Probability of throwing a six on each of the two dice, or the results combined?

Theres only one outcome on each out of six that is a six, in which case, it'd be 1/6 x 1/6, or 1/36 chance of getting a pair of sixes. Its not the same as 2/6 or the equivalent as one outcome is dependent on another.

its basically shorthand for saying "The probability of getting a six given that you already have gotten one", kind of.

 

The probability of getting one 6 on any of the dice. A 1 and a 6 would be a desirable result. A 4 and a 6 would also be good. I want to know the probability of getting a 6 if you throw 2 dice.

 

A six on either? 2/12.

 

A six on either? 2/12.

Which pretty much breaks down to 1/6 again doesn't it?

There is still a 1/6 chance of each dice rolling a 6, but because you are rolling 2 dice you are you are doubling your chance of rolling a 6.....

..... which is kind of 2/6.....


.......I think!!! :huh:

 

I would not trust this. Last night I was owed in a 400 pts combat patrol with my Necrons. I would have one but I seriously rolled at least 4 1 a shooting phase. When the Space Wolves were in range of Rapid Fire I got 18 shots and only 3 hit, so I dont relay on probaility to see if I will win.

 

I would not trust this. Last night I was owed in a 400 pts combat patrol with my Necrons. I would have one but I seriously rolled at least 4 1 a shooting phase. When the Space Wolves were in range of Rapid Fire I got 18 shots and only 3 hit, so I dont relay on probaility to see if I will win.

Sensible..... the only thing about probability is that it probably won't happen when you want it to.

 

A six on either? 2/12.

May I ask how you calculate this?

 

Here's the actual maths......

The chance of at just one 6 coming up when you roll 2 dice is 10/36

the chance of rolling a double 6 is 1/36

How?........

If you work through the possible combinations you will see that the number 6 appears on a single dice in 10 of the 36 possibilites (it also appears as a double 6, but this is not then a sigle 6 is it?).... hence there is a 10/36 chance of rolling a single 6.

In all 36 possibilities the chance of two 6's appearing only occurs once, obviously as a double 6.... therefore there is only a 1/36 chance of rolling two 6's.

Here's the table...

Hope that helps,

Wolfkin

 

Wolfkin is correct. And that is a nice picture of the dice Wolfkin... *snatch* ...got a copy of it now.

Anyhow, you take the total number of times the combination you want to occur and divide that by the total number of possible combinations.

Examples:

If you are looking for ONLY one six to appear on 1d6 then:

total number of single sixes possible = 1
total number of possible combinations = 6

So your odds of throwing a single six on a single d6 is 1/6 or about 17%

If you are looking for ONLY one six to appear, then:

total number of single sixes possible on 2d6 = 10
total number of possible dice combinations = 36

So your odds of throwing one six on 2d6 is 10/36 or about 28%

Now, if you are looking for a six in any combination to appear then:

total number of sixes possible on 2d6= 11
total number of possible combinations = 36

So your odds of throwing a 6 (including the double six) on 2d6 is 11/36, or about 31%.

This should be VERY apparent by the chart put up by Wolfkin.

 

I took the probabilities of rolling sixes, not numbers which add up to six.

Theyre 2/12 since theyre independent of each other, have no affect on the outcome of each other. In effect, its rolling one dice twice, in which case its 1/6, as said.

 

Thanks alot for all your answers

 

There is an faster way to calculate the probability to come up with at least one six on two dice, you don't need to count all possible combinations of dice throws.

Take a look at the opposite. What's the chance not to score any sixes with two dice?
The chance not to roll a six with one is 5/6. Therefore the chance not to score at least one six on two dice is 5/6 * 5/6. That's 25/36.

So if you don't roll any sixes with a 25/36, the remaining (36-25=11) 11/36 are the chance to roll at least one six.

 

I took the probabilities of rolling sixes, not numbers which add up to six.

Theyre 2/12 since theyre independent of each other, have no affect on the outcome of each other. In effect, its rolling one dice twice, in which case its 1/6, as said.

The probability of rolling sixes is 11/36. ( The probability of numbers which add up to 6 is different, 10/36, or 5/18 )


Think of it this way.

There are 36 possibilities all together (6 sides x 6 sides).

If you want AT LEAST ONE "6" out of your two dice, the possibility is 11/36.
The possibility you roll a certain number "x" on one dice is 1/6. So if you roll the X on your first die, you don't roll again for the second. (1/6) However, if you don't roll an X on your first die (5/6), then you roll the second die for an X (1/6), so the possibility of getting an X on your second dice is 5/6 * 1/6 = 5/36. You have to add the probability you get an X on your first try (1/6) with the probability you get an X on your second try (5/36).

1/6 + 5/36 = 11/36


If you only want ONE and ONLY ONE "6" out of your two dice, the possibility is 10/36.
Follow the same procedure as above, however, the result of (6, 6) doesn't satisfy the conditions(it did however in the above set of rules, so we have to subtract it out). The chance that you get (6, 6) is 1/36, so you have to subtract this from 11/36.

11/36 - 1/36 = 10/36



Note:

I'm good at math, but we all make mistakes. This is a simple problem, so I might have rushed through it.

 

LOL...

I somehow feel we are just repeating ourselves now.....

wolfkin

 

LOL...

I somehow feel we are just repeating ourselves now.....

wolfkin

Yeah, we are, but some people keep changing it.

 

Yes.

I'm a math wonder at times. Deus Mechanicus and eatmydarkapostle are typically correct.

All of the other ones at least got some errors, most times only 1. So look at Deus Mechanicusses or eatmydarkapostle's post to find the correct answer.

Cheers,
-MvC

 

While we are on the subject of math, how many of you guys have heard of MathCounts?