Monty Hall Problem... For Newbies

[ynot]

I don't know what you're asking. Why don't you just state the full rules of the game, and I'll tell you the odds?

'If I toss two fair coins, examine both of them and truthfully tell you that at least one of them landed heads. What is the probability both coins landed heads?'

 

[Brian-M]

In scenario B the setup does not match the logic. If the selected coin was not X, but the other coin was,

There's your problem. X is set to whatever the selected coin is, I made that perfectly clear. It's impossible for the selected coin to not be X.

 

[Red Baron Farms]

'If I toss two fair coins, examine both of them and truthfully tell you that at least one of them landed heads. What is the probability both coins landed heads?'

1 in 3

 

[JohnnyG]

There's your problem. X is set to whatever the selected coin is, I made that perfectly clear. It's impossible for the selected coin to not be X.

OK, I see that I misread you case B. But there is still a problem with it. The problem statement was clear that both coins were examined. Case B only requires one coin to be examined. I understand that it is a subset of examining both coins, but to arrive at the odds of 1/2 requires the assumption of information we were not given.

I could just as easily define the game to say "the two coins are different" if only one is heads and "at least one is heads" if both are heads, making the odds 100%. But if we limit ourselves to only what is stated in the original problem I see no basis to claim the correct odds are 1/2 and not 1/3 (as the OP claimed).

 

[dlorde]

'If I toss two fair coins, examine both of them and truthfully tell you that at least one of them landed heads. What is the probability both coins landed heads?'

There are two possible scenarios here, each with different odds. The description of what's going on is ambiguous, it could apply for either situation. The answers people give going to depend on which scenario they assume is being talked about.

Let's say X represents either heads or tails.

Scenario A:

You pick a value for X at random
You flip a pair of coins until at least one coin lands X
You then say "At least one coin is X"
The odds of both coins being X is 1/3​

Scenario B:

You flip a pair of coins.
You set X as the value of an arbitrarily selected coin
You then say "At least one coin is X"
The odds of both coins being X is 1/2​

Yes. I assumed scenario A which interprets the statement as an instance of the general problem where heads or tails is specified in advance, whereas scenario B interprets it as an instance of the general problem where heads or tails is specified according to the outcome of each toss.

So yes, either 1/2 or 1/3 may be correct, depending on the interpretation.

 

[Alferd_Packer]

If I remember the format correctly, Monty would often offer "incentives" to trade or not to trade.

 

[GlennB]

Yes. I assumed scenario A which interprets the statement as an instance of the general problem where heads or tails is specified in advance, whereas scenario B interprets it as an instance of the general problem where heads or tails is specified according to the outcome of each toss.

So yes, either 1/2 or 1/3 may be correct, depending on the interpretation.

Yes, and it's taken a while for this to sink into my skull. Along the way I was able to accept both solutions because it was viewing it differently each time.

 

[dlorde]

Yes, and it's taken a while for this to sink into my skull. Along the way I was able to accept both solutions because it was viewing it differently each time.

The logic is really so simple it quickly became clear we must have been arguing at cross-purposes, but I just couldn't see the alternative interpretation.

 

[Modified]

'If I toss two fair coins, examine both of them and truthfully tell you that at least one of them landed heads. What is the probability both coins landed heads?'

How hard is this? State the full rules. What must you say if they both come up tails? Must you always say "at least one of them landed heads" if at least one of them landed heads?

 

[Stamuel]

'If I toss two fair coins, examine both of them and truthfully tell you that at least one of them landed heads. What is the probability both coins landed heads?'

The answer will depend on how likely you are to tell me that at least one landed heads, given the different possible outcomes. Without those likelihoods, no answer can be determined.



To see why, suppose I know you will only tell me that at least one landed heads if both landed heads.* Then, the probability that both are heads is 1. This is consistent with your description of the rules.



On the other hand, suppose I know you will only tell me that at least one landed heads when exactly one landed heads. Then the probability that both landed heads is 0. This is also consistent with your description of the rules. I need the likelihood of your statement in order to get a determinate answer.



The description given is analogous to describing the MH problem as, "A car is randomly placed behind one of three doors. You pick a door, then Monty opens one of the other doors, revealing a goat." Without knowing how Monty decided to reveal a goat, we are adrift.



*In other words, the likelihood you will tell me that at least one landed heads, given HH, is 1, while the likelihood you will tell me that at least one landed heads, given any other outcome, is 0.

 

[Red Baron Farms]

The description given is analogous to describing the MH problem as, "A car is randomly placed behind one of three doors. You pick a door, then Monty opens one of the other doors, revealing a goat." Without knowing how Monty decided to reveal a goat, we are adrift.

Ahem....no
Actually it is irrelevant how or why Monty revealed a goat. The simple fact that he did reveal a goat makes the odds 2/3rds to switch, 1/3rd to stay.

 

[Michael C]

Ahem....no
Actually it is irrelevant how or why Monty revealed a goat. The simple fact that he did reveal a goat makes the odds 2/3rds to switch, 1/3rd to stay.

These will be the odds if we know that Monty always reveals a goat, which is the case for the problem as originally stated. They will not be the odds, for instance, if Monty had opened one of the two remaining doors at random and had happened to reveal a goat.

 

[Meed]

The answer will depend on how likely you are to tell me that at least one landed heads, given the different possible outcomes. Without those likelihoods, no answer can be determined.

Agreed. I (personally) think if that information is not given then the natural interpretation would be to assume that the flipper would tell you at least one landed heads for any combination of two flips where that was the case. But it's always best to set these questions up as unambiguously as possible.

Ahem....no
Actually it is irrelevant how or why Monty revealed a goat. The simple fact that he did reveal a goat makes the odds 2/3rds to switch, 1/3rd to stay.

It's not irrelevant. For instance, if Monty only opens a door in cases where the initial pick was correct, then it's 1 to stick, 0 to stay. If Monty just opens a random door that isn't the initial pick (e.g. he is willing to open the door with the car behind it) and it's a goat, then it's 1/2 stick, 1/2 switch. If Monty always opens a random goat door that CAN include the initial pick, but in this instance is not the initial pick, then it's 1/2 stick 1/2 switch.

The movie 21 got it wrong in this scene:


Micky Rosa: "How do you know he's not playing a trick on you, trying to use reverse psychology to get you to pick a goat?"
Ben: "I wouldn't really care."

The Monty Hall problem assumes that Monty is not playing a trick on you. On the contrary, he is following the standard rules of the game in which he must always open a goat door that you did not pick and you, the strategist, are well aware of this rule.

 

[Red Baron Farms]

These will be the odds if we know that Monty always reveals a goat, which is the case for the problem as originally stated. They will not be the odds, for instance, if Monty had opened one of the two remaining doors at random and had happened to reveal a goat.

Whether he reveals a goat at random or not, he revealed a goat. Same odds. 2/3 1/3 switch stay

It's not irrelevant.

It is actually. All you are doing with your long post is changing the game by giving players of the game omniscience of one form or another, reading Monty's mind and trying to cheat the system. Sorry, but it is fail. Odds means the probability lacking foreknowledge. If you know various things, just pick the car and be done with it. If you don't know, the odds are 1/3rd to stay and 2/3rds to switch.

 

[boooeee]

Whether he reveals a goat at random or not, he revealed a goat. Same odds. 2/3 1/3 switch stay

Clearly it does matter what rules Monty is following. What if Monty only shows a goat if you initially picked a car? In that situation, it's now 0:switch 1:stay.

 

[Michael C]

Whether he reveals a goat at random or not, he revealed a goat. Same odds. 2/3 1/3 switch stay

No. If Monty opens one of the two remaining doors at random, in 1/3 of all games he will reveal a car, in which case the game is finished. The fact that a goat was revealed at random limits us to 2/3 of all possible games. In this subset of all possible games, your initial choice will be correct 50% of the time.

Have a look at this article by a which explains in some detail how different strategies for revealing the goat affect the odds.

 

[Meed]

Whether he reveals a goat at random or not, he revealed a goat. Same odds. 2/3 1/3 switch stay

No, if he just randomly opened one of the two remaining doors and it happened to be a goat, then the odds are 1/2 switch 1/2 stay. Try it...

It is actually. All you are doing with your long post is changing the game by giving players of the game omniscience of one form or another, reading Monty's mind and trying to cheat the system.

Actually you are changing the game. The standard Monty Hall problem is:

-The car is behind one of three doors
-You pick a door
-After you pick a door, Monty Hall will always open a door from the remaining two with a goat behind it

Knowledge of Monty's rules for whether or not to open a door and which door to open are required to arrive at the probabilities of 1/3 for sticking and 2/3 for switching. You can call it "omniscience" or "cheating the system", but it's how the problem is set up. If you don't know his rules, then you have no way of calculating the probabilities for switching and sticking.

 

[GlennB]

No, if he just randomly opened one of the two remaining doors and it happened to be a goat, then the odds are 1/2 switch 1/2 stay. Try it...

Actually you are changing the game. The standard Monty Hall problem is:

-The car is behind one of three doors
-You pick a door
-After you pick a door, Monty Hall will always open a door from the remaining two with a goat behind it

Knowledge of Monty's rules for whether or not to open a door and which door to open are required to arrive at the probabilities of 1/3 for sticking and 2/3 for switching. You can call it "omniscience" or "cheating the system", but it's how the problem is set up. If you don't know his rules, then you have no way of calculating the probabilities for switching and sticking.

Yep. And the rules were laid out clearly way back in the thread.

Once again - 'revealing a goat at random' is only available to Monty in 1/3 trials, the one where the contestant guessed the car at the outset. In 2/3 trials he's obliged to open the non-car door. That isn't random, it's forced. In those same 2/3 trials the car must be behind Monty's unopened door.

 

[Red Baron Farms]

No, if he just randomly opened one of the two remaining doors and it happened to be a goat, then the odds are 1/2 switch 1/2 stay. Try it...

Tried it, the odds are 2/3 1/3 switch stay. Which has been proven many times in many ways above.

Actually you are changing the game. The standard Monty Hall problem is:

-The car is behind one of three doors
-You pick a door
-After you pick a door, Monty Hall will always open a door from the remaining two with a goat behind it

Knowledge of Monty's rules for whether or not to open a door and which door to open are required to arrive at the probabilities of 1/3 for sticking and 2/3 for switching. You can call it "omniscience" or "cheating the system", but it's how the problem is set up. If you don't know his rules, then you have no way of calculating the probabilities for switching and sticking.

Nope. Not true. No matter how Monty opens the door, your initial choice is still 1/3...meaning after Monty removes one goat, you are 2/3 chances by switching. You don't need to read Monty's mind. He removed a negative choice from the 2/3rds side. That means the remaining door on the 2/3rds side has the full 2/3rds odds of containing the car.

 

[JohnnyG]

No, if he just randomly opened one of the two remaining doors and it happened to be a goat, then the odds are 1/2 switch 1/2 stay. Try it...

The odds at that point are 1/2 for staying only because you are throwing out the 1 out of 3 times when Monty reveals the car. If you just select a door (no switching), you have a 1/3 chance of winning the car no matter what Monty does with the doors.