Kevin's Watch

Thanks for the correction re Monty Hall, ali... blame my UK nationality for that one!

Okay here goes, and there is no linguistic trickery here.

You get through to the grand final of a gameshow - you're the only remaining contestant. The grand final format is always the same - you're presented with three closed doors. Behind two of the three doors, there's a goat, and behind the remaining door, there's a Ferrari.

The host invites you to pick a closed door, so you do. The host then opens one of the other two closed doors to reveal a goat and asks you "Okay, do you want to stick with your original choice of door, or do you want to change your mind and swap to the other closed door?" To remind you, there are now only two closed doors left, behind one of which is a goat and behind the other of which is a Ferrari.

The question is simple... presuming you want to win the Ferrari, should you:-

a) Stick with the closed door you originally chose?

b) Change your mind and switch to the other closed door?

c) It's irrelevant - there are only two closed doors left, concealing one Ferrari and one goat, so it must be a 50/50 chance? (A statistician will correctly tell you that past results cannot affect future outcomes - a coin that you've tossed 10 times and that has come down heads on those 10 times still has a 50/50 chance of being heads or tails on that 11th toss.)

So, what should you do?

PS I found it helped my understanding that I have next to no knowledge of statistics. Friends far cleverer than I in the laws of probability drove themselves next to mad trying to work this out.

aliantha wrote:The only coherent thing I can add to this discussion is that it's Monty Hall, not Monty Haul. He was the emcee of a game show called "The Price Is Right".

Monty Hall hosted Let's Make a Deal; Bob Barker hosted The Price is Right.

TheFallen wrote:You get through to the grand final of a gameshow - you're the only remaining contestant. The grand final format is always the same - you're presented with three closed doors. Behind two of the three doors, there's a goat, and behind the remaining door, there's a Ferrari.

The host invites you to pick a closed door, so you do. The host then opens one of the other two closed doors to reveal a goat and asks you "Okay, do you want to stick with your original choice of door, or do you want to change your mind and swap to the other closed door?" To remind you, there are now only two closed doors left, behind one of which is a goat and behind the other of which is a Ferrari.

The question is simple... presuming you want to win the Ferrari, should you:-

a) Stick with the closed door you originally chose?

b) Change your mind and switch to the other closed door?

c) It's irrelevant - there are only two closed doors left, concealing one Ferrari and one goat, so it must be a 50/50 chance? (A statistician will correctly tell you that past results cannot affect future outcomes - a coin that you've tossed 10 times and that has come down heads on those 10 times still has a 50/50 chance of being heads or tails on that 11th toss.)

So, what should you do?

The correct answer is b--you change you mind and pick the other door. This raises the overall probability of winning the car to 2/3 rather than 1/3.

This is because your chance of winning is 1/3 (had you chosen the correct door the first time and you stick with it) + [1/2 * 2/3] = 1/3 (the chance of winning the car by switching door, as if the first door had never been an option).

The Monty Hall Problem, with explanation and generalization, from the Wolfram site (the best place for mathematical information).

Hashi Lebwohl wrote:

TheFallen wrote:You get through to the grand final of a gameshow - you're the only remaining contestant. The grand final format is always the same - you're presented with three closed doors. Behind two of the three doors, there's a goat, and behind the remaining door, there's a Ferrari.

The host invites you to pick a closed door, so you do. The host then opens one of the other two closed doors to reveal a goat and asks you "Okay, do you want to stick with your original choice of door, or do you want to change your mind and swap to the other closed door?" To remind you, there are now only two closed doors left, behind one of which is a goat and behind the other of which is a Ferrari.

The question is simple... presuming you want to win the Ferrari, should you:-

a) Stick with the closed door you originally chose?

b) Change your mind and switch to the other closed door?

c) It's irrelevant - there are only two closed doors left, concealing one Ferrari and one goat, so it must be a 50/50 chance? (A statistician will correctly tell you that past results cannot affect future outcomes - a coin that you've tossed 10 times and that has come down heads on those 10 times still has a 50/50 chance of being heads or tails on that 11th toss.)

So, what should you do?

The correct answer is b--you change you mind and pick the other door. This raises the overall probability of winning the car to 2/3 rather than 1/3.

This is because your chance of winning is 1/3 (had you chosen the correct door the first time and you stick with it) + [1/2 * 2/3] = 1/3 (the chance of winning the car by switching door, as if the first door had never been an option).

The Monty Hall Problem, with explanation and generalization, from the Wolfram site (the best place for mathematical information).

The correct answer is indeed b - you should always change your mind and switch to the remaining closed door and this does increase your chances to 2 in 3, rather than 1 in 3.

However, I don't understand your equation at all, Hashi - no doubt because I'm no mathematician. The way I've always got my head around this is as follows:-

When there are 3 doors, your chance of randomly selecting the door with the car behind it first time around is 1 in 3.

Equally, your chances of randomly selecting a door with a goat behind it first time around are 2 in 3.

So that means that there must be a 2 in 3 chance of the host revealing to you the other goat - because there's a 2 in 3 chance that you've picked a door with a goat - so you should always switch doors. There's only a 1 out of 3 chance that the host has the luxury of revealing to you either goat (because you've picked the door with the car).

Remember - the host only ever reveals to you a goat behind a door that you didn't pick first time around...

That's not an equation, but it's clear to me at least.

A die roll, or a lotto pick-5 is calculatable precisely because all possible outcomes are known in advance. There is no finite number of outcomes from which to calculate whether or not a sack of money will be thrown from a moving car to your feet.
You would have to make a ridiculously large number of up-front assumptions to calculate this. What city? Population? Time of day? How many cars on road? How many people have access to large sacks of money? How many of them would be driving around and simultaneously practicing throwing on a given day? How many of these would be inclined to do so with their sacks of money? And how many of these events would coincide with your being in the right location to receive this sending? How many of these transactions would end up being legal to the point that you would actually be allowed to keep the money thrown at your feet?
This would combine to become a very, very low probability, I think. You might actually have a better chance of winning the lottery twice.

dw

Hashi Lebwohl wrote:Overall p(winning the car *if* you change your mind) = 2/3. It is as if you got to pick two doors when he first asked you the question...because you do get to pick two doors.

The key is changing your mind. If you don't change, your odds of winning fall back to the original 1 in 3.

But, the way the game is played out, you actually have a 1 in 2 chance of getting the car, right? Regardless of which door you initially choose, if Monty always shows you a goat then asks if you want to keep or choose the other door, then it still comes down to an either/or situation. The stagecraft of showing you a goat just adds to the excitement of the unknown. *You* don't know which door has the car behind it, so from my POV only the final choice you make has the odds that actually count, and they are 1 in 2.

dw

Hashi Lebwohl wrote:

aliantha wrote:The only coherent thing I can add to this discussion is that it's Monty Hall, not Monty Haul. He was the emcee of a game show called "The Price Is Right".

Monty Hall hosted Let's Make a Deal; Bob Barker hosted The Price is Right.

You're right, of course. So much for my ability to be coherent.

DukkhaWaynhim wrote: This would combine to become a very, very low probability, I think. You might actually have a better chance of winning the lottery twice.

dw

This would seem to be supported by the fact that I recall at least 3 times seeing headlines about people who had actually won lottery jackpots twice, and never heard of anyone walking down the street and randomly having a big sack of cash hurled at them.

DukkhaWaynhim wrote:But, the way the game is played out, you actually have a 1 in 2 chance of getting the car, right? Regardless of which door you initially choose, if Monty always shows you a goat then asks if you want to keep or choose the other door, then it still comes down to an either/or situation. The stagecraft of showing you a goat just adds to the excitement of the unknown. *You* don't know which door has the car behind it, so from my POV only the final choice you make has the odds that actually count, and they are 1 in 2.

dw

Well...you could pick the correct door the first time, thus winning the car.

The Monty Hall Problem, as defined, is the situation in which you pick a door then the host opens the door and reveals the goat. You still have a chance to win here.

If the host opens a door and the car is sitting there, you win a goat.

DukkhaWaynhim wrote:But, the way the game is played out, you actually have a 1 in 2 chance of getting the car, right? Regardless of which door you initially choose, if Monty always shows you a goat then asks if you want to keep or choose the other door, then it still comes down to an either/or situation. The stagecraft of showing you a goat just adds to the excitement of the unknown. *You* don't know which door has the car behind it, so from my POV only the final choice you make has the odds that actually count, and they are 1 in 2.

dw

If all your choices are random, then (capitalised is the one you selected):

Ggc - Gc
Ggc - gC
gGc - Gc
gGc - gC
ggC - Gc
ggC - gC

Overall your chance of winning is 50%, because of the second choice. But, out of the three winning conditions above, two of them involve changing your original choice, while of the three losing conditions, two of them involve keeping your original choice. Because you're given a choice in the second portion, and not just flipping a coin, you increase your odds by changing your decision.

Murrin wrote:Overall your chance of winning is 50%, because of the second choice. But, out of the three winning conditions above, two of them involve changing your original choice, while of the three losing conditions, two of them involve keeping your original choice. Because you're given a choice in the second portion, and not just flipping a coin, you increase your odds by changing your decision.

Precisely. This is why, in a Monty Hall situation, your best strategy is to change your mind.

Murrin wrote:Ggc - Gc
Ggc - gC
gGc - Gc
gGc - gC
ggC - Gc
ggC - gC

So now I've been chanting "Goat goat car - goat Car..... goat Goat car - Goat car...." for the past few seconds, which wins me strange looks from my coworkers. Great fun!

dw

Ok - now show me the opposite, and I will be *mightily* impressed.

dw

Hashi Lebwohl wrote:goat in a car

Everybody wins!


(Except the person who has to get the goat out of the car.)

This is the beginning of the process...