Probabilities Question

[Gerry]

This came from a discussion about the Monty Hall problem (more later). Someone postulated that in "Who Want to Be a Millionaire" if you were picking an answer at random and then chose to use your 50/50 IF your original "guess" was still there it was statistically more likely you'd be right if you stuck with your guess rather than switch the one other remaining answer.

This then got expanded to a slightly different example. Imagine you have 101 paper cups, numbered 1 to 101. I place a coin under one of the cups at random. You then randomly pick which cup the coin is under. I then remove 99 cups which do not have a coin. IF your original guess is still there (which 99% of the time it wouldn't be) this person is claiming that it is 99% likely that the coin is under the cup he chose rather than the other remaining cup.

His words:

"Think about it, my cup remain when either I'm right or you've removed the other 99 cups and not mine. Therefore it is far more likely that I have got the correct cup than it is that you have removed all the other cups bar mine. If I pick a cup at random and you remove all the wrong answers but one at random then probability says I've picked the pound.

It would be a long game though with 101 cups as 99 times in 100 my cup disappears. "

Now I'm claiming it's still a 50/50 probability - indeed I'm sure it is. I've even modelled the Millionaire scenario on a PC and proven it's still a 50% chance either way. Anyone good at probabilities and more importantly explaining the way they work like to take a stab at explaining why he's wrong (or me, if that the case).

Incidentally: The Monty Hall problem is this. You're on a game show. You're presented with 3 doors - 2 of which have a goat behind them, 1 of which has a car. You pick a door at random. The presenter will then always open one of the other doors and reveal a goat. You then get the option to stick with your original pick or switch to the other unopened door. Which option gives you the best chance of winning the car?

 

[Sxotty]

At that moment in time of course it is 50-50 since there are two cups left and one has a coin under it.


Just tell him you are starting afresh at that moment. Which is what you just said.

 

[Simon F]

This then got expanded to a slightly different example. Imagine you have 101 paper cups, numbered 1 to 101. I place a coin under one of the cups at random. You then randomly pick which cup the coin is under. I then remove 99 cups which do not have a coin. IF your original guess is still there (which 99% of the time it wouldn't be) this person is claiming that it is 99% likely that the coin is under the cup he chose rather than the other remaining cup.

Surely the way you should formulate this is "should you change your choice"? Given the Monty Hall problem, the answer is a resounding "yes".

Spoiler

A) You correctly choice the cup with the coin: Probability: 1/101.
B) You choice an empty cup: Probability 100/101.

If you were to always stick with your original choice, your chance of winning is 1/101.

If you were to always swap, then your chance of winning is 100/101

For WWTBAM, it should be irrelevant. The "computer" is always meant to randomly remove two wrong answers irrespective of your choice.

 

[Zaphod]

I don't think so. The probability transference in the Monty Hall problem happens because it's assumed that Monty knows which door holds the price and will never pick that one to open. There is no such stipulation in this game. It's just a description of a situation that'll rarely occur in a fair game.

 

[Cheezdoodles]

Incidentally: The Monty Hall problem is this. You're on a game show. You're presented with 3 doors - 2 of which have a goat behind them, 1 of which has a car. You pick a door at random. The presenter will then always open one of the other doors and reveal a goat. You then get the option to stick with your original pick or switch to the other unopened door. Which option gives you the best chance of winning the car?

---
I remember this from statistics in college.

Spoiler

Switching the door allways yields the best chance of winning the car (2\3 vs 1\3). Lets name the doors A B C. Lets say we pick door A.

All doors have 1\3 chance of having a car. Thus, doors B and C have a 2\3 probability of having the car.

Lets say the goat is in C. Since B + C had a 2\3 chance of being correct, (and we know that C doesn't have a goat), the likelyhood of B having the car is 2\3 (because C is 0%).

Thus, you should allways switch in such a situation as the probability of being correct is 66% if you switch.

The same should apply the your "Cups" example, switching would yield 100\101 chance rather than 1\101. Same applies because we know that the 99 cups that are removed are wrong.

So safe to say that the friend your discussing with is wrong..

 

[Gerry]

Surely the way you should formulate this is "should you change your choice"? Given the Monty Hall problem, the answer is a resounding "yes".

Spoiler

A) You correctly choice the cup with the coin: Probability: 1/101.
B) You choice an empty cup: Probability 100/101.

If you were to always stick with your original choice, your chance of winning is 1/101.

If you were to always swap, then your chance of winning is 100/101

For WWTBAM, it should be irrelevant. The "computer" is always meant to randomly remove two wrong answers irrespective of your choice.

Well you can certainly formulate the question that way, which is indeed what the Monty Hall problem is asking.

The cup game is the same as WWTBAM. I'm removing 99 cups without any knowledge of what your selected cup is. If it helps, just think of it as 4 cups, and me removing two randomly. I only upped the cups to make a point to the original person I was arguing with!

 

[Simon F]

Makes me wonder why I bothered with the spoiler tag

 

[Cheezdoodles]

Happy now?

 

[Simon F]

The cup game is the same as WWTBAM. I'm removing 99 cups without any knowledge of what your selected cup is.

You didn't make that very clear! You said "Monty Hall" and "I then remove 99 cups which do not have a coin" so I assumed you weren't removing the selected cup.

 

[Gerry]

I don't think so. The probability transference in the Monty Hall problem happens because it's assumed that Monty knows which door holds the price and will never pick that one to open. There is no such stipulation in this game. It's just a description of a situation that'll rarely occur in a fair game.

Exactly. The two problems aren't the same. Monty cannot open your original door whether there's a goat behind it or not. He can only open one of other doors. The cup example is different because I don't know what your choice is, so I'll remove your original pick most of the time. It's only when I happen not to that the question of probabilities kicks in.

It's funny that two people are arguing that it's better to switch, when the original person I'm arguing with is saying it's better to not switch.

 

[Simon F]

Happy now?

Deliriously so.

 

[Gerry]

You didn't make that very clear! You said "Monty Hall" and "I then remove 99 cups which do not have a coin" so I assumed you weren't removing the selected cup.

Sorry. Thought the bit about "99% of the time the original guess would be removed" made the clear. Obviously not! Yes, we're talking about the 1% of the time your cup remains on the table, so you're left with a choice between your original cup and one other.

I probably shouldn't have mentioned Monty Hall, but that's where the discussion original started. Then somebody chucked in the WWTBAM example as another scenario.

 

[Simon F]

Sorry. Thought the bit about "99% of the time the original guess would be removed" made the clear.

Sorry ... brain filtered that bit out.

 

[Sxotty]

Well I guess we established it is 50-50 so switching or not is irrelevant unless you have x-ray vision and made your decision based on something other than chance.

Perhaps the original argument is b/c of the confusion that occurred here.

 

[Gerry]

Well I guess we established it is 50-50 so switching or not is irrelevant unless you have x-ray vision and made your decision based on something other than chance.

Perhaps the original argument is b/c of the confusion that occurred here.

Unfortunately not. I've double/triple checked that we weren't just getting confused over the semantics of the "problem" - as you can see from the quote in the original post.

And indeed - I think we've decided it remain a 50/50 shot. The issue is how to explain it to him so that he understands. Perhaps the best solution would be for me to give up trying.

 

[Cheezdoodles]

Wait, you allways pick 99 cups right? And all of them are "wrong" correct?

I dont see how his cup getting throught that elimination (which is purely by chance) gains him any higher probability of having the right one, rather than the other non selected cup that is left out.

What he is trying to is argue some retard math based on the probability of the cup being removed. Which is wrong, the other cup left is just as likely to be there because it has the coin, as cup X has.

 

[arjan de lumens]

Unfortunately not. I've double/triple checked that we weren't just getting confused over the semantics of the "problem" - as you can see from the quote in the original post.

And indeed - I think we've decided it remain a 50/50 shot. The issue is how to explain it to him so that he understands. Perhaps the best solution would be for me to give up trying.

Try a somewhat different approach: you have 101 cups. You place a coin under one cup, and a dung beetle under another one. The remaining 99 cups are left empty.

Now, let your friend choose a cup. You then remove the 99 cups that have neither coins nor dung beetles under them. Now, if your friend chose a cup that wasn't removed, what is the probability that he will get the coin? And just as importantly: what is the probability that he will be facing the dung beetle?

This maps to the WWTBAM 50/50 situation as follows:

• The cup with the coin under it corresponds to the correct option
• The empty cups correspond to the options that the computer will take away from you.
• The cup with the dung beetle corresponds to the one wrong answer that the computer will present to you.

Not sure how it can be made any clearer than that.

 

[Mintmaster]

And indeed - I think we've decided it remain a 50/50 shot. The issue is how to explain it to him so that he understands. Perhaps the best solution would be for me to give up trying.

Think of it this way: The cups chosen to be removed are independent of your guess. Thus you can classify it the other way around.

99/101 chance that your guess is removed and incorrect.
1/101 chance that your guess is not removed and is correct.
1/101 chance that your guess is not removed and is incorrect.

Now, for WWTBAM, it depends on whether you're taking a random guess or are leaning towards a choice for a particular reason.

Say you had 100 such questions where you had a feeling about which was correct, and after ranking your choices, you had a distribution of 35-30-20-15 (in terms of which choice was correct). If your original choice was still there after removing two wrong ones, then sticking with your choice gives you percentages ranging from 54% to 70%.

Of course, the choices could be designed to trick a typical person's intuition, so that distribution could be entirely different and you actually wind up with a less than 50% chance.

 

[Fred]

When I was in high school the class was given the Monty Hall problem to solve. I got the first version right (always swap, b/c you double your odds).

But.. There is a subtlety and a twist, and I got the following version wrong.

You pick a door, and then the gameshow host randomly picks a door to open (not knowing whats behind it).

If the prize is in the door he opens, the game is annuled, and a new game starts where you repeat the same procedure.

You do this until such time as the door the Gameshow host opens is empty.

Now do you switch or do you stay and whats the probability that you win?

 

[Mintmaster]

When I was in high school the class was given the Monty Hall problem to solve. I got the first version right (always swap, b/c you double your odds).

I wasn't paying attention to the Monty Hall part of the OP, but it's easy to see that Simon F is right in dismissing the analogy. In WWTMAM and the cup problem, there's no guarantee that the eliminated options are different from your initial choice, so you don't get any new information about the relative probabilities between that and the remaining choice.

When I read the first version of the MH problem on Wikipedia, I was perplexed because I thought the host would pick the door with the car. I guess that would be one lame host, though...

But.. There is a subtlety and a twist, and I got the following version wrong.

You pick a door, and then the gameshow host randomly picks a door to open (not knowing whats behind it).

If the prize is in the door he opens, the game is annuled, and a new game starts where you repeat the same procedure.

You do this until such time as the door the Gameshow host opens is empty.

Now do you switch or do you stay and whats the probability that you win?

Okay, here's my answer:

Spoiler

Switching doesn't help, and your odds of getting the prize are 50%. For each round, you have a 1/3 chance of picking the correct door (host will end the game for sure), (2/3)*(1/2) chance of winding up with the wrong door and ending the game, and (2/3)*(1/2) chance of restarting. Therefore the chance of being right is the same that of being wrong, so switching does nothing.

I like the variation though, because in the end you still do know that the host's door is empty whenever your choice matters, so it looks similar. However, the restarting changes the relative weighting of the probability tree.