Probability of 6 or 1 with Two Dice: An Infinite Experiment

[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?

 

[kai_sikorski]

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

 

[HallsofIvy]

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

 

[Bacle2]

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

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

 

[kai_sikorski]

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

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

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.

 

[Bacle2]

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.

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

 

[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?

 

[aaaa202]

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

 

[skiller]

By the way, there is no such word (in English) as "dices".

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

 

[kai_sikorski]

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

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.

 

[checkitagain]

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

"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."


http://www.answers.com/topic/dice


http://www.answers.com/topic/die-manufacturing

 

[HallsofIvy]

Thanks, guys, I bow to superior knowledge.

 

[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.

 

[kai_sikorski]

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

 

[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.

 

[kai_sikorski]

yup! ;)

 

[alan2]

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

 

[kai_sikorski]

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

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