. *It isn't if I explained it well, but my optimism on that point is limited. ![]() I haven't actually watched them, but they probably handle this nicely. Fortunately, I did a quick check, and Khan Academy has a series of videos () for the binomial distribution. The explanation above is a bit short, and might be confusing*. With this question, the latter is much easier. If you need multiple added up, you can either manually add them all up, or subtract the ones you don't have from 1, which must be the sum of all the probabilities. That is just multiplied by the result you want. You then need to compensate for the duplicates. The reasons behind that should be pretty intuitive. ![]() Ignoring duplicates, you have the probability of PkQ(n-k). With the already answered example question k is 0. The reason is that for the first you just have HHH and TTT for the second you have the combination of HTT, THT, TTH and THH, HTH, HHT.Ĭoming back to dice, lets call the number of successes you're looking for k. With a three coin example, there's a 1/8 chance of all heads or all tails, and a 3/8 chance of two heads and a tail or two tails and a head. However, duplicates make things more likely. Both success and failure have a 0.5 probability, so for any combination of heads and tails, the probability before considering duplicate results is 0.5^n. For a simplified example, consider flipping coins. The way calculating binomial probabilities works is basically that you first calculate the probability of getting the combination of successes and failures as if there is one way to do it, then multiply by the number of distinct combinations that produce the same combination. We also have a number corresponding to the number of trials (rolls), which by convention is called n. It'll also depend on how you hold it in your hand, the nature of the surface it's rolled on, how hard it's rolled, whether it hits anything on the way, how it's released from the hand, whether you shake it in your hand (and how you shake it) or just pick it up and roll it, the height between your hand and the surface.Įxpanding on the answer a bit: What we're looking at here is a binomial distribution, where you either get a success (rolling a 20) or a failure (rolling anything else). This chance only changes if you know what the imperfections are. Thus, if you take a random die and roll it, you have a 1/20 chance of getting a 20. Technically true, but irrelevant if you don't know how the die is imbalanced, there's an equally high chance it could be biased towards twenty as any other number. Technically even computer based dice-rollers aren't PERFECTLY even between the 20 options - but the imbalance is statistically insignificant. A machined d20 would be far more expensive if someone ever did build a machine to make them. Those casino d6s are machined - not created in a mold - and they're pretty expensive. The closest that physical dice come is the casino d6 - and those they replace from time to time because the wear of rolling them wears them down slightly.Ī plastic mold d20? Not even close to as well balanced. If you are rolling a physical die - it WILL NOT be perfectly balanced. While all of the % answers above are true - assuming that you have 1/20 chance of rolling a 20 on a d20 - it's just that. So you subtract the 36% likelihood of no 20s from 100%, and get 64% - the likelihood that you will roll at least one 20. The likelihood of rolling no 20s and the likelihood of rolling at least one 20 should add up to 100% since they encompass every possible result any trial where you rolled at least one 20 is a trial in which you did not roll no 20s. Since you actually want to roll a 20 at least once, you need the opposite probability. This is equal to slightly less than 0.36 (36%). The odds that you do not roll 20 when you roll thrice are 95%*95%*95% or 0.953, and so on until you get to the odds that you do not roll a 20 after rolling 20 times, 0.9520. Thus, the odds that you do not roll a 20 when you roll twice are 95%*95% or 0.952 (the odds that you do not roll a 20 the first time, and the odds that you do not roll a 20 the second time). The odds that you do not roll a 20 whenever you roll a d20 are 95% (0.95). Say you had a d20, and you rolled it exactly twenty times, what are the odds that you roll a 20 at least once?īRC posted the math, but let me walk you through it.
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