Joke answer: consume none. Open boxes, dump out contents, look for prizes, eat none of it.
Kobyashi Maru.

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Capitalize on this good fortune, one word can bring you round ... changes.

Not very mathematically rigorous (many years since I covered this ground) but my punt is 16 boxes.

F.

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Sarge--I need to give this more rigor instead of the intuitive stuff I like which is usually wrong. On the problem of the 4 colors. I think about the following.

The chance (probability) of having an acceptable draw at trial 1 is 1.00
And that having been done the chance of having a successful draw on trial 2 is 3/4 or .75 and that having been done the chance of having a successful draw on the third trial is .5 since only half of the color choices would be successful. On the fourth trial the probability of success is .25
.
Therefore .75 x .5 x .25 =.0975 Taking the inverse of this is 1/.0975 =10.25 so I say that the expected number of trials would be 11.

Of course I seem to remember something about conditional probability that is missing here in my analysis. I don't even know if Bayes theory needs to be applied here --I think not


Anyhow: My answer is 11 trials

here: let's think out load--Actually the .0975 is the probability of doing it in four boxes. So out of a bunch of trials of four only 9.75 % of those trials would be successful. So if you did 10 trials your probability would be close to one or .975. so 11 trials would put you over 1.00 (fuzzy thinking ?) Now the question is : what is the meaning of expected number of trials/ The problem is not how many trials of four boxes-it is how many boxes to get 4 different colors.

Another way is that any result of a draw of four is equally likely ie 1234 is just as likely as 1111.

There would be 4x4x4x4 combinations or 256 possible outcomes of which 4 x 3 x2 x 1 would be a success or 24. so 24/256 = only 9.375 % of the outcomes are favorable --this also suggests 11 tries to be at a certainty if everything is random.

But I think it has to be done on the basis of conditional probability and may even be more complicated than that. I will rest and sleep on it for now.

Sorry Bill. Wrong Answer. Try Again!

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- Luke.

my guess at cereal problem is (4 * 3 * 2 * 1) - (3 * 2 * 1) = 18.

But I'm useless at these prediction problems, probably because like statistics, I never have to use them, so I'm probably wrong again.

Q26: There is a free gift in my breakfast cereal. The manufacturers say that the gift comes in four different colors, and encourage one to collect all four (thus eating lots of cereal). Assuming there is an equal chance of getting any one of the colors, what is the expected number of boxes I must consume to get all four?

The answer to the problem can be expressed as the sum of the following:

Number of tries to get a first toy.
Number of tries to get a second toy once you have one toy.
Number of tries to get a third toy once you have two toys.
Number of tries to get the final toy once you have three toys.

The number of tries to get one toy is obviously one.
Once you have one the probability of getting a different toy in the next box is 3/4, thus the expected number of tries is 1/(3/4) = 4/3 to get the second toy.

By the same logic the number of tries to get the third is 1/(1/2)=2 and 1/(1/4)=4 for the final toy.

Adding these up yields 1 + 4/3 + 2 + 4 = 25/3 or 8.3333333333etc.

I'd say that actually you'd need 9 boxes.

Then again if you're really lucky you only need 4

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Q26: There is a free gift in my breakfast cereal. The manufacturers say that the gift comes in four different colors, and encourage one to collect all four (thus eating lots of cereal). Assuming there is an equal chance of getting any one of the colors, what is the expected number of boxes I must consume to get all four?

The answer to the problem can be expressed as the sum of the following:

Number of tries to get a first toy.
Number of tries to get a second toy once you have one toy.
Number of tries to get a third toy once you have two toys.
Number of tries to get the final toy once you have three toys.

The number of tries to get one toy is obviously one.
Once you have one the probability of getting a different toy in the next box is 3/4, thus the expected number of tries is 1/(3/4) = 4/3 to get the second toy.

By the same logic the number of tries to get the third is 1/(1/2)=2 and 1/(1/4)=4 for the final toy.

Adding these up yields 1 + 4/3 + 2 + 4 = 25/3 or 8.3333333333etc.

I'd say that actually you'd need 9 boxes.

Then again if you're really lucky you only need 4

Dominique is correct...
Q26 Answer: 8.333334 Boxes

Standings:
1. Mr. Kevvy - 7 Points
2. William Rothamel - 6 1/2 Points
3. WinterKnight - 5 Points
4. Sarge - 5 Points
5. Dominique - 4 1/2 Points
6. Fred W - 1 1/2 Points
7. John McLeod VII - 1 Point
8. Scary Capitalist - 1/2 Point
9. TBD......

Dominique is 1/2 a point out of range of Sarge and WinterKnight. But has created a big gap between 5th and 6th.....

Q29, 30, 31 coming soon.....

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- Luke.

I've got 3 more for everybody out there....

Q29: You are to roll a die over and over until you get a six.
What is the expected total of all throws before you throw a six?
By the "total" I mean the sum of the die faces, for example the sequence 1-3-5-6 would have a total of 9.

Q30: A cubic piece of cheese has been subdivided into 27 subcubes.
(so that it looks like a Rubik's Cube). A mouse starts to eat a corner subcube.
After eating any given subcube it goes on to another adjacent subcube. Is it possible for the mouse to eat all 27 subcubes and finish with the center cube?

Q31: On average how often does the minute hand pass the hour hand on an ordinary clock?

Best Regards,
Luke.

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- Luke.

Q31 seems obvious that it is once per hour or every 60 minutes on average--although the hour hand would move 1/12 of the way around so that this would be 65 minutes on average.

Q31: every 65 minutes.

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Calm Chaos Forum...Join Calm Chaos Now

I've got 3 more for everybody out there....

Q29: You are to roll a die over and over until you get a six.
What is the expected total of all throws before you throw a six?
By the "total" I mean the sum of the die faces, for example the sequence 1-3-5-6 would have a total of 9.

Q30: A cubic piece of cheese has been subdivided into 27 subcubes.
(so that it looks like a Rubik's Cube). A mouse starts to eat a corner subcube.
After eating any given subcube it goes on to another adjacent subcube. Is it possible for the mouse to eat all 27 subcubes and finish with the center cube?

Q31: On average how often does the minute hand pass the hour hand on an ordinary clock?

Best Regards,
Luke.

I still have trouble with the notion of "Expected" value or number.

The roll of two dice has 36 combinations of which a six can be rolled in the following ways:

1 and 5
5 and one
2 and 4
4 and two
3 and 3
3 and 3
so 1/6 of the rolls would be a six so i say that the expected number of rolls would be 6

Answer = 6

Bill Wins Q31. Sorry Labbie! You were 2 Minutes Late!

Q31 Answer: 65 Minutes

And we have a new problem solver champion.... Congratulations to Bill!!! May you have a long and prosperous reign.

Problem Solver Champions:
1. WinterKnight
2. Mr. Kevvy
3. William Rothamel - Current

Standings:
1. William Rothamel - 7 1/2 Points
2. Mr. Kevvy - 7 Points
3. WinterKnight - 5 Points
4. Sarge - 5 Points
5. Dominique - 4 1/2 Points
6. Fred W - 1 1/2 Points
7. John McLeod VII - 1 Point
8. Scary Capitalist - 1/2 Point
9. TBD......

Bill grabs the crown of Number One. Just....

Also Bill- just to clear the confusion around Question 29...

Example: What would be the average total of all die rolls until a six is rolled
i.e.
1,3,5 - 6 = 1+3+5=9. 9 is the total sum of all the rolls until a six. What would be the average total sum?

Q29 & 30 still up for grabs.

Best Regards,
Luke.

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- Luke.

I am looking at a Rubik's Cube right now.
There's a little ambiguity as to what you are taking to be adjacent.

I was going write a few different versions, based on different interpretations of adjacent, first assuming you mean they share an edge on the visible outside. But that in itself would prevent getting to what I call the core piece (your center cube).

Even then, there's some amiguity. For example, is a corner piece adjacent to a face center piece?



Let us assume the answer is no. Then we have the following.

6 face center pieces, 4 edges pieces and the core piece, for a total of 5 pieces.
8 corner pieces, each adjacent to 3 edges pieces.
12 edge (connecting to face center pieces) pieces, adjacent to 2 face center pieces and 2 corner pieces, for a total of 4 pieces.
1 core piece, adjacent to 6 face center pieces.

Rephrasing, two sub-cubes are being considered adjacent if they share a face, even if that face is not visible on the outside of the Rubik's Cube. Put yet another way, one cannot travel from one sub-cube to another if the two sub-cubes share only a point or an edge.

Now, I want to use up all the sub-cubes, ending at the core piece. Do I want the sub-cubes to be the nodes or the arcs? (I am using these terms as opposed to vertices and edges to avoid conflicting geometrical terms about the cubes and graph theoretic/combinatorics terminology.)

I am currently working on viewing the pieces as the arcs, so that I can consider an Euler path (using all the arcs), for which there are necessary and sufficient conditions to determine if an Euler path exists. "A connected, undirected multigraph has an Euler path but not an Euler circuit if and only if it has exactly two nodes of odd degree."

A nice thing about this is that I can look at 5 + 3 + 4 + 6 = 18 nodes, rather than 27.

I guess that means there are 27 arcs then, as well. Therefore, the sum of the degrees is 54. At least it is not odd, but it does not narrow anything out yet.

Here's the arc set, then: {core,blue,{blue,red,orange},{blue,orange},{blue,orange,green},{blue,green},{blue,green,white},{blue,white},{blue,white,red},{blue,red},white,red,orange,green,{green,white},{white,red},
{red,orange},{orange,green},yellow,{yellow,red,white},{yellow,white},{yellow,white,green},{yellow,green},{yellow,green,orange},{yellow,orange},{yellow,orange,red},{yellow,red}}.
Besides, the core, a single color is used for a face center piece, two colors for an edge piece and 3 colors for a corner piece.

(More editing underway.)

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Capitalize on this good fortune, one word can bring you round ... changes.

Sarge, in this context "adjacent" is up,down,left or right of the square. Not Diagonal.

Q29 & Q30 are still here:
Q29: You are to roll a die over and over until you get a six.
What is the expected total of all throws before you throw a six?
By the "total" I mean the sum of the die faces, for example the sequence 1-3-5-6 would have a total of 9.

Q30: A cubic piece of cheese has been subdivided into 27 subcubes.
(so that it looks like a Rubik's Cube). A mouse starts to eat a corner subcube.
After eating any given subcube it goes on to another adjacent subcube. Is it possible for the mouse to eat all 27 subcubes and finish with the center cube?

Best Regards,
Luke.

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- Luke.

ok I think I understand the DIE problem? Just thinking out loud. A six occurs one sixth of the time so: I guess that the expected value for the number of rolls would be six and the total of all the spots would be expected to be 6+5+4+3+2+1=21. Another way to look at it is that the 6 could occur at any one of the rolls and that on average you would expect it to occur after (1+2+3+4+5+6)/6=3.5 rolls the problem here is that any other number could appear during the other 2.5 rolls. The expected value on a single roll would; also be 3.5

so 3.5 x 3.5 =12.25 is my answer


I need to read up on what EXPECTED VALUE means. My last probability course was in 1969.

Q30: The answer is No.

This may be a Maths thread but I don't know how to put my reasoning into mathematical terms; however the logic goes as follows:

The cube has 3 layers, 1, 2 and 3 where the mouse is starting at the corner of layer 1 and the final piece to be eaten is at the centre of layer 2.

If it were possible then the following would have to be true:

(a) The penultimate piece to be eaten would have to be the centre of layer 3 since the final one is adjacent only to the centre pieces on each of the faces of the cube.

(b) For (a) to be true, the mouse would have to have eaten round the edge of layer 3 ending in the middle of one side before consuming the centre piece on layer 3. Thus the first piece on layer 3 must be corner piece.

(c) To start layer 3 on a corner piece, the previous piece must have been a corner piece on layer 2. Thus the mouse has eaten round layer 2, leaving the centre piece on this layer, having started in the middle of one of the sides of layer 2.

(d) To have started layer 2 in the middle of one side, it must have finished layer one in the middle of a side of the square.

But there is no sequence that will allow it to start at the corner, consume all 9 pieces on layer 1 and not finish at a corner piece.

Consequently, the pesky rodent cannot succeed (and a good thing too, as I like my cheese ;)

F.

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If my previous answer to Q30 is wrong then the answer is Yes ;)

F.

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Bill Wins Q31. Sorry Labbie! You were 2 Minutes Late!

Q31 Answer: 65 Minutes

Yeah, check his edit time, 4 minutes after my post.

[EDIT] He added everything following the "--" during that edit.

But hey, if you want to play favorites and Bill needs the satisfaction of winning badly enough to cheat, more power to you both. [/EDIT]

See ya.

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Bill Wins Q31. Sorry Labbie! You were 2 Minutes Late!

Q31 Answer: 65 Minutes

Yeah, check his edit time, 4 minutes after my post.

[EDIT] He added everything following the "--" during that edit.

But hey, if you want to play favorites and Bill needs the satisfaction of winning badly enough to cheat, more power to you both. [/EDIT]

See ya.

You are right --Labbie should get the point--I presumed he would.

Bill Wins Q31. Sorry Labbie! You were 2 Minutes Late!

Q31 Answer: 65 Minutes

Yeah, check his edit time, 4 minutes after my post.

[EDIT] He added everything following the "--" during that edit.

But hey, if you want to play favorites and Bill needs the satisfaction of winning badly enough to cheat, more power to you both. [/EDIT]

See ya.

When I first saw Bills post, It was 2 Minutes in front of yours. I got there before he edited it.
Hey, I'll happily give you the point and a half for any trouble I may of coursed.

Bill, your answer to Q29 is wrong.
Fred, you got Q30 right first time!
Q30 answer is No.

Q29 Remains:
Q29: You are to roll a die over and over until you get a six.
What is the expected total of all throws before you throw a six?
By the "total" I mean the sum of the die faces, for example the sequence 1-3-5-6 would have a total of 9.

Standings:
1. Mr. Kevvy - 7 Points
2. William Rothamel - 6 1/2 Points
3. WinterKnight - 5 Points
4. Sarge - 5 Points
5. Dominique - 4 1/2 Points
6. Fred W - 2 1/2 Points
7. Labbie - 1 1/2 Points
8. John McLeod VII - 1 Point
9. Scary Capitalist - 1/2 Point
10. TBD.....

Best Regards,
Luke.

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- Luke.