10 fair 6-sided dice are rolled. The outcome is as follows:
111 22 33 456(This orderly arrangement is AFTER the roll.) What is the probability of this configuration showing up? Thanks for help.
I think this is correct.....but someone else might know better than me !!!
On any roll of one die...the chance of any number appearing is (1/6)
So...the probability of any particular arrangement showing up by rolling 10 dice is just (1/6)^10 ≈ 0.0000000165381717
CPhill: I think your answer would be true IF all 10 were one of a kind, i.e., 10 ones, 10 threes, 10 sixes...etc.
I think this approach is more accurate:
10C3 * 7C2 * 5C2 * 3^3. The last 3^3 means that there are only 3 dice left, but each can take only 3 values(4,5,6) since the other 3(1, 2, 3) are excluded!. So, then the probability should be:
[10C3 * 7C2 * 5C2 * 3^3] / 6^10 =680,400 / 6^10 =0.0112525720 =1.125%.
That is my take on this!!.
A Dicey Commentary.
I’m sure CPhill’s answer is right.
Even if I wasn’t sure, I know this answer is wrong. This is one of the BBs that GingerAle trolls.
After all the years he’s been on here, he is still a dumbshit! He can screwup a one spaceship Martian funeral.
I asked GingerAle what she thought of the intellectual thought process of the forum’s dickheads.
Here’s her reply.
While it’s not unusual to find a guy who thinks with his d i c k, most reserve the pecker-based thought processes for the primeval social and mating imperatives, and use their cranial brain for intellectual reasoning. Such behavior may have helped give credence to the "two brains" myth of the Sauropod dinosaurs. However, the BBs take it to a new level, by reversing the roles of their brains.
Add, that the BBs may have damaged their brains by repetitive carelessness of zipping up before following proper procedures; it’s not a wonder that their thought process results in mental chaos, and compels a psychosexual attraction to intellectual scat. A base thought is “It was good for me, so it must be great for you!”
GingerAle’s comments may not pass scientific rigor, but she gets high marks for intellectual satire and humor.
This intellectual scat should have been directed to the sewage processing plant, but instead it gets a “Best Answer” stamp. Either it must have been good for one of the mods, or he didn’t follow proper procedures before zipping-up.
See this answer #6 by a certain "GA" !!! https://web2.0calc.com/questions/rolling-dice
See also answer #5 by a certain "GA"!!! https://web2.0calc.com/questions/conditional-probability_7#r9
Neither of those questions are the same as this one.
It’s interesting that you link to Ginger’s troll post. https://web2.0calc.com/questions/conditional-probability_7#r9 Did you notice your screwedup answer supports her position?
The number of permutations for screwing up a simple statistics question is truly mind boggling. I’ve always wondered what the number actually is. It may not actually have a ceiling --it seems as vast as space, and the time to do it is limited only by the imagination and cogitative lifespan of humankind.
It was this post that prompted my question about the intellectual thought process of the forum’s dickheads, prompting Gingerale’s very droll, satirical reply. This started because Ginger offered me a $50 (Bit Coin) wager on how that thread would conclude. (I took the wager and lost.)
Here is my jab at this!!
10C3 * 7C2 * 5C2 * 3! =151,200 / 6^10 =0.00250057 =0.250057%.
Because the last "456" are fixed, they can only appear in 3! ways =456, 465, 546, 564, 645, 654.
P.S. There is bound to be a "screw up" somewhere! After all, it is based on "Probability Theory" invented by some Italian and a couple of French mathematicians for the purpose of gambling, according to my understanding of it!.
Here’s another dumb answer in the queue. “There is bound to be a "screw up" somewhere!” This has little to do with probability theory, because whenever YOU make a post it’s a certainty there will ALWAYS be a SCREWUP.
OK.....here's was my logic....
Suppose...instead of dice, we have 2 coins.....what is the probaility that, after flipping both of them, we want to know the probability of this outcome : H T
We have 2 ways the first coin can land and 2 ways that the second coin can land..so...the total outcomes = 2 * 2 = 4
And the probability of any outcome is just 1 / 4 = (1/2) * (1/2) = (1/2)^2 =
(1 / all possible outcomes on one flip )^2