Find the Coin puzzle

[pdurrant]

This is an adaption of a classic puzzle, in two parts and with three questions.

You're in the Bazaar of Honest Conmen. Yes, really — here there are numerous stall where you can play the shell game in various forms, but the showmen don't cheat. It's actually possible to find the lady, pick the right shell, or, as in the stall we're going to look at, find the coin. There's no sleight of hand or trickery, the games work just as they appear to do on the surface.

You wander through looking for a game to play. Obviously, you're only going to be risking money you'd intended to spend on entertainment, but you're still keen to win if you can.

You come up to a stall, where on the table top are three identical, upside-down, tall, hollow cylinders. You watch for a while, and you see that the game is to find a pound coin hidden under one of the cylinders. The showman hides a pound coin under one of the cylinders while the punter isn't looking. The punter places a pound coin bet by one of the cylinders. The showman lifts one of the other cylinders (that he knows isn't over the hidden coin) to reveal that the coin isn't under that cylinder, and offers the punter a chance to swap to the other remaining cylinder by doubling their bet. Once the punter has decided to double and move, or to stick with the original choice, both remaining cylinders are lifted. If the punter's coins are by the correct cylinder, he keeps the coins on the table, otherwise the showman takes them.

You decide to have a go. You place your bet against the right-hand cylinder. The showman lifts the centre cylinder to reveal nothing. You now have to choose whether to double your stake and swap to the left hand cylinder, or leave your bet where it is.

Question 1: On average, are you going to be better off moving your bet and doubling it, staying put, or does it not make any difference which you do?

As it happens, this time you stay put, and the showman lifts the cylinders to reveal that the hidden coin was indeed under the right-hand cylinder. You've won. The showman sets up the game again. This time, just after you've placed your bet on the middle cylinder, someone stumbles against the table, and the left-hand cylinder falls over, revealing —— nothing. The showman offers to continue the game anyway, as usual also offering you the choice of staying put, or moving (this time to the right-hand cylinder) and doubling your stake.

Question 2: Should you agree to continue with this game after this accident? (Yes, it was really a genuine accident. It wasn't some nefarious plan by the showman and an accomplice. "Bazaar of Honest Conmen", remember?)

Question 3: If you do continue with this game, are you more likely to be better off moving your bet and doubling it, staying put, or does it not make any difference which you do?


As always, answers in spoiler tags please. And an explanation of your answers would be appreciated. After all, you've a one in eighteen chance of getting them all right just by chance!

[HarryT]

Originally Posted by pdurrant

You decide to have a go. You place your bet against the right-hand cylinder. The showman lifts the centre cylinder to reveal nothing. You now have to choose whether to double your stake and swap to the left hand cylinder, or leave your bet where it is.

Question: When the showman lifts the initial cylinder, will he always choose to lift one that he knows the coin is not under? On the assumption that the answer to this is "Yes", then I'm going to tackle question 1 as follows:

Spoiler:

Let us call the three cylinders A, B, and C. You guess "A" and have (obviously) a 1/3 probability of being right. The chance of the coin being under "B" or "C" is 2/3. Now, the showman lifts "C" to reveal nothing. This leaves us still with a 1/3 chance of A being the correct guess, but now a 2/3 chance of B being correct. So, all being equal, you should select "B".

However, you will have to double your stake to do so. Double the stake for double the chance. If you stay where you are, you have a 1/3 chance of winning £1, and a 2/3 chance of losing £1. If you change your bet, you have a 2/3 chance of winning £1, but a 1/3 chance of losing £2.

So, stay where you are and, over the long term, the result will be:

+1/3*1 - 2/3*1 = -1/3

Change your bet: +2/3*1 - 1/3*2 = 0

So I reckon you should always change your bet. The long-term results of doing so will be that you break even.

Now, for question 2:

Spoiler:

The fact that now the cylinder has been lifted at random to reveal nothing changes the odds to 50:50 between the coin being under "A" or "B".

So stick with "A" - a 50:50 chance of winning £1 or losing £1. Change to "B" and the odds become a 50:50 chance of winning £1, or losing £2. Clearly you should leave your bet where it is, and again in the long term you should break even.

I've answered Q3 in part 1.

[omk3]

Here's my take:

Spoiler:

Question 1:
We have cylinders A, B, C. Let's say A is the one with the coin.
If I bet on A, the honest conman will randomly reveal either B or C, so the possible combinations are now AB, AC
If I bet on B, the honest conman will reveal C, so we only get AB.
If I bet on C, the honest conman will reveal B, so we only get AC.
So, four possible outcomes. Out of the four, in two I have the right cylinder from the start, in two I don't. It seems like a 1/2 chance.

Question 2:
Why not?

Question 3:
Let's say again that A is the one with the coin.
I may have chosen A,B,C. Let's call Ax, Bx, Cx the cylinder I've bet on.
We get the following possible combinations:
AxBC, ABxC, ABCx
Now one cylinder is randomly knocked over. We have the following possible combinations:
AxB, AxC, BC, ABx, AC, BxC, AB, ACx, BCx
As no coin is revealed, we know that BC, BxC, BCx are out. AC and AB are also out, because the cylinder I bet on is still standing.
So we are left with AxB, AxC, ABx, ACx. We have 2/4(1/2) chance to have bet on the right cylinder, so nothing changed.

This seems a little strange to me, so I may well be wrong....

[pdurrant]

Originally Posted by HarryT

Question: When the showman lifts the initial cylinder, will he always choose to lift one that he knows the coin is not under? On the assumption that the answer to this is "Yes", then I'm going to tackle question 1 as follows:

Spoiler:

Let us call the three cylinders A, B, and C. You guess "A" and have (obviously) a 1/3 probability of being right. The chance of the coin being under "B" or "C" is 2/3. Now, the showman lifts "C" to reveal nothing. This leaves us still with a 1/3 chance of A being the correct guess, but now a 2/3 chance of B being correct. So, all being equal, you should select "B".

However, you will have to double your stake to do so. Double the stake for double the chance. If you stay where you are, you have a 1/3 chance of winning £1, and a 2/3 chance of losing £1. If you change your bet, you have a 2/3 chance of winning £1, but a 1/3 chance of losing £2.

So, stay where you are in, over the long term, the result will be:

+1/3*1 - 2/3*1 = -1/3

Change your bet: +2/3*1 - 1/3*2 = 0

So I reckon you should always change your bet. The long-term results of doing so will be that you break even.

Now, for question 2:

Spoiler:

The fact that now the cylinder has been lifted at random to reveal nothing changes the odds to 50:50 between the coin being under "A" or "B".

So stick with "A" - a 50:50 chance of winning £1 or losing £1. Change to "B" and the odds become a 50:50 chance of winning £1, or losing £2. Clearly you should leave your bet where it is, and again in the long term you should break even.

I've answered Q3 in part 1.

I've added some text to the question to make it explicit that the showman lifts a cylinder that he knows isn't covering the coin.

Both the answers you've given are right. I can't see your explicit answer to Q2, but given that you've got Q1 and Q3 right, the answer to Q2 is easy.

[HarryT]

Sorry, you're right, I've not answered Q2:

Spoiler:

Should you continue with the game? Answer: it doesn't matter - you still have an option which gives you the probability of breaking even; it's just that the option has changed from changing your bet to leaving it where it is.

[pdurrant]

Originally Posted by omk3

Here's my take:

Spoiler:

Question 1:
We have cylinders A, B, C. Let's say A is the one with the coin.
If I bet on A, the honest conman will randomly reveal either B or C, so the possible combinations are now AB, AC
If I bet on B, the honest conman will reveal C, so we only get AB.
If I bet on C, the honest conman will reveal B, so we only get AC.
So, four possible outcomes. Out of the four, in two I have the right cylinder from the start, in two I don't. It seems like a 1/2 chance.

Question 2:
Why not?

Question 3:
Let's say again that A is the one with the coin.
I may have chosen A,B,C. Let's call Ax, Bx, Cx the cylinder I've bet on.
We get the following possible combinations:
AxBC, ABxC, ABCx
Now one cylinder is randomly knocked over. We have the following possible combinations:
AxB, AxC, BC, ABx, AC, BxC, AB, ACx, BCx
As no coin is revealed, we know that BC, BxC, BCx are out. AC and AB are also out, because the cylinder I bet on is still standing.
So we are left with AxB, AxC, ABx, ACx. We have 2/4(1/2) chance to have bet on the right cylinder, so nothing changed.

This seems a little strange to me, so I may well be wrong....

Your answer to Q1 is wrong.

Hint:

Spoiler:

When you initially choose, you have a 1 in 3 chance of being right. The showman lifting a cylinder that doesn't have the coin under it doesn't change your odds, as he can lift such a cylinder every time. Splitting the first case (where you've chosen the right one initially) into two is erroneous.

[omk3]

Yeah I peaked into Harry's answer and saw what I got wrong.
Probabilities are frustratingly confusing

[HarryT]

Yes, the key is that:

Spoiler:

the showman lifting a cylinder he knows is empty doesn't alter your initial 1/3 chance of being right, so you're left with 2/3 for the other cylinder, whereas revealing whether or not a cylinder is empty at random does change it.

[pdurrant]

Originally Posted by HarryT

Sorry, you're right, I've not answered Q2:

Spoiler:

Should you continue with the game? Answer: it doesn't matter - you still have an option which gives you the probability of breaking even; it's just that the option has changed from changing your bet to leaving it where it is.

Exactly right! Congratulations to HarryT - the first with a complete set of correct answers.

I suspect you might have guessed which famous problem I adapted for this puzzle?

[pdurrant]

Originally Posted by omk3

Yeah I peaked into Harry's answer and saw what I got wrong.
Probabilities are frustratingly confusing

They are indeed. There is some similarity between this and the Tuesday's boy problem in certain aspects.

[HarryT]

Spoiler:

It's a variant of the classic "which door is the lady behind" game, isn't it? Where the showman knows in advance where the lady is, and opens doors one at a time to reveal them to be empty. At the end, the contestant is left with two doors and has the choice of either sticking with their selection, or changing it. The winning strategy is always to change: the greater the number of initial doors, the more certainly he should change! Eg, if there are 10 initial doors, there's a 1 in 10 chance of winning by sticking with the initial selection, and a 9 in 10 chance of winning by changing selection.

[pdurrant]

Originally Posted by HarryT

Spoiler:

It's a variant of the classic "which door is the lady behind" game, isn't it? Where the showman knows in advance where the lady is, and opens doors one at a time to reveal them to be empty. At the end, the contestant is left with two doors and has the choice of either sticking with their selection, or changing it. The winning strategy is always to change.

Spoiler:

Yes, also known as the "Monty Hall" puzzle.

[dreams]

Congratulations, HarryT! You are quick with these.

**putting away paper and crayons, as I try to answer.. I don't do well with probability ones. **

Q1 :

Spoiler:

Cups .. L - C - R 1/3 chance when placing bet

L Cup empty so now it's 2/3 that my choice was correct.

For making the decision to stay or change the chance is 1/2.

So 1 coin at 2/3 or 2 coins at 1/2

Some how I'm lost already, but it looks like that if I stay with the 1 coin on my original guess of 2/3, then I will end up with spending less and earning more over time?

Q2:

Spoiler:

Someone knocks a cup over and it may not be the one that the showman was going to show you was empty.

You started with a 1/3 chance, but the unexpected cover lifting changed it for the showman, but I don't think it changed it for you. You still have the 2/3 chance with one cup gone, which is now 1/2.

I would still continue and not double my bet, because of the reasoning I used in Q1.

Q3:

Spoiler:

I would still continue the game as my known odds aren't different.

I still wouldn't double my bet as my odds of coming out with more money over the long run is better by staying with the original bet.

I think this is the same reasoning I used for all the Q's...

[pdurrant]

Originally Posted by dreams

Congratulations, HarryT! You are quick with these.

**putting away paper and crayons, as I try to answer.. I don't do well with probability ones. **

Q1 :

Spoiler:

Cups .. L - C - R 1/3 chance when placing bet

L Cup empty so now it's 2/3 that my choice was correct.

For making the decision to stay or change the chance is 1/2.

So 1 coin at 2/3 or 2 coins at 1/2

Some how I'm lost already, but it looks like that if I stay with the 1 coin on my original guess of 2/3, then I will end up with spending less and earning more over time?

Q2:

Spoiler:

Someone knocks a cup over and it may not be the one that the showman was going to show you was empty.

You started with a 1/3 chance, but the unexpected cover lifting changed it for the showman, but I don't think it changed it for you. You still have the 2/3 chance with one cup gone, which is now 1/2.

I would still continue and not double my bet, because of the reasoning I used in Q1.

Q3:

Spoiler:

I would still continue the game as my known odds aren't different.

I still wouldn't double my bet as my odds of coming out with more money over the long run is better by staying with the original bet.

I think this is the same reasoning I used for all the Q's...

Sorry Dreams, not right yet.

Q1: Your first line is right, but after that you seem to go astray
Q2: Until you have Q1 right, and an idea about Q3, you can't realy hope to get Q2 right. (I probably should have put it last - sorry)
Q3: You make the right decision, but the reasoning is wrong because you don't have the calculations for Q1 or Q3 right yet.

[dreams]

Originally Posted by pdurrant

Sorry Dreams, not right yet.

Q1: Your first line is right, but after that you seem to go astray
Q2: Until you have Q1 right, and an idea about Q3, you can't realy hope to get Q2 right. (I probably should have put it last - sorry)
Q3: You make the right decision, but the reasoning is wrong because you don't have the calculations for Q1 or Q3 right yet.

Well, that's what I'm usually told when I make a right decision, but completely confuse people with how I actually do it. It drives my totally logical son crazy.

Time to get the crayons and paper out again.