What are the odds of winning?

[nomadking]

Originally Posted by daveeb

I told my sister this one years ago and she gave exactly your reply. You're both wrong. Years ago a maths professor in the US explained it and she was ridiculed by many of her peers until she proved it mathematically. Mathematical odds aren't always intuitive.

You CAN'T win the 1st round. There is NO outcome. Win or lose, you get a chance with just 2 doors.

Instead of opening the door you have chosen, the host then opens one of the other doors. And behind this door is a booby prize.

No mention of what if you got it right 1st time.
3 doors, A, B, & C. Prize is behind A.
If you can win 1st time around, then if you pick B, the host will open C. Pick C and B is opened. Either way, you would then know it was A and would win. That is why you can't win 1st time around.


There are always n-2 booby doors to open.


If the host opened one of the other doors at random, then yes the 1st round choice would matter. If you didn't make the right 1st choice, then there is the risk the host would open the prize door instead. But the host knows where the prize is, so it can't be at random.

[spiderplant]

Originally Posted by Paul

I dont think so.

You have two doors, and no idea which is the correct one, your chances are equal, how can one possibly be more likely then the other ?

But I do have an idea, because the host has shown me a box that definitely contains a goat.

Originally Posted by nomadking

There is always just the one other door.

Yes there is, and you should pick it.

Originally Posted by spiderplant

But I do have an idea, because the host has shown me a box that definitely contains a goat.

No you dont have any idea at all.
Knowing whats behind door number 3 gives zero clue as to whats behind door 1 or door 2. Pick one, its 50/50.

You could have 10 doors, and be told whats behind 8 of them, that still doesnt tell you whats behind the other two, you still have a 50/50 chance of picking the right one out of the remaining two.

[Halcyon]

Seems this puzzle is still keeping us guessing. I like that.


Everyone seems to keep telling me I'm wrong but I see each round as individual rounds.
It doesnt matter what you started with as by the time of your final choice you will only have two doors left.
Both of these remaining doors could have a goat or car behind them. It is 50/50


You will leave with one or the other.


So what if the host has showed me a door has a goat behind it? I still have no idea what is behind the other two.

Originally Posted by Halcyon

Seems this puzzle is still keeping us guessing. I like that.


Everyone seems to keep telling me I'm wrong but I see each round as individual rounds.
It doesnt matter what you started with as by the time of your final choice you will only have two doors left.
Both of these remaining doors could have a goat or car behind them. It is 50/50


You will leave with one or the other.


So what if the host has showed me a door has a goat behind it? I still have no idea what is behind the other two.

And therein lies the problem. You can’t accurately calculate a probability if you don’t know what it is you’re calculating the probability of. In this case the rules of the game are unclear (whether there are ‘rounds’, how each round is played and whether things done in one round affect the next).

It is possible to produce a definitive answer, but only if there’s a definitive question.

[nomadking]

Is there an inaccuracy in the description of the problem?
As the host knows where the prize is, they can only open all but one of the booby prize doors. They can't ever open the prize door. It would be unfair.
It's can't be a real type of game show scenario, and that is where the confusion lies. As there is no win or lose in the 1st round, you add the win options and the lose options together, giving 3 in 3 in odds for that 1st round.

[daveeb]

You have three doors to pick from. Suppose you pick door A, there is a 1/3 chance of it having the prize. So there is a 2/3 chance of it being behind B or C. If the host then tells you it isn't behind door B there is a 2/3 chance of it being behind C. So double the chance of that of it being behind A. Rather counter intuitive I know but numbers can be weird.

[spiderplant]

Originally Posted by nomadking

It's can't be a real type of game show scenario, and that is where the confusion lies. As there is no win or lose in the 1st round, you add the win options and the lose options together, giving 3 in 3 in odds for that 1st round.

As a game show it's pretty rubbish. Just an interesting thought experiment.

There again. this one's even simpler, yet brilliant:
https://www.youtube.com/watch?v=0UGuPvrsG3E

[daveeb]

Originally Posted by Chris

And therein lies the problem. You can’t accurately calculate a probability if you don’t know what it is you’re calculating the probability of. In this case the rules of the game are unclear (whether there are ‘rounds’, how each round is played and whether things done in one round affect the next).

It is possible to produce a definitive answer, but only if there’s a definitive question.

The definitive question is, should you stay with your first choice or swap to the other unknown.

[Huxie]

Originally Posted by daveeb

The definitive question is, should you stay with your first choice or swap to the other unknown.

So it’s 50/50 then? 👍

[daveeb]

Originally Posted by Huxie

So it’s 50/50 then? 👍

If you like

[Halcyon]

Or maybe a more worthwhile question to ask would be:


With petrol and electricity prices on the constant rise, Would it be better to win the goat!

I mentioned this to a friend of mine who teaches maths.

His reply (word for word) was "ah, that old chessnut, always change your mind" [ ie switch your choice ].

So I asked him to explain.

He said its a poor example to be able to understand, its far easier to understand how it works with a more extreme example.
The important points are you made you initial choice when all options were available, but you second choice is after the [wrong] options have been reduced.

His example ;

Take a pack of cards, and spread them all out on a glass table.
Then make you choice for which you think is the Ace of Spades (AOS).
Then get a friend to look under the table, and tell you 50 wrong cards to remove (but not yours).

This leaves you with your choice, and one other card.
However, the odds are not even, the chances the other card is the AOS are huge.

The odds you initially chose the AOS correctly is 1 in 52 [low], so the odds of it being in the other cards is 51 in 52 [high].
Now your friend has removed 50 incorrect cards, the chances of the remaining card being the AOS are still 51 in 52, so always change your mind, and pick that card.

It would only be 50/50 if your friend had removed the 50 cards before you made your first choice.

I understand now.

[nomadking]

Originally Posted by Paul

I mentioned this to a friend of mine who teaches maths.

His reply (word for word) was "ah, that old chessnut, always change your mind" [ ie switch your choice ].

So I asked him to explain.

He said its a poor example to be able to understand, its far easier to understand how it works with a more extreme example.
The important points are you made you initial choice when all options were available, but you second choice is after the [wrong] options have been reduced.

His example ;

Take a pack of cards, and spread them all out on a glass table.
Then make you choice for which you think is the Ace of Spades (AOS).
Then get a friend to look under the table, and tell you 50 wrong cards to remove (but not yours).

This leaves you with your choice, and one other card.
However, the odds are not even, the chances the other card is the AOS are huge.

The odds you initially chose the AOS correctly is 1 in 52 [low], so the odds of it being in the other cards is 51 in 52 [high].
Now your friend has removed 50 incorrect cards, the chances of the remaining card being the AOS are still 51 in 52, so always change your mind, and pick that card.

It would only be 50/50 if your friend had removed the 50 cards before you made your first choice.

I understand now.

The 1st pick could be AOS or any of the other 51 cards, makes no difference to the outcome. 2 cards will always be left, the AOS and another card.

Any of the 52 cars is a "winner", as there's no losing choice. It's a complete redo with 2 cards instead of 52.
As a game show idea the 1st round would be meaningless. You could just go straight to the 2nd round.

[spiderplant]

Originally Posted by nomadking

The 1st pick could be AOS or any of the other 51 cards, makes no difference to the outcome. 2 cards will always be left, the AOS and another card.

But it's far more likely to be the card you didn't pick than the one that you did. That's why you should always swap. Get yourself a pack of cards and try it.

Originally Posted by nomadking

As a game show idea the 1st round would be meaningless. You could just go straight to the 2nd round.

But then you wouldn't have a card to swap.