# Throwing a dice

by aaaa202
Tags: dice, throwing
 P: 1,005 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?
 PF Gold P: 162 If you count the two die 6-6 case as 2 then yes.
 Math Emeritus Sci Advisor Thanks PF Gold P: 39,564 By the way, there is no such word (in English) as "dices". "Dice" is the plural of "die".
 Sci Advisor P: 1,170 Throwing a dice You can try using the expected value for a random variable. Halls of Ivy: Congratulations on your 2^15 posts!.
PF Gold
P: 162
 Quote by Bacle2 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.
P: 1,170
 Quote by 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.
You're right that it's overkill; I just thought the OP asked to have a formal proof.
 P: 1,005 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?
 P: 1,005 btw... It seems like you guys think of it as intuitive. I don't - please give me an example that makes it intuitive :)
P: 235
 Quote by HallsofIvy By the way, there is no such word (in English) as "dices".
Be careful; you're dicing with death with that remark.
PF Gold
P: 162
 Quote by aaaa202 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.
P: 99
 Quote by HallsofIvy 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."

 Math Emeritus Sci Advisor Thanks PF Gold P: 39,564 Thanks, guys, I bow to superior knowledge.
 P: 206 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.
 PF Gold P: 162 No he was asking how many times 6s would come up, so you count the 6 - 6 case as 2
 P: 206 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.
 PF Gold P: 162 yup! ;)
 P: 206 Thanks, it helps to be talking about the same problem.
PF Gold
P: 162
Ups sorry, noticed a typo in my other post should say:
 Quote by kai_sikorski E[X1 + X2] = E[X1] +E[X2] = 1*1/6 + 1*1/6 = 1/3

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