Riddle me this!

[z4vdBt]

acronym recognition...

ottffssent

?

one two three...

[PStone]

acronym recognition...

ottffssent

?

one two three...

That was a good one.

[jpm2]

I was given the hockey puck problem by my Dad when I was 12, figured it out 4 years later. There is more than one method, but they arrive at the same destination. He used cue balls instead of hockey pucks.
After I figured it out, I gave it to a few friends, of which only one figured it out.... in 2 days. Made me feel pretty dumb at the time.

The Monty Hall problem is strictly a matter of odds.

[Monty]

It took me a while but I think I figured out the hockey puck riddle.

Wife walked in while I was scribbling on the notepad, I explained the riddle. She said, "I'd just hold them in each hand and judge the weights, don't need the scale." Beaten again.

[Surfingringo]

It took me a while but I think I figured out the hockey puck riddle.

Wife walked in while I was scribbling on the notepad, I explained the riddle. She said, "I'd just hold them in each hand and judge the weights, don't need the scale." Beaten again.

Shoot me a pm if you want to know. :spyder:

[tangent]

Oooh I love these. My favourite one is the Monty Hall problem:

Suppose you're on a game show, and you're given the choice of three doors: Behind one door is a car; behind the others, goats. You pick a door, say No. 1, and the host, who knows what's behind the doors, opens another door, say No. 3, which has a goat. He then says to you, "Do you want to pick door No. 2?" Is it to your advantage to switch your choice?

Simple answer: Yes, always switch your choice...you'll be right 2/3 of the time. Trust me...I'm a math prof.

[Surfingringo]

Oooh I love these. My favourite one is the Monty Hall problem:

Suppose you're on a game show, and you're given the choice of three doors: Behind one door is a car; behind the others, goats. You pick a door, say No. 1, and the host, who knows what's behind the doors, opens another door, say No. 3, which has a goat. He then says to you, "Do you want to pick door No. 2?" Is it to your advantage to switch your choice?

Simple answer: Yes, always switch your choice...you'll be right 2/3 of the time. Trust me...I'm a math prof.

That’s correct but I find the problem interesting in its subtleties. It’s also interesting to consider both why it makes sense to switch and why it is confusing.

Ultimately, the additional information changes the statistics. For example, were I playing the game I would have a 66% chance of winning if I switched my answer and a 33% chance if I stuck with my original choice. If another contestant came along AFTER one of the three doors was eliminated and was presented the same two choices then he would have a 50/50 chance. The extra information i received changed the odds.

[tangent]

Yeah...I agree. It's an interesting problem. What information you have and WHEN you have it is important.
BTW, Your hockey puck problem is super interesting...and challenging. I know we have to start by weighing 4 pucks vs 4 pucks...and then go from there. Anyway...

[Pelagic]

If it's between doors 1 and 2, you have a 50% chance regardless. What am I missing?

Just like if you flip a coin 10 times and get heads every time. The probability of getting heads on the 11th toss is still 50%.

[jpm2]

If it's between doors 1 and 2, you have a 50% chance regardless. What am I missing?

Just like if you flip a coin 10 times and get heads every time. The probability of getting heads on the 11th toss is still 50%.

Explained 2 posts above yours.

[tangent]

If it's between doors 1 and 2, you have a 50% chance regardless. What am I missing?

Just like if you flip a coin 10 times and get heads every time. The probability of getting heads on the 11th toss is still 50%.

When you choose a door initially, you have a 1/3 chance of selecting the winning door. Once a losing door (of the other two) has been eliminated, when you change your choice to the remaining door, you have in essence selected both doors...giving you a 2/3 chance of selecting the winning door.

[Pelagic]

If it's between doors 1 and 2, you have a 50% chance regardless. What am I missing?

Just like if you flip a coin 10 times and get heads every time. The probability of getting heads on the 11th toss is still 50%.

When you choose a door initially, you have a 1/3 chance of selecting the winning door. Once a losing door (of the other two) has been eliminated, when you change your choice to the remaining door, you have in essence selected both doors...giving you a 2/3 chance of selecting the winning door.

With only 2 doors remaining, how is there a 2/3 chance of anything?

[tangent]

If it's between doors 1 and 2, you have a 50% chance regardless. What am I missing?

Just like if you flip a coin 10 times and get heads every time. The probability of getting heads on the 11th toss is still 50%.

When you choose a door initially, you have a 1/3 chance of selecting the winning door. Once a losing door (of the other two) has been eliminated, when you change your choice to the remaining door, you have in essence selected both doors...giving you a 2/3 chance of selecting the winning door.

With only 2 doors remaining, how is there a 2/3 chance of anything?

The only way to make you a believer is to ask you to do a simulation...label three red plastic cups (non-see-through) with numbers 1,2 and 3. Have someone place an object under one of the cups (without you looking). Then, you select a cup (without revealing what is or is not under it). Then, the person who knows where the object lies reveals that under one of the remaining cups there is no prize and asks you if you want to switch. Switch your choice and record the results (whether you have won or not). Repeat this 100 times and I guarantee you will select the winning cup around 66% of the time.

[Pelagic]

If it's between doors 1 and 2, you have a 50% chance regardless. What am I missing?

Just like if you flip a coin 10 times and get heads every time. The probability of getting heads on the 11th toss is still 50%.

When you choose a door initially, you have a 1/3 chance of selecting the winning door. Once a losing door (of the other two) has been eliminated, when you change your choice to the remaining door, you have in essence selected both doors...giving you a 2/3 chance of selecting the winning door.

With only 2 doors remaining, how is there a 2/3 chance of anything?

The only way to make you a believer is to ask you to do a simulation...label three red plastic cups (non-see-through) with numbers 1,2 and 3. Have someone place an object under one of the cups (without you looking). Then, you select a cup (without revealing what is or is not under it). Then, the person who knows where the object lies reveals that under one of the remaining cups there is no prize and asks you if you want to switch. Switch your choice and record the results (whether you have won or not). Repeat this 100 times and I guarantee you will select the winning cup around 66% of the time.

So you're saying the person's knowledge of the location of the prize influences statistics?

[SkullBouncer]

(Might get in a wee bit o' trouble with this one.... :eek: )

What's the difference between the Panama Canal and Nancy Pelosi??

...One is a Busy Ditch... :p














If this offends, my friends, I'll delete it.
It's Humor Folks!! :D :D :D

:spyder: :spyder: Stay Sharp --
-- SB / BRUCE :) :)

[tangent]

When you choose a door initially, you have a 1/3 chance of selecting the winning door. Once a losing door (of the other two) has been eliminated, when you change your choice to the remaining door, you have in essence selected both doors...giving you a 2/3 chance of selecting the winning door.

With only 2 doors remaining, how is there a 2/3 chance of anything?

The only way to make you a believer is to ask you to do a simulation...label three red plastic cups (non-see-through) with numbers 1,2 and 3. Have someone place an object under one of the cups (without you looking). Then, you select a cup (without revealing what is or is not under it). Then, the person who knows where the object lies reveals that under one of the remaining cups there is no prize and asks you if you want to switch. Switch your choice and record the results (whether you have won or not). Repeat this 100 times and I guarantee you will select the winning cup around 66% of the time.

So you're saying the person's knowledge of the location of the prize influences statistics?

It's counter-intuitive and hard to explain via messages like this, but I assure you it's correct.

[Pelagic]

With only 2 doors remaining, how is there a 2/3 chance of anything?

The only way to make you a believer is to ask you to do a simulation...label three red plastic cups (non-see-through) with numbers 1,2 and 3. Have someone place an object under one of the cups (without you looking). Then, you select a cup (without revealing what is or is not under it). Then, the person who knows where the object lies reveals that under one of the remaining cups there is no prize and asks you if you want to switch. Switch your choice and record the results (whether you have won or not). Repeat this 100 times and I guarantee you will select the winning cup around 66% of the time.

So you're saying the person's knowledge of the location of the prize influences statistics?

It's counter-intuitive and hard to explain via messages like this, but I assure you it's correct.

I'm just trying to wrap my head around it. I know riddles and answers aren't about being 100% mathematically or statistically correct, but correct in their own unique and clever way.

[SkullBouncer]

...Rolling the Bones, could go either way...


What's the difference between a cornhusker with epilepsy, and a prostitute with diahhereah?

One shucks between fits... :eek:



:spyder: :spyder: SB / BRUCE :p :p

[Surfingringo]

The only way to make you a believer is to ask you to do a simulation...label three red plastic cups (non-see-through) with numbers 1,2 and 3. Have someone place an object under one of the cups (without you looking). Then, you select a cup (without revealing what is or is not under it). Then, the person who knows where the object lies reveals that under one of the remaining cups there is no prize and asks you if you want to switch. Switch your choice and record the results (whether you have won or not). Repeat this 100 times and I guarantee you will select the winning cup around 66% of the time.

So you're saying the person's knowledge of the location of the prize influences statistics?

It's counter-intuitive and hard to explain via messages like this, but I assure you it's correct.

I'm just trying to wrap my head around it. I know riddles and answers aren't about being 100% mathematically or statistically correct, but correct in their own unique and clever way.

This is what makes this such a beautiful little riddle. It is indeed 100% mathematically sound. It’s not an overly complicated concept either but it’s one of those scenarios where once the mind conceptualizes something a certain way, it is very difficult to switch the way you see it. This mental illusion is the very essence of a good riddle and why I like this one so much. Let me give you a similar problem and see if this helps.

Let’s say you are on a similar show but there are 50 doors instead of 3. Only one one of them has a prize and the host knows which. You select 1 door. What are the chances that you will win the prize? 1 in 50. Before receiving any other information, if you were able to switch from your 1 door to the other 49 would you do it? Of course you would, because then you would have a 49 out of 50 chance instead of 1 out of 50. The point is, even if the host (remember, he knows where the prize is) eliminates 48 of those 49 doors, you still have the opportunity to choose the set of 49 doors over your 1. It doens’t matter that the guy who knew where the prize was revealed 48 doors that didn’t have the prize. You can trade your door for the 49...before or after he does his thing, it doesn’t matter. Just like in the original problem you can trade your 1-3 chance for the 2-3 chance of the other two doors. Before or after the host has revealed anything, doesn’t matter.

I’m sure it’s clear now, but here’s one other way to think of it. The “trick” is that once one of the doors is Revealed, it looks as if we are given the option of switching one door for the other. This is the illusion. What we are really being given is a chance to switch one door for two.

[Pelagic]

So you're saying the person's knowledge of the location of the prize influences statistics?

It's counter-intuitive and hard to explain via messages like this, but I assure you it's correct.

I'm just trying to wrap my head around it. I know riddles and answers aren't about being 100% mathematically or statistically correct, but correct in their own unique and clever way.

This is what makes this such a beautiful little riddle. It is indeed 100% mathematically sound. It’s not an overly complicated concept either but it’s one of those scenarios where once the mind conceptualizes something a certain way, it is very difficult to switch the way you see it. This mental illusion is the very essence of a good riddle and why I like this one so much. Let me give you a similar problem and see if this helps.

Let’s say you are on a similar show but there are 50 doors instead of 3. Only one one of them has a prize and the host knows which. You select 1 door. What are the chances that you will win the prize? 1 in 50. Before receiving any other information, if you were able to switch from your 1 door to the other 49 would you do it? Of course you would, because then you would have a 49 out of 50 chance instead of 1 out of 50. The point is, even if the host (remember, he knows where the prize is) eliminates 48 of those 49 doors, you still have the opportunity to choose the set of 49 doors over your 1. It doens’t matter that the guy who knew where the prize was revealed 48 doors that didn’t have the prize. You can trade your door for the 49...before or after he does his thing, it doesn’t matter. Just like in the original problem you can trade your 1-3 chance for the 2-3 chance of the other two doors. Before or after the host has revealed anything, doesn’t matter.

I’m sure it’s clear now, but here’s one other way to think of it. The “trick” is that once one of the doors is Revealed, it looks as if we are given the option of switching one door for the other. This is the illusion. What we are really being given is a chance to switch one door for two.

That specific scenario is clear, but not the 3 door situation. Once a door is eliminated, it's eliminated. There are 2 remaining doors, yet chances of 1/3 and 2/3? I have given up completely on that and frankly am not very interested in understanding it.

The 50 door example did highlight a separate concept I wasn't noticing though, which is cool. I appreciate you taking the time to demonstrate that.