Jason,

Maybe you and I could become rich acting as expert witnesses for all of the sports folk who test positive for performance enhancing drugs. We could warm up on the upcoming Olympics and then move on to the Tour-d' France teams.

regards,

Bill

Jason,

Maybe you and I could become rich acting as expert witnesses for all of the sports folk who test positive for performance enhancing drugs. We could warm up on the upcoming Olympics and then move on to the Tour-d' France teams.

regards,

Bill

Great !, they tell me the parties are fantastic and the party favours even better ;D

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"Living by the wisdom of computer science doesn't sound so bad after all. And unlike most advice, it's backed up by proofs." -- Algorithms to live by: The computer science of human decisions.

Back to bed --should be sleeping now

Fred is right about his answer


The answer to 75 is only 50%. Bayes' theorem is a little hard to understand.

Extra puzzle for Fred for fun : If you came to a 5-way split in a road and saw that the signpost that pointed the way and mileage to all of the distant towns had been knocked over and was lying down--broken entirely from its moorings. How could you determine which road to take?

Assuming the roads are at 72 degrees to each other so that the angles give no clue, then there is either an arm of the signpost, or no arm, pointing to where you have just come from. Line that one up and the others are correct.

F.

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nice job Fred !!

Well done to Bill and Jason Gee for solving Question 75. Bill gave the correct answer, Jason Gee gave the solution.

I will also award Fred one point for his effort in Bill's problem....

Congratulations to all three of you!!!

Microsoft Windows Standings XP
1. William Rothamel - 19 1/2 Points
2. Fred W - 19 Points
3. Guido.Man - 10 Points
4. Mr. Kevvy - 7 Points
5. Dominique - 6 1/2 Points
6. WinterKnight - 6 Points
7. Sarge - 5 1/2 Points
8. Philadelphia - 5 1/2 Points
9. Jason gee - 3 Points
10. John McLeod VII - 1 Point
11. Labbie - 1 Point
12. Scary Capitalist - 1/2 Point
13. TBD...

And about 60 posts remain...

Question 76 (1 Point) : Two boats on opposite sides of a river head towards each other at different speeds. When they pass each other the first time they are 700 yards from one shoreline. They continue to the opposite shoreline, turn around, and move towards each other again. When they pass the second time they are 300 yards from the other shoreline. How wide is the river? Assume both boats travel at a constant speed and ignore factors such as turn-around time and the current of the river.

Luke.

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- Luke.

Well done to Bill and Jason Gee for solving Question 75. Bill gave the correct answer, Jason Gee gave the solution.

I will also award Fred one point for his effort in Bill's problem....

Congratulations to all three of you!!!

Microsoft Windows Standings XP
1. William Rothamel - 19 1/2 Points
2. Fred W - 19 Points
3. Guido.Man - 10 Points
4. Mr. Kevvy - 7 Points
5. Dominique - 6 1/2 Points
6. WinterKnight - 6 Points
7. Sarge - 5 1/2 Points
8. Philadelphia - 5 1/2 Points
9. Jason gee - 3 Points
10. John McLeod VII - 1 Point
11. Labbie - 1 Point
12. Scary Capitalist - 1/2 Point
13. TBD...

And about 60 posts remain...

Question 76 (1 Point) : Two boats on opposite sides of a river head towards each other at different speeds. When they pass each other the first time they are 700 yards from one shoreline. They continue to the opposite shoreline, turn around, and move towards each other again. When they pass the second time they are 300 yards from the other shoreline. How wide is the river? Assume both boats travel at a constant speed and ignore factors such as turn-around time and the current of the river.

Luke.

Who said these were "simple" math problems, lol.

After wearing out my brain, my guestimate is 1,800.

[edit]talking about math problems, check this out Large Hadron Collider

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That's the correct answer, Philly! Congratulations!!! Also, very good article, can't wait for it to be up and running!

Question 76 Answer: 1800.

Microsoft Windows Standings Vista
1. William Rothamel - 19 1/2 Points
2. Fred W - 19 Points
3. Guido.Man - 10 Points
4. Mr. Kevvy - 7 Points
5. Dominique - 6 1/2 Points
6. Philadelphia - 6 1/2 Points
7. WinterKnight - 6 Points
8. Sarge - 5 1/2 Points
9. Jason gee - 3 Points
10. John McLeod VII - 1 Point
11. Labbie - 1 Point
12. Scary Capitalist - 1/2 Point
13. TBD...

Philly is quickly racing up the table...

Q77 (1 Point) : You have a box of matches and six 64-minute fuses. Each fuse burns exactly 64 minutes but the rate of burning is inconsistent. A half a fuse would not necessarily burn in 32 minutes. The fuses are also not equally inconsistent. Two cut fuses of the same length would not necessarily burn in the same time. Without the use of a clock, scissors, or anything other than the matches and the fuses how can you create a 60-minute fuse?

Luke.

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- Luke.

Hint: I Think: You can establish an accurate time base by lighting a fuse at both ends. This will give an accurate measure of 32 minutes, Do this one at a time while lighting one end of the remaining ones at the right times and then the other ends of various fuses at key time points. Remember total burn time is 64 minutes total but count when both ends are burning to add both ends' burn times. The key is that they always burn through -end to end in 64 minutes. The trick is then to establish these time bases and light the last fuse when there is four minutes left on the previous one and snuff it out when the previous one burns out--you therefore have 60 minutes left on your last fuse.

So to clarify you go from 64 to 36 to 16 to 8 to 4 by lighting the second ends at appropriate times when the previous one burns out--all you need is one with 4 minutes left and then light your last 64 minute fuse and blow it out when the 4 minute one finally burns out completely.

Also that's my answer ??

DADDIO

Nice one, Bill. Recognising the power of 2.

To me the sequence is:

Light one end of 5 fuses and the other end of one of those 5.
When the first fuse has burned through, there will be 32 minutes left on the other 4. Light the other end of the second fuse.
When the second fuse has burned through, there will be 16 minutes left on the other 3. Light the other end of the third fuse.
When the third fuse has burned through, there will be 8 minutes left on the other 2. Light the other end of the fourth fuse.
When the fourth fuse has burned through there will be 4 minutes left on the other 1. Light the other end of the fifth fuse AND ONE END OF THE SIXTH FUSE.
When the fifth fuse has burned through, snuff out the sixth fuse and there will be 60 minutes burning time left.

But the points belong to Bill.

F.

[edit]
Wish I'd noticed Bill's clarification before wearing my fingers out typing all that ;P
BTW - Bill's 36 should be a 32
[/edit]

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This question will give me a headache, lol :)

I'm gonna sit back and watch on this one :)

What happened to "Simple",(teasing, great questions) lol. :)

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Hey, congratulations to Bill & Fred for solving Q77! I will award Bill 1 point for first solve, but I think Fred deserves a half-point at least for the detailed solution to Q77!!!

Both of you, congratulations!

And Philly - Well they started out as simple... and as I advanced the thread, i continually found harder problems. Hopefully the (possibly...) Second thread will be easier :)

Microsoft Windows Standings 7
1. William Rothamel - 20 1/2 Points
2. Fred W - 19 1/2 Points
3. Guido.Man - 10 Points
4. Mr. Kevvy - 7 Points
5. Dominique - 6 1/2 Points
6. Philadelphia - 6 1/2 Points
7. WinterKnight - 6 Points
8. Sarge - 5 1/2 Points
9. Jason gee - 3 Points
10. John McLeod VII - 1 Point
11. Labbie - 1 Point
12. Scary Capitalist - 1/2 Point
13. TBD...

Question 78 (1 Point) - A casino offers to refund half your net losses if you bet only on red or black in roulette. The roulette wheel is the standard American variety that has both a zero and double zero and pays even money. You must bet the same amount every time. How many times should you bet to maximize your expected return?

Luke.

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- Luke.

Ramblings on the Casino Problem. Normally You would have to know the size of the gambler's stash to predict how long he could stave off Gambler's Ruin.

The game is only slightly in favor of the house. The wheel has 36 numbers 19 red and 19 black lplus two greens 0 and 00. So your chance of winning is 18/38 or 9/19 or .4737. If half your losses are refunded then you will always come out ahead in the long run.

So my guess is that he should play an infinite number of times until he breaks the bank or they throw him out. but I must be interpreting wrong --it's not refunding each time --refunding only after you cash in and are down for sure--Therefore he should bet until he loses the first time--New answer.

But this is probably wrong I will have to try some real math when I wake up tomorrow --it's 3 AM and Ijust got up for a minute.

Having never encountered a live roulette wheel or studied the design of the wheel, I'll sit this one out as I wouldn't have a clue where to start.

F.

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Question 78 (1 Point) - A casino offers to refund half your net losses if you bet only on red or black in roulette. The roulette wheel is the standard American variety that has both a zero and double zero and pays even money. You must bet the same amount every time. How many times should you bet to maximize your expected return?

Luke.

(Edit)Given the probability of a first spin win is below 50% then statistically you should not play, and leave with the money you have, However, if you are forced to:
All down, Once and Once only, on either colour. As the casino is half refunding net losses, this presumably means no effect on funds during play. As the initial highest probability of a win is below 50%, successive wins become less likely very rapidly. (e.g. two wins in a row = ~0.47x~0.47 = ~22.1%). forget it, Highest probability is first spin, then leave with your winnings if you won, or half pot refund if you lost. :D

Jason

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"Living by the wisdom of computer science doesn't sound so bad after all. And unlike most advice, it's backed up by proofs." -- Algorithms to live by: The computer science of human decisions.

Hint and guess:

The flaw in our thinking lies in the fact that half of the losses are recovered. So if you lose early there may not be disincentive to play again since rewards await further down the string of trials.. if you lose on the next turn early on it is compensated to a high degree for that turn. But the more losses you have in the string the less compensation overall (percentage wise) for another loss. So I would guess that a handful of tries would be the best with the benefit falling off rapidly after a dozen or so tries.

Probably I should take the possibility of all win combinations out of a number of tries and multiply those by the joint probability of loss and winning at each trial --Too much calculating and factorials for me. I would need a calculator or better yet a program to carry the decimals --I would think that you might have to carry six decimal places or so since these probabilities would be multiplied by themselves many times.

I would guess the answer to be as few as five and maybe less than ten bets.

Make a table and see where the extra wins allowed by favorably compensating the early losses lose out to the smaller overall compensation on average as the number of trials goes up. (Huh ?) I do have a BASIC complier for my MAC but am a bit rusty and lazy now.

Regards,

Daddio

I know the answer but I can't take credit for it, a friend figured it out, lol. So, good luck guys :)

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Think out of the box guys.... :)

Question 78 (1 Point) - A casino offers to refund half your net losses if you bet only on red or black in roulette. The roulette wheel is the standard American variety that has both a zero and double zero and pays even money. You must bet the same amount every time. How many times should you bet to maximize your expected return?

Luke.

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- Luke.

Okay, a 'proper' crack at it (sortof):
A mess of factorials that I always did badly with, then I get best payoff it seems at 7 bets, though I had to mess with variances and standard deviations and stuff to get that. The payoff wasn't huge, somewhere around 25 to 30%. Now I have a headache and I'm no more certain of that answer anyway :S

Jason

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"Living by the wisdom of computer science doesn't sound so bad after all. And unlike most advice, it's backed up by proofs." -- Algorithms to live by: The computer science of human decisions.

Well Done Jason! Correct Answer....

Q78 Answer: 7 Bets

Standings:
1. William Rothamel - 20 1/2 Points
2. Fred W - 19 1/2 Points
3. Guido.Man - 10 Points
4. Mr. Kevvy - 7 Points
5. Dominique - 6 1/2 Points
6. Philadelphia - 6 1/2 Points
7. WinterKnight - 6 Points
8. Sarge - 5 1/2 Points
9. Jason gee - 4 Points
10. John McLeod VII - 1 Point
11. Labbie - 1 Point
12. Scary Capitalist - 1/2 Point
13. TBD...

About 40 posts remain...

Q79 (1/2 Point) : Take these group of numbers:
0,6,7,4,3,2,5,7,5,3,2,4,6,8,2,1,3,4,1,1,2,3,4,4,7,4,4

Determine the:
Mode
Median
Mean
Range (...of the group)

Luke.

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- Luke.