Poll

8 votes (47.05%)
6 votes (35.29%)
3 votes (17.64%)
2 votes (11.76%)
6 votes (35.29%)
2 votes (11.76%)
3 votes (17.64%)
2 votes (11.76%)
8 votes (47.05%)
6 votes (35.29%)

17 members have voted

charliepatrick
charliepatrick
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Gialmere
August 11th, 2020 at 12:44:05 PM permalink
One of the dice has an 8
I've seen this before but worked it out remembering the top were 4 and 8. 1 3 4 5 6 8 ; 1 2 2 3 3 4.
Wizard
Administrator
Wizard
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Gialmere
August 11th, 2020 at 2:16:07 PM permalink
My answer agrees with Charlie's, but I know he gets the credit for being first.

My solution was basically trial and error. It took about 20 minutes to stumble on the answer.
It's not whether you win or lose; it's whether or not you had a good bet.
Gialmere
Gialmere
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August 11th, 2020 at 5:06:50 PM permalink
Quote: charliepatrick

I've seen this before but worked it out remembering the top were 4 and 8. 1 3 4 5 6 8 ; 1 2 2 3 3 4.


Quote: Wizard

My answer agrees with Charlie's, but I know he gets the credit for being first.

My solution was basically trial and error. It took about 20 minutes to stumble on the answer.


Correct!

Some versions of this puzzle will send you on a snipe hunt by asking you to find all pairs of dice that qualify. In fact, the only alternative dice that meet the conditions are {1,2,2,3,3,4} and {1,3,4,5,6,8}.

-----------------------------

Dice. My favorite Dice rhyme: "Jack 'n Jill went up the hill, both wit a buck 'n a quarter..."

... Jill came down wit two fifty
Have you tried 22 tonight? I said 22.
rsactuary
rsactuary
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August 11th, 2020 at 6:59:08 PM permalink
silly me tried to do it without a number going above 6.... that probably should have been stated in the question....
ThatDonGuy
ThatDonGuy
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Gialmeregordonm888
August 12th, 2020 at 6:57:39 AM permalink
Quote: Gialmere

Some versions of this puzzle will send you on a snipe hunt by asking you to find all pairs of dice that qualify.



In order to have exactly one way to roll a 2, each die needs exactly one 1.
Because of this, in order to have exactly two 3s, either both dice have one 2, or one has two 2s and the other has none.

At the other end, in order to have exactly one 12, the highest numbers on the two dice have to be (6,6), (5,7), or (4,8), and they have to be unique on their respective dice, as otherwise there would be multiple ways to roll 12.
(3,9) is not possible as, on the die where 3 is the highest, there can only be one 3 and one 1, so the die would be {1,2,2,2,2,3}, but that would have four ways to roll a 3.
This leaves eight possibilities (well, nine, but {1 x x x x 6} {1 2 2 x x 6} is the same as {1 2 2 x x 6} {1 x x x x 6}):
{1 2 2 x x 6} {1 x x x x 6}
{1 2 x x x 6} {1 2 x x x 6}
{1 2 2 x x 5} {1 x x x x 7}
{1 2 x x x 5} {1 2 x x x 7}
{1 x x x x 5} {1 2 2 x x 7}
{1 2 2 x x 4} {1 x x x x 8}
{1 2 x x x 4} {1 2 x x x 8}
{1 x x x x 4} {1 2 2 x x 8}
In each case, x is between 3 and (the highest number on that die - 1)
This means the last three are:
{1 2 2 3 3 4} {1 x x x x 8}
{1 2 3 3 3 4} {1 2 x x x 8}
{1 3 3 3 3 4} {1 2 2 x x 8}
However, the last two are impossible as each has (at least) four ways to roll a 4, and (at least) three ways to roll an 11.

With {1 2 2 3 3 4} {1 x x x x 8}, there are already 2 ways to roll 10 (28 28), so there can be only one more; this has to be (46)
{1 2 2 3 3 4} {1 x x x 6 8} - there are 3 ways to roll 9 (18 36 36), so there can be only one more; this has to be (45)
{1 2 2 3 3 4} {1 x x 5 6 8} - there are 4 ways to roll 8 (26 26 35 35), so there can be only one more; this has to be (44)
{1 2 2 3 3 4} {1 x 4 5 6 8} - there are 5 ways to roll 7 (16 25 25 34 34), so there can be only one more; this has to be (43)
{1 2 2 3 3 4} {1 3 4 5 6 8} is a solution
Only six other combinations to go...

Also note that, in order for the probabilities to match, the sum of the numbers on the two dice must be 42.

Gialmere
Gialmere
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August 13th, 2020 at 7:33:31 AM permalink


Twenty-seven identical cubes are painted white. They are assembled into a larger 3x3x3 cube, the outside of which is then painted black. Next, the cube is disassembled and the smaller cubes thoroughly shuffled in a bag. A blindfolded man then reassembles the pieces back into a 3x3x3 cube.

What is the probability that the outside of this cube is completely black?
Have you tried 22 tonight? I said 22.
ThatDonGuy
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Gialmere
August 13th, 2020 at 8:29:22 AM permalink
Quote: Gialmere

Twenty-seven identical cubes are painted white. They are assembled into a larger 3x3x3 cube, the outside of which is then painted black. Next, the cube is disassembled and the smaller cubes thoroughly shuffled in a bag. A blindfolded man then reassembles the pieces back into a 3x3x3 cube.

What is the probability that the outside of this cube is completely black?



The 27 cubes consist of one "center" cube with no black sides, 6 "face center" cubes with one black side, 12 "edge center" cubes with two black sides, and 8 "corner" cubes with three black sides.
The center cube has to be in the center as otherwise it will expose at least one white face.
The corner cubes have to be in the corners as otherwise a cube in the corner will expose at least one white face.
The edge center cubes have to be in the edge centers as, since the corner cubes have to be in the corners, any other cube in an edge center will expose at least one white face.
This leaves the face center cubes for the face center positions.

Of the 27! possible ways to arrange the cubes, there are 12! 8! 6! ways to arrange them so that each cube is in a correct space.
Each corner cube has 8 corners, one of which has to be in the corner of the large cube in order for all 3 black faces to be exposed. Note that the orientation of the cube is irrelevant.
Each edge center cube has 12 edges, one of which has to be the edge of the two exposed faces; again, orientation is irrelevant.
Each face center cube has 6 faces, one of which has to be the exposed face.
The probability is (12! 8! 6!) / (27! 12^12 8^8 6^6) = (12! 8! 6!) / (27! 2^54 3^18) = (12! 8! 6!) / (27! 24^18).
According to the JB calculator, this is 1 / 5,465,062,811,999,459,151,238,583,897,240,371,200

Gialmere
Gialmere
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August 13th, 2020 at 10:57:02 AM permalink
Quote: ThatDonGuy


The 27 cubes consist of one "center" cube with no black sides, 6 "face center" cubes with one black side, 12 "edge center" cubes with two black sides, and 8 "corner" cubes with three black sides.
The center cube has to be in the center as otherwise it will expose at least one white face.
The corner cubes have to be in the corners as otherwise a cube in the corner will expose at least one white face.
The edge center cubes have to be in the edge centers as, since the corner cubes have to be in the corners, any other cube in an edge center will expose at least one white face.
This leaves the face center cubes for the face center positions.

Of the 27! possible ways to arrange the cubes, there are 12! 8! 6! ways to arrange them so that each cube is in a correct space.
Each corner cube has 8 corners, one of which has to be in the corner of the large cube in order for all 3 black faces to be exposed. Note that the orientation of the cube is irrelevant.
Each edge center cube has 12 edges, one of which has to be the edge of the two exposed faces; again, orientation is irrelevant.
Each face center cube has 6 faces, one of which has to be the exposed face.
The probability is (12! 8! 6!) / (27! 12^12 8^8 6^6) = (12! 8! 6!) / (27! 2^54 3^18) = (12! 8! 6!) / (27! 24^18).
According to the JB calculator, this is 1 / 5,465,062,811,999,459,151,238,583,897,240,371,200


Correct! Down to the last digit.

I wondered if the JB calculator might get used for this one.

-------------------------

I placed a sugar cube on the kitchen floor for an ant.
When he left to get the rest of the ants, I quickly removed the sugar cube.

Now everyone thinks he lied.
Have you tried 22 tonight? I said 22.
Gialmere
Gialmere
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August 13th, 2020 at 1:09:06 PM permalink


An older brother was sent to find his younger brother who had gone bicycle riding at the park. When the older brother entered the park, he spotted his younger brother's tire treads on a dirt trail crossing his path. (See above image.)

Was the younger brother traveling left to right, or right to left?

How do you know?
Have you tried 22 tonight? I said 22.
gordonm888
gordonm888
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August 13th, 2020 at 1:45:31 PM permalink
Here's a puzzle that is probably too easy, even for this thread. But I was playing with numbers and stumbled on this.

Decompose the cube of an integer, n, into a sum of two squares - one of which must be n2.

Ex: 17*17*17 = 4913. And 4913 = 522 + 472 but also = 682 + 172

a) Give a few examples of other integers (other than the trivial example of "1") that meet this criteria
b) Give a generic mathematical formula for identifying such integers.
So many better men, a few of them friends, were dead. And a thousand thousand slimy things lived on, and so did I.

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