Question on probability...

[jman19]

Originally posted by: Kev

Originally posted by: LegendKiller

Originally posted by: blackdogdeek

Originally posted by: Viper GTS
The correct answer is switch, yet some people still haven't figured that out.

Viper GTS

this is correct.

the easiest way i've heard the solution explained so that it made sense was this:

imagine there are 100 cups. you choose #1. the person then flips over all the other cups except for cup #42. wouldn't you change to cup #42?

technically, your original cup #1 still has a 1/100 chance of having the dollar bill but cup #42 has a 99/100 chance of having the dollar bill.

This is a great way of putting it, after doing some research.

Yay, great. Please no responses after this. If this thread goes past 100 posts I will be very upset.

The question has been answered - mods, please lock this before any (more) morons chime in with their "opinion" on the matter.

 

[OS]

Originally posted by: LegendKiller

Originally posted by: blackdogdeek

Originally posted by: Viper GTS
The correct answer is switch, yet some people still haven't figured that out.

Viper GTS

this is correct.

the easiest way i've heard the solution explained so that it made sense was this:

imagine there are 100 cups. you choose #1. the person then flips over all the other cups except for cup #42. wouldn't you change to cup #42?

technically, your original cup #1 still has a 1/100 chance of having the dollar bill but cup #42 has a 99/100 chance of having the dollar bill.

This is a great way of putting it, after doing some research.

I don't think it works that way, as mentioned in the very beginning, after each action the probability is redistributed, because the set of unknowns is smaller.

You can't open up 99 empty cups, and say the last cup had 100% chance the whole time.

It would be pretty easy to write a small program that tallied this and after enough trials it should be apparent. I'm pretty convinced it is a wash at the last two.

 

[b0mbrman]

Originally posted by: Kev

Originally posted by: LegendKiller

Originally posted by: blackdogdeek

Originally posted by: Viper GTS
The correct answer is switch, yet some people still haven't figured that out.

Viper GTS

this is correct.

the easiest way i've heard the solution explained so that it made sense was this:

imagine there are 100 cups. you choose #1. the person then flips over all the other cups except for cup #42. wouldn't you change to cup #42?

technically, your original cup #1 still has a 1/100 chance of having the dollar bill but cup #42 has a 99/100 chance of having the dollar bill.

This is a great way of putting it, after doing some research.

Yay, great. Please no responses after this. If this thread goes past 100 posts I will be very upset.

Why do you get to be the last one?

 

[jjones]

Originally posted by: Vic
The plane takes off.

The only correct answer in this thread.

 

[UCDAggies]

The probability of winning if you switch is 2/3, the odds are NOT 2/3. The odds are 2/1.

 

[jman19]

Originally posted by: OS

Originally posted by: LegendKiller

Originally posted by: blackdogdeek

Originally posted by: Viper GTS
The correct answer is switch, yet some people still haven't figured that out.

Viper GTS

this is correct.

the easiest way i've heard the solution explained so that it made sense was this:

imagine there are 100 cups. you choose #1. the person then flips over all the other cups except for cup #42. wouldn't you change to cup #42?

technically, your original cup #1 still has a 1/100 chance of having the dollar bill but cup #42 has a 99/100 chance of having the dollar bill.

This is a great way of putting it, after doing some research.

I don't think it works that way, as mentioned in the very beginning, after each action the probability is redistributed, because the set of unknowns is smaller.

You can't open up 99 empty cups, and say the last cup had 100% chance the whole time.

It would be pretty easy to write a small program that tallied this and after enough trials it should be apparent. I'm pretty convinced it is a wash at the last two.

And like I said, another idiot chimes in. PLEASE LOCK THIS THREAD!

 

[esun]

Originally posted by: UCDAggies
Does the person who removed the second cup know that the dollar isn't under it, or is it random.

If the person always removes a cup with no dollar than it is 2/3, if they are picking cups at random than it 1/2.

This is correct.

 

[JujuFish]

Originally posted by: esun

Originally posted by: UCDAggies
Does the person who removed the second cup know that the dollar isn't under it, or is it random.

If the person always removes a cup with no dollar than it is 2/3, if they are picking cups at random than it 1/2.

This is correct.

Ditto. Switch if the cup without a dollar was knowingly and deliberately chosen. Don't bother if the revealed cup was shown at random.

 

[OS]

Originally posted by: jman19

Originally posted by: OS

Originally posted by: LegendKiller

Originally posted by: blackdogdeek

Originally posted by: Viper GTS
The correct answer is switch, yet some people still haven't figured that out.

Viper GTS

this is correct.

the easiest way i've heard the solution explained so that it made sense was this:

imagine there are 100 cups. you choose #1. the person then flips over all the other cups except for cup #42. wouldn't you change to cup #42?

technically, your original cup #1 still has a 1/100 chance of having the dollar bill but cup #42 has a 99/100 chance of having the dollar bill.

This is a great way of putting it, after doing some research.

I don't think it works that way, as mentioned in the very beginning, after each action the probability is redistributed, because the set of unknowns is smaller.

You can't open up 99 empty cups, and say the last cup had 100% chance the whole time.

It would be pretty easy to write a small program that tallied this and after enough trials it should be apparent. I'm pretty convinced it is a wash at the last two.

And like I said, another idiot chimes in. PLEASE LOCK THIS THREAD!

the only idiot is you with your baseless bullsh8.

 

[jman19]

Originally posted by: OS

Originally posted by: jman19

Originally posted by: OS

Originally posted by: LegendKiller

Originally posted by: blackdogdeek

Originally posted by: Viper GTS
The correct answer is switch, yet some people still haven't figured that out.

Viper GTS

this is correct.

the easiest way i've heard the solution explained so that it made sense was this:

imagine there are 100 cups. you choose #1. the person then flips over all the other cups except for cup #42. wouldn't you change to cup #42?

technically, your original cup #1 still has a 1/100 chance of having the dollar bill but cup #42 has a 99/100 chance of having the dollar bill.

This is a great way of putting it, after doing some research.

I don't think it works that way, as mentioned in the very beginning, after each action the probability is redistributed, because the set of unknowns is smaller.

You can't open up 99 empty cups, and say the last cup had 100% chance the whole time.

It would be pretty easy to write a small program that tallied this and after enough trials it should be apparent. I'm pretty convinced it is a wash at the last two.

And like I said, another idiot chimes in. PLEASE LOCK THIS THREAD!

the only idiot is you with your baseless bullsh8.

You said it yourself. After you show a cup, the number of unknowns is smaller. The model is updated with information - in your example, that WOULD NOT MEAN the last cup had a "100% chance the whole time."

 

[tfinch2]

But what if there is 0.999...9 dollars under the cup. Is there a dollar bill under there?

 

[jman19]

Originally posted by: tfinch2
But what if there is 0.999...9 dollars under the cup. Is there a dollar bill under there?

Is the cup in a plane on a treadmill?

 

[OS]

btw, the fact you picked a cup without opening it has has no bearing on the probability, because no information was revealed by only picking a cup.

 

[tfinch2]

Originally posted by: OS

Originally posted by: jman19

Originally posted by: OS

Originally posted by: LegendKiller

Originally posted by: blackdogdeek

Originally posted by: Viper GTS
The correct answer is switch, yet some people still haven't figured that out.

Viper GTS

this is correct.

the easiest way i've heard the solution explained so that it made sense was this:

imagine there are 100 cups. you choose #1. the person then flips over all the other cups except for cup #42. wouldn't you change to cup #42?

technically, your original cup #1 still has a 1/100 chance of having the dollar bill but cup #42 has a 99/100 chance of having the dollar bill.

This is a great way of putting it, after doing some research.

I don't think it works that way, as mentioned in the very beginning, after each action the probability is redistributed, because the set of unknowns is smaller.

You can't open up 99 empty cups, and say the last cup had 100% chance the whole time.

It would be pretty easy to write a small program that tallied this and after enough trials it should be apparent. I'm pretty convinced it is a wash at the last two.

And like I said, another idiot chimes in. PLEASE LOCK THIS THREAD!

the only idiot is you with your baseless bullsh8.

You're wrong. Look up the Monty Hall Problem.

 

[GeneValgene]

WIKIPEDIA EXPLAINS WHY YOU SHOULD SWITCH CUPS

 

[Rob9874]

I have another probability question, but didn't want to waste the board on a new thread.

They say that probability is not affected by previous attempts. For example, when you flip a coin, you have a 50/50 chance of it being heads every time you flip it, regardless of what happened before. However, aren't the odds of getting heads twice in a row less than 50%? And aren't the odds of flipping heads 3 times in a row less than that? So if I flipped a coin 5 times, and got heads 5 times, wouldn't it be smart to bet that the next flip will be tails? I understand that isolated flip has a 50/50 chance of being heads again, but the series of 6 flips has very low odds of being 6 heads in a row, and if the first 5 were heads, the 6th will probably be tails. Or am I stupid?

 

[crownjules]

Originally posted by: Rob9874
I have another probability question, but didn't want to waste the board on a new thread.

They say that probability is not affected by previous attempts. For example, when you flip a coin, you have a 50/50 chance of it being heads every time you flip it, regardless of what happened before. However, aren't the odds of getting heads twice in a row less than 50%? And aren't the odds of flipping heads 3 times in a row less than that? So if I flipped a coin 5 times, and got heads 5 times, wouldn't it be smart to bet that the next flip will be tails? I understand that isolated flip has a 50/50 chance of being heads again, but the series of 6 flips has very low odds of being 6 heads in a row, and if the first 5 were heads, the 6th will probably be tails. Or am I stupid?

No. Each coin flip is completely independent of any previous or following flip. The coin has no bias, it does not know how many times it has come up heads or tails in the past. The fact that it has flipped heads 6 times in a row doesn't matter, it is still exactly a 50/50 chance on the seventh flip.

Over a large enough sampling size, you will see the ratio of heads to tails flips balance out to 50/50. In a small sampling, you are subject to variation.

 

[jman19]

Originally posted by: Rob9874
I have another probability question, but didn't want to waste the board on a new thread.

They say that probability is not affected by previous attempts. For example, when you flip a coin, you have a 50/50 chance of it being heads every time you flip it, regardless of what happened before. However, aren't the odds of getting heads twice in a row less than 50%? And aren't the odds of flipping heads 3 times in a row less than that? So if I flipped a coin 5 times, and got heads 5 times, wouldn't it be smart to bet that the next flip will be tails? I understand that isolated flip has a 50/50 chance of being heads again, but the series of 6 flips has very low odds of being 6 heads in a row, and if the first 5 were heads, the 6th will probably be tails. Or am I stupid?

You are asking about conditional probability. The prob. of any given coin flip being heads is .5, the probability of getting heads on a flip given that your previous flip resulted in heads is .5, but the probability of flipping heads twice with no prior information is .25.

 

[OS]

Originally posted by: GeneValgene
WIKIPEDIA EXPLAINS WHY YOU SHOULD SWITCH CUPS

ugh if i wasn't swamped with work i would read it all for comprehension but right away i notice;

"If the host were to open one of the two remaining doors at random and happen to reveal a goat, there would indeed be a 50/50 choice between the other two. However, in the problem as originally presented, the player's initial choice does influence the host's available choices."

 

[dullard]

Switch. I'm not getting into why, but switching is better. The new information is irrelevant in this situation, you still have a 33% chance if you keep your choice. Thus switching is 100%-33% = 67%.

Remember, the person revealing the empty cup knows which one has the dollar and thus manipulates the odds in his favor. This ISN'T a random game.

 

[OS]

taken from the wiki article again;

"Asked whether the contestant should switch, vos Savant correctly replied, "If the host is clueless, it makes no difference whether you stay or switch. If he knows, switch" (vos Savant, 2006)."

So the OP scenario as posed, implies both players do not know what is in any of the cups, in which case, even the wiki article says it is a wash at the last two.

The different scenario later put in is if the other player does in fact know what is in each.

 

[crownjules]

Originally posted by: OS
taken from the wiki article again;

"Asked whether the contestant should switch, vos Savant correctly replied, "If the host is clueless, it makes no difference whether you stay or switch. If he knows, switch" (vos Savant, 2006)."

So the OP scenario as posed, implies both players do not know what is in any of the cups, in which case, even the wiki article says it is a wash at the last two.

The different scenario later put in is if the other player does in fact know what is in each.

Did you not read the example I posted at the top of page 2? Let me do it again because you obviously didn't.

Assume that Cup 1 contains the object.

SCENARIO 1: You pick Cup 1. Cup 2 or 3 is shown. If you KEEP the cup, you WIN. If you SWITCH the cup, you LOSE.

SCENARIO 2: You pick Cup 2. Cup 3 is shown to you. If you KEEP the cup, you LOSE. If you SWITCH, you WIN.

SCENARIO 3: You pick Cup 3. Cup 2 is show to you. If you KEEP the cup, you LOSE. If you SWITCH, you WIN.

In 2 of the 3 scenarios, you win when switching and lose by keeping. That's why it's better to switch.

 

[dullard]

Originally posted by: OS
"Asked whether the contestant should switch, vos Savant correctly replied, "If the host is clueless, it makes no difference whether you stay or switch. If he knows, switch" (vos Savant, 2006)."

That is what I was getting at. The way this is usually played, the host knows which cup has the prize and thus it isn't truely random. You should switch.

Lets do it for the 100 cup version.

Random game that no one ever plays:
[*]You have a 1% chance of picking the correct cup.
[*]There is a 98% chance that the 98 opened cups have the prize. In this case, who wins the dollar? Lets claim it is a tie and they split the dollar.
[*]There is a 1% chance that the last remaining cup has the prize.

You tie 98% of the time, and the rest is random: there is a 50% chance you win and 50% chance you lose. Switching doesn't make sense. You might as well switch just because you don't want to practice the wrong way in case you play the real game (below).

Normal game:
[*]You have a 1% chance of picking the correct cup.
[*]There is a 100% chance that the 98 opened cups do NOT have the prize. The host knows these cups are empty and chooses them on purpose.
[*]There is a 99% chance that the last remaining cup has the prize.

Only a person who doesn't understand the game would keep the origional pick. You should always switch in this case. 99% vs 1%, I'll take the 99%.

 

[OS]

Originally posted by: crownjules

Originally posted by: OS
taken from the wiki article again;

"Asked whether the contestant should switch, vos Savant correctly replied, "If the host is clueless, it makes no difference whether you stay or switch. If he knows, switch" (vos Savant, 2006)."

So the OP scenario as posed, implies both players do not know what is in any of the cups, in which case, even the wiki article says it is a wash at the last two.

The different scenario later put in is if the other player does in fact know what is in each.

Did you not read the example I posted at the top of page 2? Let me do it again because you obviously didn't.

Assume that Cup 1 contains the object.

SCENARIO 1: You pick Cup 1. Cup 2 or 3 is shown. If you KEEP the cup, you WIN. If you SWITCH the cup, you LOSE.

SCENARIO 2: You pick Cup 2. Cup 3 is shown to you. If you KEEP the cup, you LOSE. If you SWITCH, you WIN.

SCENARIO 3: You pick Cup 3. Cup 2 is show to you. If you KEEP the cup, you LOSE. If you SWITCH, you WIN.

In 2 of the 3 scenarios, you win when switching and lose by keeping. That's why it's better to switch.

The key difference in your scenario is that the cup shown to you is always empty. That implies the person picking the cup knows what is in all the cups. That is a stacked scenario and not the same as the original one.

 

[dullard]

Originally posted by: Rob9874
They say that probability is not affected by previous attempts. For example, when you flip a coin, you have a 50/50 chance of it being heads every time you flip it, regardless of what happened before. However, aren't the odds of getting heads twice in a row less than 50%? And aren't the odds of flipping heads 3 times in a row less than that? So if I flipped a coin 5 times, and got heads 5 times, wouldn't it be smart to bet that the next flip will be tails? I understand that isolated flip has a 50/50 chance of being heads again, but the series of 6 flips has very low odds of being 6 heads in a row, and if the first 5 were heads, the 6th will probably be tails. Or am I stupid?

The odds of getting 6 heads in a row is small. That is true. But you ALREADY did the hard part of getting 5 in a row. That 6th is still a 50/50 chance. That is, you already did the rare and difficult feat of getting 5 in a row and you don't need to consider that part again. The last flip is always 50/50 in a random coin toss.

Here is another way of saying the same thing. Suppose you bought a lottery ticket and won, is the chance of your next coin flip being heads 1 in 10,000,000,000 just because you were lucky last time? Nope, being lucky in the past doesn't affect the future.