Challenge your brains! Can you solve this problem?

[Fan from VT]

Suppose you had eight billiard balls, the recruiter began. One of them is slightly heavier, but the only way to tell is by putting it on a scale against the others. What's the fewest number of times you'd have to use the scale to find the heavier ball?

4


weight 4 balls, then 4 balls. (2 weighs). record those weights. throw out the lighter 4. weigh 2 balls. (3rd weigh). if they weigh exactly half of the lighter weighing group of 4 throw them out. if not, save them. weigh one ball from the pair that contains the heavy ball (4th weigh). if that ball is exactly 1/4 of the weight of the original lighter 4, then the remaining unweighed ball is the heavy one. if it ways a little more than 1/4 of the lighter 4, then it's the heavy one.

[Rondo2287]

Suppose you had eight billiard balls, the recruiter began. One of them is slightly heavier, but the only way to tell is by putting it on a scale against the others. What's the fewest number of times you'd have to use the scale to find the heavier ball?

3

1 time.  You pick up two balls one is heavier you know thats the heavy ball. 

[Fan from VT]

Suppose you had eight billiard balls, the recruiter began. One of them is slightly heavier, but the only way to tell is by putting it on a scale against the others. What's the fewest number of times you'd have to use the scale to find the heavier ball?

3

1 time.  You pick up two balls one is heavier you know thats the heavy ball. 

so it's a trick question? you have to be lucky?

[Rondo2287]

Suppose you had eight billiard balls, the recruiter began. One of them is slightly heavier, but the only way to tell is by putting it on a scale against the others. What's the fewest number of times you'd have to use the scale to find the heavier ball?

3

1 time.  You pick up two balls one is heavier you know thats the heavy ball. 

so it's a trick question? you have to be lucky?

The question is whats the minimum amount of times.  So ya its kinda a trick question.  It's used during job interviews as a question.  I think as a test of clear thinking under pressure.  Count me as one that would get that wrong during an interview though haha.

[fairweatherfan]

Suppose you had eight billiard balls, the recruiter began. One of them is slightly heavier, but the only way to tell is by putting it on a scale against the others. What's the fewest number of times you'd have to use the scale to find the heavier ball?

3

1 time.  You pick up two balls one is heavier you know thats the heavy ball. 

so it's a trick question? you have to be lucky?

The question is whats the minimum amount of times.  So ya its kinda a trick question.  It's used during job interviews as a question.  I think as a test of clear thinking under pressure.  Count me as one that would get that wrong during an interview though haha.

Then why not 0?  Just guess at random and be right, without ever using the scale. 

[Rondo2287]

Suppose you had eight billiard balls, the recruiter began. One of them is slightly heavier, but the only way to tell is by putting it on a scale against the others. What's the fewest number of times you'd have to use the scale to find the heavier ball?

3

1 time.  You pick up two balls one is heavier you know thats the heavy ball. 

so it's a trick question? you have to be lucky?

The question is whats the minimum amount of times.  So ya its kinda a trick question.  It's used during job interviews as a question.  I think as a test of clear thinking under pressure.  Count me as one that would get that wrong during an interview though haha.

Then why not 0?  Just guess at random and be right, without ever using the scale.

Cause the question says the  only way you can tell is by using the scale. 

[Fan from VT]

Suppose you had eight billiard balls, the recruiter began. One of them is slightly heavier, but the only way to tell is by putting it on a scale against the others. What's the fewest number of times you'd have to use the scale to find the heavier ball?

3

1 time.  You pick up two balls one is heavier you know thats the heavy ball. 

so it's a trick question? you have to be lucky?

The question is whats the minimum amount of times.  So ya its kinda a trick question.  It's used during job interviews as a question.  I think as a test of clear thinking under pressure.  Count me as one that would get that wrong during an interview though haha.

now that i think about it, i don't even think 1 is the right answer. say you pick up 2 balls, and you're lucky enough to grab the heavy one. well according to the question, you can only tell one is heavier by weighing! so you would have to still weigh twice in order to tell that one ball is heavier than the other. So I think lowest possible is 2, and only if you are super lucky. And if you weren't lucky the first time, you're locked into doing 1 at a time. But if you do it systematically, 4 weighs is the worst you can do.

[fairweatherfan]

Suppose you had eight billiard balls, the recruiter began. One of them is slightly heavier, but the only way to tell is by putting it on a scale against the others. What's the fewest number of times you'd have to use the scale to find the heavier ball?

3

1 time.  You pick up two balls one is heavier you know thats the heavy ball. 

so it's a trick question? you have to be lucky?

The question is whats the minimum amount of times.  So ya its kinda a trick question.  It's used during job interviews as a question.  I think as a test of clear thinking under pressure.  Count me as one that would get that wrong during an interview though haha.

Then why not 0?  Just guess at random and be right, without ever using the scale.

Cause the question says the  only way you can tell is by using the scale. 

Yeah, but you don't have to know you're right to actually be right.  You've still found the heavy ball even though you didn't get any evidence to prove it.

Seems kind of like an unfair question, since most people would assume "find the heavier ball" = "be guaranteed to find the heavier ball every time".

PS - VT - I think the kind of scale they're talking about is a balance scale, where you can weigh items against each other.

[mgent]

Suppose you had eight billiard balls, the recruiter began. One of them is slightly heavier, but the only way to tell is by putting it on a scale against the others. What's the fewest number of times you'd have to use the scale to find the heavier ball?

3

1 time.  You pick up two balls one is heavier you know thats the heavy ball. 

so it's a trick question? you have to be lucky?

The question is whats the minimum amount of times.  So ya its kinda a trick question.  It's used during job interviews as a question.  I think as a test of clear thinking under pressure.  Count me as one that would get that wrong during an interview though haha.

Then why not 0?  Just guess at random and be right, without ever using the scale.

Cause the question says the  only way you can tell is by using the scale. 

Yeah, but you don't have to know you're right to actually be right.  You've still found the heavy ball even though you didn't get any evidence to prove it.

Seems kind of like an unfair question, since most people would assume "find the heavier ball" = "be guaranteed to find the heavier ball every time".

PS - VT - I think the kind of scale they're talking about is a balance scale, where you can weigh items against each other.

Yeah but you didn't "find" the heavier ball if you didn't prove it on the scale.  You're just completely guessing.

But I agree that the question can be interpreted both ways.

[Fan from VT]

Suppose you had eight billiard balls, the recruiter began. One of them is slightly heavier, but the only way to tell is by putting it on a scale against the others. What's the fewest number of times you'd have to use the scale to find the heavier ball?

3

1 time.  You pick up two balls one is heavier you know thats the heavy ball. 

so it's a trick question? you have to be lucky?

The question is whats the minimum amount of times.  So ya its kinda a trick question.  It's used during job interviews as a question.  I think as a test of clear thinking under pressure.  Count me as one that would get that wrong during an interview though haha.

Then why not 0?  Just guess at random and be right, without ever using the scale.

Cause the question says the  only way you can tell is by using the scale. 

Yeah, but you don't have to know you're right to actually be right.  You've still found the heavy ball even though you didn't get any evidence to prove it.

Seems kind of like an unfair question, since most people would assume "find the heavier ball" = "be guaranteed to find the heavier ball every time".

PS - VT - I think the kind of scale they're talking about is a balance scale, where you can weigh items against each other.

aaah, ok. so yeah, then 1 if lucky, 3 if you want to be right every time.

[wjb3000]

Hey math people, I need to a come up with a math problem to present to my class.  Something along the same level as the last couple I posted on here (nothing insanely difficult).  Anyone have any favorites?

In the mean time here's one that took me a couple of minutes to figure out, but was a big DERRR once I did.

During the game show "Pick A Door" the contestant gets to choose one of the doors on the stage. Behind two of the doors is a bucket of dirt, behind the 3rd is some fabulous prize (make up your own) Bucko, the host, opens one of the two doors the contestant did not choose.  The door he opens always reveals a bucket of dirt.  The contestant is then given the option of sticking with his original choice, or switching his pick to the unrevealed door.

If you were the contestant would you stick with your original pick or switch?  Why?

You switch.  When you made your original pick, you had a 1/3 chance of being correct, and a 2/3 chance of being wrong. Because the door that's eliminated is always a wrong response, that 2/3 chance now only applies to the remaining door that you didn't pick.  

This one is tricky from an interpretation perspective.  I understand the thought behind your answer, but I view each guess as an independent event.  If there are two choices then there are two choices, regardless of the previous outcome.  If I were playing the mega millions and I got the first five numbers right, I wouldn't be thinking to myself "there WAS a 1 in 312,500,000 chance that I would get this point, and because of that I HAVE to change my last number to one of the 44 still available, other than the ones already chosen and whatever my original choice was."  In reality, you have a precisely equal shot at each of the remaining 44 balls...the chance of getting that last number are 1 in 44, not 1 in 312 million, irregardless of what the chances of actually getting to that point were.

[fairweatherfan]

Hey math people, I need to a come up with a math problem to present to my class.  Something along the same level as the last couple I posted on here (nothing insanely difficult).  Anyone have any favorites?

In the mean time here's one that took me a couple of minutes to figure out, but was a big DERRR once I did.

During the game show "Pick A Door" the contestant gets to choose one of the doors on the stage. Behind two of the doors is a bucket of dirt, behind the 3rd is some fabulous prize (make up your own) Bucko, the host, opens one of the two doors the contestant did not choose.  The door he opens always reveals a bucket of dirt.  The contestant is then given the option of sticking with his original choice, or switching his pick to the unrevealed door.

If you were the contestant would you stick with your original pick or switch?  Why?

You switch.  When you made your original pick, you had a 1/3 chance of being correct, and a 2/3 chance of being wrong. Because the door that's eliminated is always a wrong response, that 2/3 chance now only applies to the remaining door that you didn't pick.  

This one is tricky from an interpretation perspective.  I understand the thought behind your answer, but I view each guess as an independent event.  If there are two choices then there are two choices, regardless of the previous outcome.  If I were playing the mega millions and I got the first five numbers right, I wouldn't be thinking to myself "there WAS a 1 in 312,500,000 chance that I would get this point, and because of that I HAVE to change my last number to one of the 44 still available, other than the ones already chosen and whatever my original choice was."  In reality, you have a precisely equal shot at each of the remaining 44 balls...the chance of getting that last number are 1 in 44, not 1 in 312 million, irregardless of what the chances of actually getting to that point were.

I understand that thinking, but look at it this way.  The prize will be behind each door 1/3rd of the time, and the door that's eliminated is always an incorrect choice.  So here's all the possible outcomes if you always choose to switch:

I pick A, prize is actually behind A.  B or C is eliminated.  I switch, I lose.

I pick A, prize is behind B.  C is eliminated, I switch, I win.

I pick A, prize is behind C.  B is eliminated, I switch, I win.

The guesses aren't independent because only wrong answers are eliminated.  All else being equal, you win 2/3rds of the time when switching, and 1/3rd if you don't.

[cdif911]

Hey math people, I need to a come up with a math problem to present to my class.  Something along the same level as the last couple I posted on here (nothing insanely difficult).  Anyone have any favorites?

In the mean time here's one that took me a couple of minutes to figure out, but was a big DERRR once I did.

During the game show "Pick A Door" the contestant gets to choose one of the doors on the stage. Behind two of the doors is a bucket of dirt, behind the 3rd is some fabulous prize (make up your own) Bucko, the host, opens one of the two doors the contestant did not choose.  The door he opens always reveals a bucket of dirt.  The contestant is then given the option of sticking with his original choice, or switching his pick to the unrevealed door.

If you were the contestant would you stick with your original pick or switch?  Why?

You switch.  When you made your original pick, you had a 1/3 chance of being correct, and a 2/3 chance of being wrong. Because the door that's eliminated is always a wrong response, that 2/3 chance now only applies to the remaining door that you didn't pick.  

This one is tricky from an interpretation perspective.  I understand the thought behind your answer, but I view each guess as an independent event.  If there are two choices then there are two choices, regardless of the previous outcome.  If I were playing the mega millions and I got the first five numbers right, I wouldn't be thinking to myself "there WAS a 1 in 312,500,000 chance that I would get this point, and because of that I HAVE to change my last number to one of the 44 still available, other than the ones already chosen and whatever my original choice was."  In reality, you have a precisely equal shot at each of the remaining 44 balls...the chance of getting that last number are 1 in 44, not 1 in 312 million, irregardless of what the chances of actually getting to that point were.

I understand that thinking, but look at it this way.  The prize will be behind each door 1/3rd of the time, and the door that's eliminated is always an incorrect choice.  So here's all the possible outcomes if you always choose to switch:

I pick A, prize is actually behind A.  B or C is eliminated.  I switch, I lose.

I pick A, prize is behind B.  C is eliminated, I switch, I win.

I pick A, prize is behind C.  B is eliminated, I switch, I win.

The guesses aren't independent because only wrong answers are eliminated.  All else being equal, you win 2/3rds of the time when switching, and 1/3rd if you don't.

ah the Monty Hall problem - the best part of it was when Marilyn Von Savant actually solved it correctly for the first time she got all kinds of criticism

the good folks at Hofstra university created a simulation to prove that it works

http://people.hofstra.edu/Steven_R_Costenoble/MontyHall/MontyHallSim.html

[ChampKind]

Hey math people, I need to a come up with a math problem to present to my class.  Something along the same level as the last couple I posted on here (nothing insanely difficult).  Anyone have any favorites?

In the mean time here's one that took me a couple of minutes to figure out, but was a big DERRR once I did.

During the game show "Pick A Door" the contestant gets to choose one of the doors on the stage. Behind two of the doors is a bucket of dirt, behind the 3rd is some fabulous prize (make up your own) Bucko, the host, opens one of the two doors the contestant did not choose.  The door he opens always reveals a bucket of dirt.  The contestant is then given the option of sticking with his original choice, or switching his pick to the unrevealed door.

If you were the contestant would you stick with your original pick or switch?  Why?

You switch.  When you made your original pick, you had a 1/3 chance of being correct, and a 2/3 chance of being wrong. Because the door that's eliminated is always a wrong response, that 2/3 chance now only applies to the remaining door that you didn't pick.  

This one is tricky from an interpretation perspective.  I understand the thought behind your answer, but I view each guess as an independent event.  If there are two choices then there are two choices, regardless of the previous outcome.  If I were playing the mega millions and I got the first five numbers right, I wouldn't be thinking to myself "there WAS a 1 in 312,500,000 chance that I would get this point, and because of that I HAVE to change my last number to one of the 44 still available, other than the ones already chosen and whatever my original choice was."  In reality, you have a precisely equal shot at each of the remaining 44 balls...the chance of getting that last number are 1 in 44, not 1 in 312 million, irregardless of what the chances of actually getting to that point were.

I understand that thinking, but look at it this way.  The prize will be behind each door 1/3rd of the time, and the door that's eliminated is always an incorrect choice.  So here's all the possible outcomes if you always choose to switch:

I pick A, prize is actually behind A.  B or C is eliminated.  I switch, I lose.

I pick A, prize is behind B.  C is eliminated, I switch, I win.

I pick A, prize is behind C.  B is eliminated, I switch, I win.

The guesses aren't independent because only wrong answers are eliminated.  All else being equal, you win 2/3rds of the time when switching, and 1/3rd if you don't.

Actually, your logic there is flawed.  You're considering the eliminations of B or C in that first case to be combined incidences.  However, they are separate, just like your last two scenarios.  it should be:

I pick A, prize is actually behind A.  B is eliminated.  I switch, I lose.

I pick A, prize is actually behind A.  C is eliminated.  I switch, I lose.

I pick A, prize is behind B.  C is eliminated, I switch, I win.

I pick A, prize is behind C.  B is eliminated, I switch, I win.

So, no matter what, at the end, you have a 50/50 shot, regarding the selection with the last two doors as a separate step.  

EDIT:  Or maybe my logic is flawed.  The simulator sure seems to support your theory.

[fairweatherfan]

I understand that thinking, but look at it this way.  The prize will be behind each door 1/3rd of the time, and the door that's eliminated is always an incorrect choice.  So here's all the possible outcomes if you always choose to switch:

I pick A, prize is actually behind A.  B or C is eliminated.  I switch, I lose.

I pick A, prize is behind B.  C is eliminated, I switch, I win.

I pick A, prize is behind C.  B is eliminated, I switch, I win.

The guesses aren't independent because only wrong answers are eliminated.  All else being equal, you win 2/3rds of the time when switching, and 1/3rd if you don't.

Actually, your logic there is flawed.  You're considering the eliminations of B or C in that first case to be combined incidences.  However, they are separate, just like your last two scenarios.  it should be:

I pick A, prize is actually behind A.  B is eliminated.  I switch, I lose.

I pick A, prize is actually behind A.  C is eliminated.  I switch, I lose.

I pick A, prize is behind B.  C is eliminated, I switch, I win.

I pick A, prize is behind C.  B is eliminated, I switch, I win.

So, no matter what, at the end, you have a 50/50 shot, regarding the selection with the last two doors as a separate step.  

EDIT:  Or maybe my logic is flawed.  The simulator sure seems to support your theory.

Yeah - the problem there is you're assuming your first two options occur the same % of the time, but in actuality A is only right 1/3rd of the time.  Your setup would require A to be right 50% of the time. Whether B or C is eliminated is random within that category, but both events are still a subset of the 1/3rd of the time that A has the prize.  They are combined incidences in that sense.