Well Done Bill! 1 Point for your Q41 answer...
Q41 Answer: 4 moves
- Also... 1/2 a point to Fred for his reasoning....

More Questions later...
can anyone answer Q42?

Q42: (continuation of 41) Also solve the above question but with different shapes:
Square?
Octahedron?
Dodecahedron?
Icosahedron? (1 Point for all shapes combined)

Have I left the thread to die, not yet.... ;)

Best Regards,
Luke.

Ermmmm - Didn't Q41 ask for the cube and Bill answer for a square??

F.

---

Well Done Bill! 1 Point for your Q41 answer...
Q41 Answer: 4 moves
- Also... 1/2 a point to Fred for his reasoning....

More Questions later...
can anyone answer Q42?

Q42: (continuation of 41) Also solve the above question but with different shapes:
Square?
Octahedron?
Dodecahedron?
Icosahedron? (1 Point for all shapes combined)

Have I left the thread to die, not yet.... ;)

Best Regards,
Luke.

---

- Luke.

For the square:

The first move will put the spider in an adjacent corner. Then there is a 1/2 probability that he will get the spider but 1/2 probability that he will go back to the original starting point so on average:

1/2*1/2=1/4 or 4 moves on average

But since the question asks for the number of "turns", by which I take it to means "corners encountered", then 4 moves would involve only 3 "turns"?

F.

---

For the square:

The first move will put the spider in an adjacent corner. Then there is a 1/2 probability that he will get the spider but 1/2 probability that he will go back to the original starting point so on average:

1/2*1/2=1/4 or 4 moves on average

Yikes. I just realized that most of the other figures are solids so here is another hint:

Octahedron:




is really

Phred et al

One more hint which I think should lead to the solution.

It isn't obvious until you draw the picture--draw the cube (put the ant and spider diagonally opposed across the cube. Now you can see that after one move the spider will be on a common face with the ant and diagonally across from it. This is state Y in my earlier post.

Important to realize that the expected value of the number of trials is the AVERAGE number of trials.

So starting from state X will always be exactly one more move than the solution than if you started from state Y (same face diagonally) A(X) =1 + A(Y) where A is average or expected number of trials

Now you want to get to an adjacent node to the ant (State Z) there is a 2/3 chance that you will do this from the same face and a 1/3 chance that you will go back to state X.

So now you have: A(Y)=(2/3)*(1+A(Z))+(1/3)*(1+A(X).

When you are in state Z there is a 1/3 chance you will get the ant and a 2/3 chance you will get back to state Y.

So you have: A(Z)=(1/3)*1+(2/3)*(1+A(Y)).

Solve for A(X) , A(Y), A(Z) It's too late here now for messy algebra so I will leave this for someone else to finish. You should now be able to solve the puzzle for the other figures as well.

The difficulty here initially is not having been told and having to realize that there are three different answers for the three different starting points that are mixed up in the words ( "OPPOSITE" side of the cube)

I also had math about 40-50 years ago. Doing these poblems has refreshed some of these notions and added new ones.

Remember your Math teacher's advice . "ALWAYS draw a picture of the problem--once you understand the problem completely you are half way there to the solution ) I have re-learned this myself from playing with Luke's problems.

Let X = to the number of trials to catch the ant from the initial position,
Let Y = to the number of trails to catch the ant from the diagonal corner on the same face as the ant.
Let Z =to the number of trials to catch the ant from one corner away from the ant.


Choose the starting position ( e. g. Assume that you are diagonally across the cube on another face of the cube from the ant) At this point and at every other point in the random walk there is a 1/3 , 2/3 probability that the spider will move to condition X, Y or Z.

Write equations for how these probabilities compare. How X Y and Z compare on average, There would be three equations. Express the relationships for these three conditions and solve the equations. You could probably find a matrix tool on the Internet or maybe one to solve the equations directly so you wont have to wade through all the substitution.

Nope, I'm still not getting it. And I have to be on-site with a client for the next 2 days so I think you'll have to enlighten us all or the master might pull the plug :(

F.

---

I also had math about 40-50 years ago. Doing these poblems has refreshed some of these notions and added new ones.

Remember your Math teacher's advice . "ALWAYS draw a picture of the problem--once you understand the problem completely you are half way there to the solution ) I have re-learned this myself from playing with Luke's problems.

Let X = to the number of trials to catch the ant from the initial position,
Let Y = to the number of trails to catch the ant from the diagonal corner on the same face as the ant.
Let Z =to the number of trials to catch the ant from one corner away from the ant.


Choose the starting position ( e. g. Assume that you are diagonally across the cube on another face of the cube from the ant) At this point and at every other point in the random walk there is a 1/3 , 2/3 probability that the spider will move to condition X, Y or Z.

Write equations for how these probabilities compare. How X Y and Z compare on average, There would be three equations. Express the relationships for these three conditions and solve the equations. You could probably find a matrix tool on the Internet or maybe one to solve the equations directly so you wont have to wade through all the substitution.

One last chance, or this thread is going to the fishes....

Post your answers to the Questions above....

Best Regards,
Luke.

Warning : if youse guys don't solve this real soon Daddio will have to come out of retirement and stumble around to the answer. I think it would involve solving a few simultaneous equations is all.

Where are the other stalwarts ??

Regards,

Bill
AKA DADDIO

Sorry but it's over 40 years since my brain was in this sort of gear. Given the equations - no problemo; it's deriving the relevant equations from the question that's giving me headaches ;(

F.

---

One last chance, or this thread is going to the fishes....

Post your answers to the Questions above....

Best Regards,
Luke.

Warning : if youse guys don't solve this real soon Daddio will have to come out of retirement and stumble around to the answer. I think it would involve solving a few simultaneous equations is all.

Where are the other stalwarts ??

Regards,

Bill
AKA DADDIO

One last chance, or this thread is going to the fishes....

Post your answers to the Questions above....

Best Regards,
Luke.

---

- Luke.

Hint--it's not too hard

at each possible configuration of ant and spider, the spider has three options.

Whether he gets the ant or moves to the same face or moves in a penultimate position and then gets the ant is also broken into different numbers of 1/3 probability.

Let's go --you can do it Bunky !!

Very true... Q41 is not that hard...

Thanks for everyone's participation!!!

Dedications today go-to:
Sarge!
Bill!
Fred!

Best Regards,
Luke.

Here again are the questions!

Q41: An ant and a blind spider are on opposite corners of a cube. The ant is stationary and the spider moves at random from one corner to another along the edges only. What is the expected number of turns before the spider reaches the ant? (1 Point)

Q42: Also solve the above question but with different shapes:
Square?
Octahedron?
Dodecahedron?
Icosahedron? (1 Point for all shapes combined)

---

- Luke.

Hint--it's not too hard

at each possible configuration of ant and spider, the spider has three options.

Whether he gets the ant or moves to the same face or moves in a penultimate position and then gets the ant is also broken into different numbers of 1/3 probability.

Let's go --you can do it Bunky !!

Re-iterating the current questions left:

Q41: An ant and a blind spider are on opposite corners of a cube. The ant is stationary and the spider moves at random from one corner to another along the edges only. What is the expected number of turns before the spider reaches the ant? (1 Point)

Q42: Also solve the above question but with different shapes:
Square?
Octahedron?
Dodecahedron?
Icosahedron? (1 Point for all shapes combined)

Best Regards,
Luke.

---

- Luke.

Q43: At McDonalds you can order Chicken McNuggets in boxes of 6, 9, and 20. What is the largest number such that you can not order any combination of the above to achieve exactly the number you want?

In the Hitchhiker's Guide to the Galaxy, 42 is the number from which all meaning ("the meaning of life, the universe, and everything") could be derived, and therefore it's 1 more than that.

Dominique solves Q43....
Q43 Answer: 43
Congratulations!!!

Anybody that wants to have a go at solving Q41 & Q42, have a look at Bill's last three posts, there could be some useful information to get everybody started in there....

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

With Dominique in 5th, WinterKnight, the original problem solving champion, has fallen to 6th....

Two Questions Remain: (To solve Q42, you need to read Q41)

Q41: An ant and a blind spider are on opposite corners of a cube. The ant is stationary and the spider moves at random from one corner to another along the edges only. What is the expected number of turns before the spider reaches the ant? (1 Point)

Q42: Also solve the above question but with different shapes:
Square?
Octahedron?
Dodecahedron?
Icosahedron? (1 Point for all shapes combined)

Best Regards,
Luke.

---

- Luke.

Q43: At McDonalds you can order Chicken McNuggets in boxes of 6, 9, and 20. What is the largest number such that you can not order any combination of the above to achieve exactly the number you want?

In the Hitchhiker's Guide to the Galaxy, 42 is the number from which all meaning ("the meaning of life, the universe, and everything") could be derived, and therefore it's 1 more than that.

TU ES RECTUS OH VALDE UNUS


Illegitimati non Carborundum


DADDIUS (PATERIUS ??)

Hello Luke et al,

I will sit out on these to let others have fun working these out. I may give a hint on the McDonald's question but probably won't need to.. I will watch the struggles here with interest.

Luke: Do we need a definition of "Opposite" here ?? probably not. Are you implying that the answer is the same if they are diagonally opposite across the cube or diagonally opposite on a given face of the cube.

Regards,

Bill
AKA DADDIO

After some scribbles--it does look like there is a difference in the number of trials if they are both on the same plane (face) as opposed to separated across the cube on another face. Maybe the winner should give both answers

Q43: At McDonalds you can order Chicken McNuggets in boxes of 6, 9, and 20. What is the largest number such that you can not order any combination of the above to achieve exactly the number you want?

In the Hitchhiker's Guide to the Galaxy, 42 is the number from which all meaning ("the meaning of life, the universe, and everything") could be derived, and therefore it's 1 more than that.

---

Hello Luke et al,

I will sit out on these to let others have fun working these out. I may give a hint on the McDonald's question but probably won't need to.. I will watch the struggles here with interest.

Luke: Do we need a definition of "Opposite" here ?? probably not. Are you implying that the answer is the same if they are diagonally opposite across the cube or diagonally opposite on a given face of the cube.

Regards,

Bill
AKA DADDIO

Today, I would like to dedicate this thread to:

Sarge - for his never ending work providing answers and solutions.
Bill - for being current Math Solving champion, and always having a go.
guido.man - for quickly climbing his way to the near top, while still facing big competition.
and last but not least...
Fred - for always popping in and always giving an answer when I least expect anyone too.

Thank you all!

And here is the next round of questions:

Q41: An ant and a blind spider are on opposite corners of a cube. The ant is stationary and the spider moves at random from one corner to another along the edges only. What is the expected number of turns before the spider reaches the ant? (1 Point)

Q42: Also solve the above question but with different shapes:
Square?
Octahedron?
Dodecahedron?
Icosahedron? (1 Point for all shapes combined)

Q43: At McDonalds you can order Chicken McNuggets in boxes of 6, 9, and 20. What is the largest number such that you can not order any combination of the above to achieve exactly the number you want?

I expect everyone that even looks at this thread to have a go!

Best Regards,
Luke.

---

- Luke.