Thread: Odds of 1 to 6 straight

What are the odds of rolling 1,2,3,4,5,6 (any order) with one throw of six standard dice?

are you tossing the die six times?
I didn't understand the one throw comment.

6 dice are thrown once?
1/6*1/6*1/6*1/6*1/6*1/6
as each outcome is independant and you have 1/6 chance of any dice obtaining the number you want

BUT then any order, hence the 6!
Or use conditional probabilities....

(1)(5/6)(4/6)(3/6)(2/6)(1/6)

where you can get any number, then any number except the first...

Matheagle and RHandford are using the fact that the probability of getting a 1 on a die is 1/6, the probability of a 2 is also 1/6, etc. so the probability of getting 1, 2, 3, 4, 5, 6, in that order is . And then because you said "in any order", matheagle multiplied by 6!, the number of different orders of 6 things.

Here is another way to get the same answer: the probability of throwing any value from 1 to 6 on the first throw is, of course, 6/6= 1. Once we have that, the probability of getting any number except that first number is 5/6 since now any of the 5 numbers left will work. The probability of throwing, on the third die, any number except those two is 4/6, etc.
Since nothing is said there about the specific numbers, the probability of throwing 1, 2, 3, 4, 5, 6 in any order is ,

To err is human- to really screw up requires a computer!

To save typing all of those fractions, I typed and the "copied and pasted" the rest. Of course, I had accidently put an extra "{" in the first fraction and then copied that error into all of them!

And I should have read Matheagles second response before posting!

Sorry. I'm having trouble with the math notation. Earlier, Plato said the odds of throwing six of any kind on a single throw of six dice is 6 to the 6th power. Does the answer of HallsofIvy mean that throwing a six-die straight (1,2,3,4,5,6) any order - is six times more unlikely than six of a kind? Thanks.

Originally Posted by wshore

Sorry. I'm having trouble with the math notation. Earlier, Plato said the odds of throwing six of any kind on a single throw of six dice is 6 to the 6th power. Does the answer of HallsofIvy mean that throwing a six-die straight (1,2,3,4,5,6) any order - is six times more unlikely than six of a kind? Thanks.

Actually the probability of tossing 123456 in any order is times the probability of the probability of tossing 123456 in that order.

The first is the second is .

Please, disregard the "order" of the dice roll! When six dice are thrown, they fall and the faces are read. There is no order. The question means that all six numbers (1,2,3,4,5,6) appear on the dice - however they lie. Also, while I understand this is a math forum, I don't understand what ! means in a formula.
Can someone PLEASE supply an answer in English prose?

hi

the final answer is 6!/6^6 where the 6! means 6*5*4*3*2*1.

btw n! (pronounced n factorial) is the product of all natural numbers from 1 to n.

Thank you. I gather that is 720 times more unlikely to throw a six-die straight than six of a kind. Do I have that right?

Originally Posted by wshore

Please, disregard the "order" of the dice roll! When six dice are thrown, they fall and the faces are read. There is no order. The question means that all six numbers (1,2,3,4,5,6) appear on the dice - however they lie. Also, while I understand this is a math forum, I don't understand what ! means in a formula.
Can someone PLEASE supply an answer in English prose?

You are the one who is mathematically challenged here.
You are receiving help from professional mathematicians we cannot be faulted for using the language of the field.

First the correct way to ask about this is to use the word probability not odds.

Secondly, the outcome space is no different in tossing six dice at one time and tossing one die six times.
So the probability of getting each of the six numbers is .

and I thought I solved this twice yesterday

Originally Posted by matheagle

and I thought I solved this twice yesterday

As did I. But see reply #7.