This thread inspired me to post the following:

Remember the old game show "Let's Make a Deal"? If not, one of the most common games played on the show involved 3 doors. Each one had a prize behind it, and you'd win whatever prize was behind the door you chose. Sometimes, though, the prize would be a "zonk," such as a goat.

Okay, so here's the deal. The host tells you that there's a car behind one of the doors, and goats behind the other two doors. You choose a door... let's say it's Door #1. Instead of opening the door, the host opens another door... let's say Door #2... which has a goat behind it, and asks if you want to change your choice to Door #3.

Should you switch your choice?

Monty Hall does not want to give me the car.

I'll take my chances. I'll stick with door #1 [Hopes I get the car edit]

Yes.

coolcat said:

Correct. If you don't switch, you have a 1/3 chance of winning. But if you do switch, then your chances of winning the car go up to 2/3.

Now, let's change things up a bit....

The State of Franklin is running a scratch-off lottery ticket game. 10,000 tickets have been printed. One of them wins you $1 million. The other 9,999 tickets are losers.

9,998 tickets have already been purchased and scratched -- and they're all losers. Of the two unscratched tickets left, you bought one, and your friend bought the other.

Is it in your interest to convince your friend to swap tickets with you?

matt said:

This is game theory, not maths!

matt said:

This had me scratching my head for a while... I'm going to say, by symmetry between you and your friend, it isn't in your interest to swap, or not swap... the probability that you have the winning ticket is 50%.
[Edited 4/17/07 2:43am]

i just came in this thread to say how much i loathe math.

evenstar3 said:

They're basically the same simple question - a wider choice whittled down to two options - one win, one lose.

So the answer to both is a 50% chance. Only swap if want to, but statistically it doesn't make a difference.

PopeLeo said:

The two situations are different... for the first question the answer is definitely to switch... if you stick with your original choice, you'll win 1/3 of the times you play the game... the rest of the times you'd win by switching... which leaves 2/3 of the times...

But the same reasoning can't be applied to the lottery tickets... but why exactly... I'm not sure

I fail to see the logic - maybe math is different in the US. But probably not.

Are you multiplying the probabilities by a US gallon or a hogshead or sth?

PopeLeo said:

No, the probabilities ARE different in that case. Let me denote the prize by P. Monty will only show a door where the prize is NOT. Suppose you choose Door 1. Here are the scenarios:

1P
2
3

If Monty shows you either door 2 or door 3 and you switch, you LOSE.

1
2P
3

Since you chose Door 1, Monty has to show you Door 3. If you switch to Door 2, you WIN.

1
2
3P

Since the prize is behind Door 3 and you have chosen Door 1, Monty must show Door 2. If you switch to Door 3, you WIN.

P(Win|Switch) = 2/3.

Now suppose you choose not to switch.

1P
2
3

If Monty shows you either Door 2 and Door 3 and you don't switch, you WIN.

1
2P
3

Since you have chosen Door 1 and the prize is behind Door 2, Monty must show you Door 3. If you don't switch, you LOSE.

1
2
3P

Since the prize is behind Door 3 and you have chosen Door 1, Monty must show you Door 2. If you don't switch, you LOSE.

P(win|don't switch) = 1/3

Moonbeam said:

THANK YOU! The universal language of maths.

I've never seen the show - makes sense now.



Now anyone want to explain the logic of imperial measurements?

Actually, you start the contest at 33% chance of winning.

Monty then changes the odds so that you have a 50% chance of winning.

There is no statistical advantage to changing your choice, or swapping your ticket, because each time a Null ticket or Door is chosen, the odds change.

Dauphin said:

That's what I was thinking - go through Moonbeam's excellent answer and you'll find where we were wrong.

Basically, "Monty will only show a door where the prize is NOT" skews things.

Moonbeam said:

Moonbeam has it right -- that is the key difference. In the lottery scenario, the 9,998 losing tickets were eliminated by chance. Of the two remaining tickets, each one has a 50/50 chance of winning, so it doesn't matter whether you swap tickets. But in the Monty Hall scenario, your odds of winning go from 1/3 to 2/3 if you switch.

matt said:

Well it would make sense that he got it right he has a Masters Degree in Math
[Edited 4/17/07 9:02am]

matt said:

This is wrong. Your chance of winning is still 50%. So what would be your advantage of switching doors?

[Edited for proper quotation. --Matt]
[Edited 4/17/07 11:50am]

jaimestarr79 said:

Check out Moonbeams post above... Suppose you always stick with your original choice and play many games... It doesn't matter that a door is revealed, since you're ignoring it and just sticking with your original choice... You'll win only 33% of the games you play (1/3)... The rest of the time you would have won if you switched... So 2/3 of the games would have been won by switching...

jaimestarr79 said:

See Moonbeam's explanation above. Also, here's another way to think of it:

Suppose we have the three doors again, one concealing the prize. You pick door #1. Now you're offered this choice: open door #1, or open door #2 and door #3. In the latter case you keep the prize if it's behind either door. You'd rather have a two-in-three shot at the prize than one-in-three, wouldn't you? If you think about it, the original problem offers you basically the same choice. Monty is saying in effect: you can keep your one door or you can have the other two doors, one of which (a non-prize door) I'll open for you.

(Source: http://www.straightdope.c...3_189.html)

matt said:

That makes sense. It's always helpful for me to think in terms of a real number of games played.

Ok. So let's suppose we're in the regular Monty Hall scenario... You play 300 games... Suppose you always choose door 1, like in Moonbeam's post. Let's see how the games divide up...

A. Car is behind door 1.
This happens in 100 games (on average). Either door 2 or door 3 will be revealed. In 50 of these games door 2 is revealed. In the other 50 games door 3 is revealed.

B. Car is behind door 2
This also happens in 100 games. Only door 3 can be revealed. So in all 100 of these games, door 3 is revealed.

C. Car is behind door 3
Again, happens in 100 games. Only door 2 can be revealed. So in all 100 of these games, door 2 is revealed.

Now suppose that door 2 has been revealed. In how many games does this occur. From Case A, you get 50 games... From Case C, you get 100 games. That's the trick... the two choices are not equal... because the Car being behind door 3 happens twice as much as the car being behind door 1... If you always switch in this scenario the probability of winning is 100/(50+100) = 2/3

NOW, we'll change the game. Again 300 games are played.. Again, you always pick door 1. But suppose one of the remaining two doors is revealed RANDOMLY. So it's possible for the car to be revealed.

A. Car is behind door 1.
This happens in 100 games. Either door 2 or door 3 will be revealed. In 50 of these games door 2 is revealed. In the other 50 games door 3 is revealed.

B. Car is behind door 2
This happens in 100 games. Either door 2 or door 3 will be revealed. In 50 of these games door 2 is revealed. In the other 50 games door 3 is revealed.

C. Car is behind door 3
This happens in 100 games. Either door 2 or door 3 will be revealed. In 50 of these games door 2 is revealed. In the other 50 games door 3 is revealed.

Suppose that you automatically win if a car is revealed... But suppose door 2 is revealed with nothing behind it... How many games does this occur... You get 50 games from Case A, and 50 games from Case C... This time, it's the same number of games. So if you switch in this scenario, you'll win 50/(50 + 50) times... 1/2 the time...
[Edited 4/17/07 13:40pm]
[Edited 4/17/07 18:36pm]

Hey I have a game to play whic door 1,2, or 3 is Matt going to run into when he his drunk. Mind you he always does this when the door is OPEN!

One incident does not equal "always."

matt said:

OMG Matt, get over there and tell me I am right! Nobody wants to confirm if I have the answer or not

Ah, the Monty Hall Paradox....

the interesting thing about this problem is that the probabilities turn on information.

If Monty, like the contestant, does not know which door the car is behind, then the contestant would have a 50/50 chance of picking the correct door (or at least would not be advantaged by changing picks). The odds change only because Monty is informed about the car's placement.

cool post
[Edited 4/17/07 18:50pm]