# Throwing a dice

1. Feb 23, 2012

### aaaa202

Suppose I throw a dice and want either 6 or 1. The probability for that is obviously 1/3. Now suppose instead, that I throw two dices but only want 6 this time. Then the probability of getting 6 is a bit lower than 2/3, but instead there is of course also a probability of getting for instance 6 on both dices.

Now say I throw the dices an infinite number of times. Will the average number of 1's and 6'1 on the first dice and average number of 6's on the 2nd two dices be the same?

And if so, how can I realize that?

2. Feb 23, 2012

### kai_sikorski

If you count the two die 6-6 case as 2 then yes.

3. Feb 23, 2012

### HallsofIvy

Staff Emeritus
By the way, there is no such word (in English) as "dices". "Dice" is the plural of "die".

4. Feb 23, 2012

### Bacle2

You can try using the expected value for a random variable.

Halls of Ivy: Congratulations on your 2^15 posts!.

5. Feb 23, 2012

### kai_sikorski

You could, but I think that's really overkill. The only difference between the two situations is the labeling of the sides. The two cases must give the same result by a principle of indifference.

6. Feb 23, 2012

### Bacle2

You're right that it's overkill; I just thought the OP asked to have a formal proof.

7. Feb 23, 2012

### aaaa202

Someone once told me it had to do with the fact, that both of them follow a "product distribution" or something like that? Is that true?

8. Feb 23, 2012

### aaaa202

btw... It seems like you guys think of it as intuitive. I don't - please give me an example that makes it intuitive :)

9. Feb 23, 2012

### skiller

Be careful; you're dicing with death with that remark.

10. Feb 23, 2012

### kai_sikorski

Okay, maybe the thing to do is to go to expectations. Let X1 be 1 if the die is 1 or 6. Due to mutual exclusion of those events and law of total expectation

E[X1 | D = 1] P(D = 1) + E[X1| D=2]P(D=2) = 1*1/6 + 1*1/6 = 1/3

So due to law of large numbers if you did this experiment N >> 1 times you would expect (~N/3) occurrences of 1 or 6. Letting X1 be 1 if the first die is 6 (0 otherwise) and X2 be 1 if the second die is 6 (0 otherwise)

E[X1 + X2] = E[X1] +E[X1] = 1*1/6 + 1*1/6 = 1/3

So again after N>>1 tries you would expect (~N/3) 6s.

11. Feb 24, 2012

### checkitagain

"Dice" is the plural of "die," as when referencing a gambling object,

but "dices" is a plural to "dice," as in "cutting up an apple into dices."

Also, "dices" is a verb.

And "die," when meaning a machine that stamps/cuts out a shape,
has the plurals "dies" and "dice."

Last edited: Feb 24, 2012
12. Feb 24, 2012

### HallsofIvy

Staff Emeritus
Thanks, guys, I bow to superior knowledge.

13. Feb 24, 2012

### alan2

Help me please. Maybe I'm misunderstanding the original question. As I read it he is asking "what's the probability of rolling a 1 or a 6 on one roll of a die?" Clearly 1/3. As I read the second question, he is asking "what is the probability of rolling at least one 6 on a roll of two dice?" The answer is the complement of getting no 6's on the roll of two dice, or 1-(5/6)(5/6)=11/36. You guys appear to be saying that they are equal. How am I misreading the question? Thanks.

14. Feb 24, 2012

### kai_sikorski

No he was asking how many times 6s would come up, so you count the 6 - 6 case as 2

15. Feb 24, 2012

### alan2

So he was asking for the expected number of 1's or 6's on a single roll of the die vs. the expected number of 6's in a roll of two dice? I didn't read that, thanks. Both expectations are 1/3.

16. Feb 24, 2012

### kai_sikorski

yup! ;)

17. Feb 24, 2012

### alan2

Thanks, it helps to be talking about the same problem.

18. Feb 24, 2012

### kai_sikorski

Ups sorry, noticed a typo in my other post should say: