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mustangsally
mustangsally
Joined: Mar 29, 2011
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April 14th, 2015 at 7:24:41 AM permalink
Quote: AlanMendelson

It is very similar to what we players go through at craps.
We see one die come to rest on, for example, the 6 -- while the second die spins.
And we say to ourselves, there is only a 1/6 chance of a 7-out.

excellent observation!
i see this many times too (2)

but only one die spins like a top, never both (unless i do not pay attention)


how does the shooter make that happen, where only one die spins from tossing two dice?

I ask and they never say how they do it (the shooter that is)

i wood pay to learn that skill
(i have only done this meself maybe twice, and did not know how i did it)

Alan, you have a video of what you talk about?
Mully
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ChesterDog
ChesterDog
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April 14th, 2015 at 7:31:51 AM permalink
Below are results of ten simulations of 3600 dice rolls each:

TrialAt least one die was a 2.Both dice were 2.
1
1119
98
2
1079
82
3
1123
83
4
1106
116
5
1110
89
6
1133
97
7
1080
97
8
1115
93
9
1068
100
10
1095
101
TOTAL
11028
956

mustangsally
mustangsally
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April 14th, 2015 at 7:38:42 AM permalink
Quote: ChesterDog

Below are results of ten simulations of 3600 dice rolls each:

i say you went after a different question

the OP question is too (2) vague
too much info missing

i could also assume the dice are rolled for a purpose to see how often they will both be the same
meaning a double
but the author of the question did not word the question correctly

i look to be right (left) too
Sally
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Kerkebet
Kerkebet
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April 14th, 2015 at 8:03:52 AM permalink
The question is like the two envelopes in that possibilities, and the intersection thereof, may become confused with expected value.

If someone said to me that at least one die were a 2, I would say prove it. The someone would then reveal the one of the dice which is a 2. But, the someone would always have the choice of which of the two dice to reveal first (coming into the scenario). Are many other non-2 possibilities for the die not revealed (but associated with the 2 of the die revealed first) than six. The die not revealed brings more possibilities into the question because of the "choice" of which die to reveal in the event of at least a 2.

When at least one die is 2, then half the time the one die is the claimed-to-exist 2, but the other half of the time the other die is the claimed-to-exist 2? So that half of six possibilities (of the other die when one die is 2) plus half of another six possibilities (of one die when other die is 2) equals six possibilities. No.
Nonsense is a very hard thing to keep up. Just ask the Wizard and company.
ThatDonGuy
ThatDonGuy
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April 14th, 2015 at 8:11:21 AM permalink
Quote: ThatDonGuy

I have a feeling the problem is, you're misunderstanding what "at least one of the dice is a 2" means.


Apparently, we can't come to a consensus on what it means.

I for one read it as, "Either one of the dice is a 2 and the other isn't, or both dice are 2," which is the interpretation used in what indignant99 eventually called his "wrong" analysis.
DJTeddyBear
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Thanks for this post from:
EdCollins
April 14th, 2015 at 9:23:40 AM permalink
Quote: AlanMendelson

It is very similar to what we players go through at craps. We see one die come to rest on, for example, the 6 -- while the second die spins. And we say to ourselves, there is only a 1/6 chance of a 7-out.


That's a different situation, and 1/6 is correct. In your example, the dice are segregated. One has settled the other is still an unknown random result. It's kinda like the prior example of a red and blue die and the person says the red die is a two.

The original question is more like a craps throw where both die are spinning and someone says "If one die settles on two, what's the odds that the other will also be a two?" Answer: 1/11.
I invented a few casino games. Info: http://www.DaveMillerGaming.com/ 覧覧覧覧覧覧覧覧覧覧覧覧覧覧覧覧覧覧 Superstitions are silly, childish, irrational rituals, born out of fear of the unknown. But how much does it cost to knock on wood? 😁
RS
RS
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April 14th, 2015 at 9:50:31 AM permalink
Quote: DJTeddyBear

That's a different situation, and 1/6 is correct. In your example, the dice are segregated. One has settled the other is still an unknown random result. It's kinda like the prior example of a red and blue die and the person says the red die is a two.

The original question is more like a craps throw where both die are spinning and someone says "If one die settles on two, what's the odds that the other will also be a two?" Answer: 1/11.



It is best to say "if either dice...". That way the Alans of the world can try to have a chance of understanding.

Alan, remember this scenario: the first die lands on a non-2 number, and the second die is spinning. If that die lands on a 2, then the roll meets the " if either of the dice land on 2..." requirement.
indignant99
indignant99
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April 14th, 2015 at 11:19:46 AM permalink
Quote: RS

...If it's "there are 2 dice, 1 red and 1 blue, if the blue one is a 2, what's the chance the red one is a 2?" then the answer is surely 1/6.


And if both dice are the same color, it doesn't change the answer from 1/6.
Yeah, I made a mistake once. I thought I was wrong, when I actually wasn't. -Indignant
ThatDonGuy
ThatDonGuy
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April 14th, 2015 at 11:32:48 AM permalink
Quote: indignant99

And if both dice are the same color, it doesn't change the answer from 1/6.


You might be right, if the original statement was, "One of the dice is a 2", instead of, "At least one of the dice is a 2."

If both dice have stopped, and both of the numbers are known, then the answer is 11/36, as you yourself discovered before tossing out the correct (in this case) solution.

If only one die has stopped, then the probability is 1/6.

However, we don't know what the probability is that both dice have stopped, so in that instance, the answer is, "It is impossible to calculate without knowing if both dice have stopped or not."
indignant99
indignant99
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April 14th, 2015 at 11:38:28 AM permalink
Quote: ThatDonGuy

...If both dice have stopped, and both of the numbers are known, then the answer is 11/36, as you yourself discovered before tossing out the correct (in this case) solution...


That is not correct. Please review this.
Yeah, I made a mistake once. I thought I was wrong, when I actually wasn't. -Indignant

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