Find the Coin puzzle

[obs20]

Here's my try.

Spoiler:

This is a very bad bet to begin with. The payoff is even money for an event that only has a 1/3 chance of occurring. Rule of thumb we only increase the wager if the odds become better than even money.
In the first instance one of the cups was over turned which were of the host's choosing this shouldn't change the odds at all. The host had insider knowledge. much. The odds are still 1/3 that we made the right choice originally but now faced with a new choice of staying or switching it is still a choice of 1/2.
However if one were to overturned by chance and the odds of that the cup on the left be overturned and that it is empty now the chance of 2 cups being overturned and both being empty is 1/6( 1/3 * 1/2) so now the odds are 5/6/2 that I've picked the right one cup. The odds are now 5:12 . Hold the bet.

[pdurrant]

Originally Posted by obs20

Here's my try.

Spoiler:

This is a very bad bet to begin with. The payoff is even money for an event that only has a 1/3 chance of occurring. Rule of thumb we only increase the wager if the odds become better than even money.
In the first instance one of the cups was over turned which were of the host's choosing this shouldn't change the odds at all. The host had insider knowledge. much. The odds are still 1/3 that we made the right choice originally but now faced with a new choice of staying or switching it is still a choice of 1/2.
However if one were to overturned by chance and the odds of that the cup on the left be overturned and that it is empty now the chance of 2 cups being overturned and both being empty is 1/6( 1/3 * 1/2) so now the odds are 5/6/2 that I've picked the right one cup. The odds are now 5:12 . Hold the bet.

Your reasoning is correct up until the second half of the fourth sentence of your second paragraph.

Spoiler:

Right:
The odds are still 1/3 that we made the right choice originally

Wrong:
but now faced with a new choice of staying or switching it is still a choice of 1/2

[pdurrant]

OK, answers time. I've put them in spoiler tags for anyone late to the puzzle.
If you're still confused after readin these spoilers, see the wikipedia article on the Monty Hall Problem, which is very similar.

Originally Posted by pdurrant

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?

Spoiler:

On average, you're better off doubling your bet and switching. That turns a 1/3 coin loss per game into break even.

Here's why.

You have a one in three chance of picking the right cylinder initially. The showman can always up a cylinder that doesn't have a coin under it. Because he can always do it, it doesn't affect your odds. After he's lifted a cylinder, your cylinder still has the one in three chance of being right. The other remaining cylinder has a two in three chance of being right.

Another way to look at it is this. You've picked a cylinder. The showman offers you the chance to switch to choosing BOTH of the other cylinders instead. This is obviously switching from a 1 in 3 to a 2 in three chance. Whether he shows you before or after switching that one of the other two cylinders doesn't hide the coin is irrelevant - you already know that at least one of the cylinders doesn't hide the coin.

So if you stick with your original cylinder, you'll win one time in three. So for every three coins you bet, you can expect to get two coins back - losing an average of 1/3 coin each go.

But if you switch you can expect to win two times out of three. But you've had to double your stake, while you're still only going to win one coin. But this is still better - over three games you'll bet six coins, but you'll also win back six coins (your two coins and the one coin from the showman, twice). So on average you break even - much better than losing 1/3 of a coin each go.

Originally Posted by pdurrant

Question 2: Should you agree to continue with this game after this accident?

It's hard to answer this question until you've answered Question 3 - I really should have put it last, not here. Sorry!

The answer is,

Spoiler:

it doesn't matter, although your strategy must change, see Q3.

Originally Posted by pdurrant

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?

Spoiler:

It seems at first sight that you're in the same situation as in Question 1 - that you should double your bet and switch to the other cylinder. Actually, if you do this, on average you'll lose 1/2 coin per game, while if you stick with your original cylinder, you'll break even.

What's the difference? After all, in both cases, a cylinder that you didn't choose has been shown to not have the coin under it.

The difference is that in this case, there was not a 100% chance that the cylinder would not reveal the coin.

Let's look at what could happen when a random cylinder is knocked over.

If it's the cylinder you bet against, the game is void. Either you or the showman would not want to continue in that circumstance, although if it showed the coin you might argue that you weren't going to switch....

But let's consider what happens when the random cylinder knocked over isn't the one next to your coin, since that's the case we're looking at. There are three cases to consider:

1. If you have bet against the cylinder hiding the coin (1/3 chance), the chance that the cylinder knocked over isn't hiding the coin is 100% (1/1). Total chance of this happening: 1/3
   
2. If you have bet against a cylinder that isn't hiding the coin (2/3 chance), the chance that the cylinder knocked over is hiding the coin is 50% (1/2). Total chance of this happening: 2/3 * 1/2 = 1/3
   
3. If you have bet against a cylinder that isn't hiding the coin (2/3 chance), the chance that the cylinder knocked over also isn't hiding the coin is 50% (1/2). Total chance of this happening: 2/3 * 1/2 = 1/3

We know that case 2 hasn't happened. The two remaining possibilities are equally likely. So in this case, the chance that the other remaining cylinder is hiding the coin is 1/2, not 2/3.

Sticking with your cylinder will win one time in two. Swapping will win one time in two. But since you have to double your stake to swap, swapping will, oover two average games, cost four coins but only win three. Sticking where you are will cost two coins and win two coins.

So sticking breaks even, and swapping loses half a coin per game.

Another way of looking at this, to understand why the odds are different: If you've chosen the right cylinder to start with, the chance of the random other cylinder revealing no coin is much higher than if you have chosen one of the wrong cylinders. Since the random cylinder didn't reveal a coin, it's more likely than it was before that you have chosen the right cylinder to start with.

[clarknova]

Originally Posted by pdurrant

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?

This seriously depends...

Spoiler:

...on whether your goal is to get more wins, or to lose less money. You always lose money (games of chance are always to the houses advantage).

If you change your bet everytime, you will win the game more often, however you will lose more money than if you stay with the original bet.

I suck with statistics, so I just ran some monte carlos of the game with both strategies (each time the player starts with 1 coin). Here are four runs (of a million games) when sticking with the original bet:
Games: 1000000 Wins: 333402 Losses: 666598 Player Money: -333195
Games: 1000000 Wins: 333292 Losses: 666708 Player Money: -333415
Games: 1000000 Wins: 333540 Losses: 666460 Player Money: -332919
Games: 1000000 Wins: 333916 Losses: 666084 Player Money: -332167

And here are four runs when changing the bet (doubling your wager, but not doubling the return).
Games: 1000000 Wins: 499503 Losses: 500497 Player Money: -501490
Games: 1000000 Wins: 499822 Losses: 500178 Player Money: -500533
Games: 1000000 Wins: 500125 Losses: 499875 Player Money: -499624
Games: 1000000 Wins: 499347 Losses: 500653 Player Money: -501958

While changing your bet allows you to win around 50 percent of the time, your losses increase. If the game was a real double or nothing (where the dealer also doubled his bet when you doubled yours) then the strategy of always changing your hand would be break even.

In other words: The only winning move is not to play. How about a nice game of chess?

[pdurrant]

Originally Posted by clarknova

This seriously depends...

Spoiler:

...on whether your goal is to get more wins, or to lose less money. You always lose money (games of chance are always to the houses advantage).

If you change your bet everytime, you will win the game more often, however you will lose more money than if you stay with the original bet.

I suck with statistics, so I just ran some monte carlos of the game with both strategies (each time the player starts with 1 coin). Here are four runs (of a million games) when sticking with the original bet:
Games: 1000000 Wins: 333402 Losses: 666598 Player Money: -333195
Games: 1000000 Wins: 333292 Losses: 666708 Player Money: -333415
Games: 1000000 Wins: 333540 Losses: 666460 Player Money: -332919
Games: 1000000 Wins: 333916 Losses: 666084 Player Money: -332167

And here are four runs when changing the bet (doubling your wager, but not doubling the return).
Games: 1000000 Wins: 499503 Losses: 500497 Player Money: -501490
Games: 1000000 Wins: 499822 Losses: 500178 Player Money: -500533
Games: 1000000 Wins: 500125 Losses: 499875 Player Money: -499624
Games: 1000000 Wins: 499347 Losses: 500653 Player Money: -501958

While changing your bet allows you to win around 50 percent of the time, your losses increase. If the game was a real double or nothing (where the dealer also doubled his bet when you doubled yours) then the strategy of always changing your hand would be break even.

In other words: The only winning move is not to play. How about a nice game of chess?

Sorry, you're not right. The chances of winning by sticking to your original choice and winning by switching to the other remaining cylinder must sum to 1. You have it adding up to 5/6ths.

[clarknova]

Originally Posted by pdurrant

Sorry, you're not right. The chances of winning by sticking to your original choice and winning by switching to the other remaining cylinder must sum to 1. You have it adding up to 5/6ths.

Damnit! I forgot I was programming in perl instead of C, and I used "break" instead of "last" to exit a for loop. It skewed my game results.

Oh well. I fixed it, and it does break even now.

#!/usr/bin/perl

$win=$lose=0;
$money=1;

for ($i=1; $i<=10;$i++) {
  @shell = (0, 1, 2);
  $deal = int(rand(3));    # The shell with the coin
  splice(@shell,$deal,1);  # The empty shells.
  $money -= 1;              # Player's wager.
  $pick = int(rand(3));     # Player's shell choice.
  if (int(rand(2))) {         # Remove an empty shell that the
   $rem=$shell[1];           # ..player hasn't chosen.
   if ($pick == $rem) {
     $rem=$shell[0];
   }
  } else {
   $rem=$shell[0];
   if ($pick == $rem) {
     $rem=$shell[1];
   }
  }
  print "Dealer gets $deal.  Player picks $pick.  Dealer removes $rem.\n";
  for ($j=0;$j<3;$j++) {    # Change the pick to the other remaing shell.
    next if ($pick == $j or $rem == $j);
    $pick = $j;
    last;                           # DAMNIT!  I used "break;" instead of last!
  }
  $money -= 1;                # Double your bet.
  print "Player changes to $pick.\n";
  if ($pick == $deal) {
    $money += 3;               # Winner! Get your 3 coins!
    $win++;
    print "Player Wins.\n";
  } else {
    $lose++;                      # Loser!
    print "Player Loses.\n";
  }
  print "\nGames: $i  Wins: $win  Losses: $lose  Player Money: $money\n\n";
}

Syntax errors will be the death of me. BTW - Thanks, perl for not catching such a basic error...

[HarryT]

Originally Posted by clarknova

In other words: The only winning move is not to play. How about a nice game of chess?

There is no winning move, but if you play correctly, you will break even. See my original analysis for details.

[obs20]

Wait a minute! Have you ever played 3 card monte?
They will always turn over an empty cup and try to suck you into a larger bet. This is not charity, once your money is on the table the odds are 3 to 1 They know that at this point even though the odds may seem to be even money the overall odds still remain at 3 to 1 because you put your money on the table in the first place.
Let's look at it this way they get your money on the table and pay 2 to 1 on a 3 to 1 chance. They have to make money. If you increase your stake then it's just good money going after bad.

[pdurrant]

Originally Posted by obs20

Wait a minute! Have you ever played 3 card monte?
They will always turn over an empty cup and try to suck you into a larger bet. This is not charity, once your money is on the table the odds are 3 to 1 They know that at this point even though the odds may seem to be even money the overall odds still remain at 3 to 1 because you put your money on the table in the first place.
Let's look at it this way they get your money on the table and pay 2 to 1 on a 3 to 1 chance. They have to make money. If you increase your stake then it's just good money going after bad.

If you ever really play it, you'll lose money, because the people running the game will be cheating.

But in this honest version, your analysis is faulty.

Spoiler:

In the first scenario, doubling your stake and switching turns it from a losing game into a break-even game. Really.

[HarryT]

Originally Posted by obs20

Wait a minute! Have you ever played 3 card monte?
They will always turn over an empty cup and try to suck you into a larger bet. This is not charity, once your money is on the table the odds are 3 to 1 They know that at this point even though the odds may seem to be even money the overall odds still remain at 3 to 1 because you put your money on the table in the first place.
Let's look at it this way they get your money on the table and pay 2 to 1 on a 3 to 1 chance. They have to make money. If you increase your stake then it's just good money going after bad.

Please look at the maths in my original answer. You will see that you are incorrect.

[pdurrant]

Next puzzle is up: https://www.mobileread.com/forums/showthread.php?t=88107

[obs20]

Originally Posted by HarryT

Please look at the maths in my original answer. You will see that you are incorrect.

I see it as being more like the prison puzzle where everyone is logical.
Lets suppose that the dealer and player are both logical. The dealer knows that he has a 2/3 advantage of winning and only has to pay out even money. Most of the time he will turn over the cup where the coin is and take the money.

However he would like to get some more money into the game so he turns over an empty cup and makes an offer to let the player switch cups for the price of doubling down.

Remember the dealer knows where the coin is. He knows if the player guessed right the first time or not. Now let us suppose that the player is ever so logical and will always switch his bet when this offer comes up because mathematically it's too his advantage. But the dealer has the knowledge of where the coin actually is. By the rules of this game the dealer is not cheating if he uses that knowledge.
The dealer being ever so logical will make the offer every time he knows that the player guessed right the first time.

[omk3]

I think the rule in this particular case is that the dealer will always make the offer to switch, regardless of whether you bet on the right cup or not.
But in real life I suppose that's how it works. The dealer has the inside knowledge, and therefore the advantage.

[pdurrant]

Originally Posted by omk3

I think the rule in this particular case is that the dealer will always make the offer to switch, regardless of whether you bet on the right cup or not.
But in real life I suppose that's how it works. The dealer has the inside knowledge, and therefore the advantage.

In this game, the dealer always has to reveal a non-winning cylinder and make the offer to switch.

That's why there's the bit about watching the game for a while in the question.

[obs20]

Originally Posted by pdurrant

In this game, the dealer always has to reveal a non-winning cylinder and make the offer to switch.

That's why there's the bit about watching the game for a while in the question.

The the dealer is wasting his time for only even money. Not very logical.