Question on probability...

[LS20]

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)."

man, that B is smart

 

[blackllotus]

Heres the way I think about it. We know that the first cup we pick has a 1/3 chance of being the correct one and we know that if the first cup is wrong then the other cup must be correct so the other cup must have a 2/3 chance of being correct.

 

[OS]

Originally posted by: LS20

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)."

man, that B is smart

seriously :Q

 

[Matthias99]

Originally posted by: OS
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.

What the OP said:

You have three cups, one has a dollar bill underneath it. You choose a cup that you think has a dollar bill. The other person opens up a cup and shows there is no money, leaving two cups. What do you do?

By saying "the other person opens up a cup and shows there is no money" implies that they must know where the money is and are deliberately removing the one that is the "wrong" choice. That is always how I have seen the 'Monty Hall' problem presented.

If they were picking at random -- what if they picked the cup with the money in it? It doesn't even specify what would happen. If you assume that case is a tie, then yes, switching is irrelevant. You don't know more about the uncovered cups than you did before. If they deliberately do not select the cup with the money (that is, they always reveal an empty cup that is not the one you chose), you should switch, since they are adding information that you can use.

 

[OS]

ugh, the title stated "question on probability..."

I was considering if the first cup was randomly selected and just happened to be empty.

If the other player already knows what is in the cups, this is no longer a true probability scenario.

 

[ElFenix]

Originally posted by: LegendKiller
Ok, a co-worker and I are discussing the following scenario

You have three cups, one has a dollar bill underneath it. You choose a cup that you think has a dollar bill. The other person opens up a cup and shows there is no money, leaving two cups. What do you do?

A. Choose another cup?

B. keep your same cup?

I maintain that you keep your cup, since the current discreet probability of you choosing the correct cup is 50/50. However, he says that you should switch cups, since the cup you didn't choose has a higher probability of being correct (50%, vs your original 33%).

Sounds like a dumb question, which we both know has several facets to it regarding probably, I understand all facets, but I still maintain that you have a 50/50 chance of choosing at that point.

see sig

 

[mugs]

You know what would complete this thread? MasonLuke trying to prove that the solution to the Monty Hall problem is incorrect.

MasonLuke vs Marilyn vos Savant... FIGHT!

 

[Matthias99]

Originally posted by: OS
ugh, the title stated "question on probability..."

I was considering if the first cup was randomly selected and just happened to be empty.

If the other player already knows what is in the cups, this is no longer a true probability scenario.

It is absolutely still a probabilistic scenario from the 'player's' perspective, just not the 'house'.

What do you mean "if the first cup" was selected? The cup the player had picked? Now you're playing a totally different game. If they reveal the cup you had already selected and it's empty, obviously you have to switch...

Looking at the OP again, it is slightly vague. The usual presentation is:

You are presented with three (cups/doors/whatever). Behind one is a reward, and behind the other two is nothing. You pick one. The 'house' reveals the contents of one of the (cups/doors/whatever) you did not pick, and it is always empty (that is, the 'house' knows where the prize is and deliberately removes a 'wrong' choice).

In this case, you should always switch.

If they open one of the other two (cups/doors/whatever) at random (and the game restarts if they reveal the prize, I guess), it doesn't matter if you switch. If they pick at random and are allowed to pick your (cup/door/whatever), you should obviously switch if the (cup/door/whatever) you had originally picked is shown to have nothing, but otherwise it doesn't matter.

 

[OS]

no, the first cup the opposing player picked, if just happened to be empty.

 

[IntrinsicValue]

As stated, if the host knows where the dollar bill is, it is best to switch.

Easiest way to think of this is : at the beginning, what are your odds of picking an EMPTY cup? 2/3, of course. So if you have a higher chance of picking an empty cup at the start, then why wouldn't you switch once one of the other empty cups were taken away? Chances are, you have the other empty cup, so you'd make the switch and be happy you did so 2/3 of the time.

Since we're on the topic of probablity, here's another one -- probably easier, but still interesting to hear people's thoughts on it:

What if you put ten pieces of paper (numbered 1 through 10) into a hat. Your goal is to guess what number is going to come out of the hat next. You are to take a guess, pull a number from the hat, and tally it if you guessed correctly. You then must put the number back in the hat and do the process again -- repeating until you've done it 100 times.

Is it better to guess the same number over and over? Or would you randomly choose a number to guess each time?

 

[JS80]

switch. Monty Hall puzzle.

 

[chuckywang]

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.

Small program you say?
http://www.cut-the-knot.org/hall.shtml

That website should be required reading for anybody who wants to post in this thread.

 

[Kev]

This thread makes me want to punch a baby.

 

[Schfifty Five]

WOOHOO...another one of these threads....reading material =P

EVERYONE WHO THINKS IT IS 50/50 IS AN IDIOT, PLAIN AND SIMPLE hahaha...j/k, but they are wrong nonetheless.

It goes from 33% chance of winning (not switching) -> 66% winning (switching)

Good luck to those who are trying to convince other people haha.