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

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Discussion Overview

The discussion revolves around the probabilities associated with rolling a single die and two dice, specifically focusing on the outcomes of rolling a 1 or a 6. Participants explore the implications of rolling these dice an infinite number of times and whether the average occurrences of these outcomes will be equivalent.

Discussion Character

  • Exploratory
  • Technical explanation
  • Debate/contested
  • Mathematical reasoning

Main Points Raised

  • One participant claims the probability of rolling a 1 or a 6 on a single die is 1/3, while the probability of rolling at least one 6 with two dice is suggested to be lower than 2/3.
  • Another participant mentions counting the double 6 case as two occurrences when discussing averages.
  • Several participants discuss the use of expected value and the principle of indifference, suggesting that the two scenarios should yield the same result.
  • A participant raises a question about the relationship between the outcomes and product distribution.
  • One participant expresses confusion about the intuitive understanding of the problem and requests an example to clarify the concept.
  • Another participant attempts to clarify the original question regarding the expected number of 1's or 6's versus the expected number of 6's when rolling two dice.
  • There is a correction regarding the plural form of "dice," with some participants debating its usage in different contexts.

Areas of Agreement / Disagreement

Participants exhibit some agreement on the probabilities involved, particularly regarding the expected values. However, there remains disagreement on the interpretation of the original question and whether the scenarios yield equivalent results.

Contextual Notes

Some participants note the need for clarity in the original question and the assumptions made about the counting of outcomes. There are also unresolved discussions about the intuitive understanding of the probabilities involved.

Who May Find This Useful

This discussion may be useful for individuals interested in probability theory, particularly in the context of dice games and expected values in random experiments.

aaaa202
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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?
 
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If you count the two die 6-6 case as 2 then yes.
 
By the way, there is no such word (in English) as "dices". "Dice" is the plural of "die".
 
You can try using the expected value for a random variable.

Halls of Ivy: Congratulations on your 2^15 posts!.
 
Bacle2 said:
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.
 
kai_sikorski said:
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.
 
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?
 
btw... It seems like you guys think of it as intuitive. I don't - please give me an example that makes it intuitive :)
 
HallsofIvy said:
By the way, there is no such word (in English) as "dices".
Be careful; you're dicing with death with that remark. :wink:
 
  • #10
aaaa202 said:
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.
 
  • #11
HallsofIvy said:
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
 
Last edited:
  • #12
Thanks, guys, I bow to superior knowledge.
 
  • #13
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
No he was asking how many times 6s would come up, so you count the 6 - 6 case as 2
 
  • #15
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
yup! ;)
 
  • #17
Thanks, it helps to be talking about the same problem.
 
  • #18
Ups sorry, noticed a typo in my other post should say:
kai_sikorski said:
E[X1 + X2] = E[X1] +E[X2] = 1*1/6 + 1*1/6 = 1/3
 

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