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Throwing a dice 
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#1
Feb2312, 12:56 PM

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? 


#3
Feb2312, 01:40 PM

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



#4
Feb2312, 01:54 PM

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Throwing a dice
You can try using the expected value for a random variable.
Halls of Ivy: Congratulations on your 2^15 posts!. 


#5
Feb2312, 01:58 PM

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#6
Feb2312, 02:04 PM

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#7
Feb2312, 02:12 PM

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?



#8
Feb2312, 02:14 PM

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 :)



#9
Feb2312, 07:07 PM

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#10
Feb2312, 07:19 PM

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E[X_{1}  D = 1] P(D = 1) + E[X_{1} 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 X_{1} be 1 if the first die is 6 (0 otherwise) and X_{2} be 1 if the second die is 6 (0 otherwise) E[X_{1} + X_{2}] = E[X_{1}] +E[X_{1}] = 1*1/6 + 1*1/6 = 1/3 So again after N>>1 tries you would expect (~N/3) 6s. 


#11
Feb2412, 01:13 AM

P: 99

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/diemanufacturing 


#12
Feb2412, 07:15 AM

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Thanks, guys, I bow to superior knowledge.



#13
Feb2412, 08:12 AM

P: 199

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
Feb2412, 09:00 AM

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



#15
Feb2412, 09:40 AM

P: 199

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.



#17
Feb2412, 10:12 AM

P: 199

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



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