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### Topic: Urgent help needed with bar puzzle (Read 14262 times)previous topic - next topic

##### Aug 12, 2013, 10:52 am
I am in a bar. The person I am with turned over three ashtrays and hid a gold krugerrand under one of them.

He knows which ashtray the coin is under, I do not.

He told me to choose one of the ashtrays and that if I picked the one which the coin was under it I could keep the coin.

I picked an astray but he did not turn it over. Instead he turned one of the other ashtrays over and showed there was no coin under it.

So the situation is that there are now two ashtrays on the table one with a krugerrand hidden under it and the other without.

He is now going to turn over the ashtray I selected but before doing so he has given me a choice; I can stick with the the ashtray I picked originally or I can change to the other one.

A 1oz gold krugerrand must be worth a fair bit.
To have the best chance of winning it what should I do?

• a) Stick with my original choice.

• b) Change my choice and choose the other ashtray.

• c) It does not make any difference which I choose.

#1
##### Aug 12, 2013, 11:07 am

Given the information presented, it makes no difference.

#2
##### Aug 12, 2013, 12:01 pm
Quote
Given the information presented, it makes no difference.

Are you sure about that? A friend is urging me to change saying I have a much better chance of winning if I do so.

#### mmcp42

#3
##### Aug 12, 2013, 12:41 pm
pick the other one
it's a classic "3 doors" problem
you can prove that your odds are increased if you choose the other
there are only 10 types of people
them that understands binary
and them that doesn't

#4
##### Aug 12, 2013, 12:55 pm
I have one person telling me it makes no difference and one person telling me to change - does anybody else have a view?

My friend tells me that because we started with three ashtrays that makes all the difference. But looking at the two ashtrays on the table it certainly feels like a 50:50 choice to me.

#### AWOL

#5
##### Aug 12, 2013, 02:11 pmLast Edit: Aug 12, 2013, 02:13 pm by AWOL Reason: 1
Quote
I have one person telling me it makes no difference and one person telling me to change

It's the classic "lying philosophers" puzzle.

I think the solution is something like asking either of them "If I asked the other guy, what would he say?" and then doing the opposite.
Or is that the "Prisoner's dilemma" ?
"Pete, it's a fool looks for logic in the chambers of the human heart." Ulysses Everett McGill.
Do not send technical questions via personal messaging - they will be ignored.
I speak for myself, not Arduino.

#6
##### Aug 12, 2013, 02:30 pm
Quote
It's the classic "lying philosophers" puzzle.

No that is not it. The person who put the krugerrand under the ashtrays knows which one its under and therfore whether or not its under my first choice. However he is not trying to psych me. Irrespective of whether or not the coin is under my choice he was always going to ask me if I wanted to change.

Although there are now only 2 ashtrays my friend says my odds of getting the valuable coin are much better than 50:50 if I swap from my original choice and chose the other one.

What is my best chance of getting the coin;
a) Stick with my original choice.
b) Change my choice and choose the other ashtray.
c) It does not make any difference which I choose.

#### AWOL

#7
##### Aug 12, 2013, 02:43 pm
Quote
Quote
It's the classic "lying philosophers" puzzle.

No that is not it

I was referring to the diverse punditry, no the original postt.
"Pete, it's a fool looks for logic in the chambers of the human heart." Ulysses Everett McGill.
Do not send technical questions via personal messaging - they will be ignored.
I speak for myself, not Arduino.

#### mmcp42

#8
##### Aug 12, 2013, 03:28 pm
here's the long form of my earlier answer: http://en.wikipedia.org/wiki/Monty_Hall_problem
there are only 10 types of people
them that understands binary
and them that doesn't

#9
##### Aug 12, 2013, 03:55 pm
@mmcp42 I was hoping for a bar brawl  ]

The answer is very counter intuitive; There before me sits two ashtrays, one of which had a gold krugerrand beneath it. It feels like the odds are the same, 50:50, as to which ashtray the coin is under.

In fact by changing my choice I double the chance of getting the coin.

The simplest way I can describe this is that initially, when there were 3 ashtays, the odds were equal 1:3 that the coin was under any particular ashtray.

I made a random choice, so there was a 1:3 chance the coin was under the ashtray I picked.

After I made my choice one of the other ashtrays was turned over to show it was empty. Doing this did not change the odds of the coin being under my ashtray. However it did change the odds of it being under the remaining unturned ashtray which then became 2:3.

As a result by changing my mind and chosing the other ashtray I will double my chance of getting the coin.

#### Palliser

#10
##### Aug 12, 2013, 04:14 pm

Barmans are like jugglers (no offense). If the barman is the person with you, I bet you the coin is in his pocket for some time now. But better, let's honor a bit Schrödinger (today's his 126th birthday). Let's start saying that the key factor here is your choice. That means you odds are not 50:50 but better. Let me be more specific. At the beginning you had 3:1 (33.33% chance to win). Now you have 83.33% (33.33 + 50). Thus, stick with your original choice and you will not only win the coin but also the cat!

Regards.

#11
##### Aug 12, 2013, 04:34 pm
Quote
But better, let's honor a bit Schrödinger (today's his 126th birthday).

So that is why my friend kept going on about Schrödinger, I though he had just had too much to drink! He was saying that, at the start, the krugerand was in a quantun superposition existing under all the ashtrays. When I chose my ashtry that caused the wave function to collapse fixing the probabilities.

I am not sure about that explanation but he was right that the odds are definitely 1:3 if I stick with my first choice and 2:3 if I change.

#12
##### Aug 12, 2013, 05:04 pm
Schrödinger, Schmödinger, I don't get all the arguments but the 1:3 then 2:3 bit sounds a little like the old 1=0 "proofs" that used to be around.

What's gone is gone and I don't see how it makes any difference after the first choice has been made. Sure overall you got two chances but now it's 50/50. Why not include the ten times you played the game before and say you have a 29:30 (or whatever it is) chance.

_____
Rob
Rob Gray aka the GRAYnomad www.robgray.com

#13
##### Aug 12, 2013, 05:29 pm
Quote
Sure overall you got two chances but now it's 50/50. Why not include the ten times you played the game before and say you have a 29:30 (or whatever it is) chance.

@Graynomad; its good to see you are going to put up a fight, but the odds are not 50:50 they are 1:3 if I don't change and 2:3 if I do.

Maybe things will be clearer if we use a million ashtrays, or perhaps not  ]

The coin is hidden under one of a million ashtrays. I choose one. The barman turns over 999,998 ashtrays showing that they are all empty, and leaving just the ashtray I chose plus one other.

When offered the chance to change my choice should I take it?
Of course I should as I have a tiny 1:1,000,000 chance with my original choice but 999,999:1,000,000 (almost a dead cert) if I change.

#### flyboy

#14
##### Aug 12, 2013, 06:28 pm
You're talking about probabilities only.  Playing the numbers, yes, your probability is much better after all of the other ashtrays have been taken away.  It doesn't matter who else knows where the coin is.  This is just added to try to embellish the story.  If the person choosing doesn't know the location of the coin, the odds are still 50:50.  Abandoning the first ashtray and selecting the other does nothing to the odds.  They're still 50:50.

After removing the previous number of ashtrays, the new conditions are these: you have 2 ashtrays, there is a coin under one of the ashtrays.  You have one ashtray under your hand and the other is not.  The location of the ashtrays has no bearing on the odds.  They are still 50:50.  The current odds are not dependent on previous odds.

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