Brain Teaser: Let's Make A Deal

[Lucky Luciano]

Try this....Suppose you are playing the game "Let's Make a Deal." There are three curtains. Behind one of the curtains is a million dollar prize and behind the other two curtains are skunks. Under the reasonable presumption that you prefer the money to the skunks, consider this:

You are asked to choose one of the curtains at random so that you may win what is behind it. Then, unexpectedly someone shows you what is behind one of the curtains that you did not choose and you see a skunk. Afterwards, you are asked if you would like to switch curtains. Should you switch or stay put? Does it matter?


Of course, you must back up your answer!

[hippiechick]

1. I would not switch.

Simply because for me anyway, my first guess/gut instinct tends to generally be very accurate.

And if it isn't, then so what ? I didn't have the $million to start with, so essentially I've lost nothing.


:hippie:

[RevolutionaryActs]

6. Thats a good point, and if you win the skunk you can aways,

threaten the host with the skunk, if he doesn't give you the money you will spray him! :evilgrin:

[Lucky Luciano]

17. see posts 12,13 for the answer!

[BlueJazz]

2. I would switch because the laws of probability say that my...

chances of winning are greater.

[Lucky Luciano]

3. Are you going by the assumption that the

question would not be asked if it did not matter or can you prove your assertion? At least give me some figures!

[CaptainCorc]

4. I'm with you....what law of probability governs that choice?

The way I see it, there's a 50-50 shot either way. There are two curtains left so your chances of hitting the million are 1 in 2 regardless of whether you change your mind or stick with your original choice. Am I missing something here?

[Lucky Luciano]

5. I'll fill everyone in later, but for now

I will avoid giving spoilers....I will say that you want to switch...It is easy to assume that is correct because otehrwise the question would not be asked....So, given that, see if you can work out the details of precisely why. In the process, you will come up with the peoper odds.

[BlueJazz]

7. The way I see it (I'm about to make a fool out of myself..?)

When I choose the first prize, I would most likely pick the wrong one.
..1/3 ...so...chances are I picked the wrong one when I picked the first time.
By changing my choice, it seems I would be increasing my chances of picking the right one..

I think..AAAAAGGGG!!!!

[CaptainCorc]

8. I can adhere to that...I own up to being mistaken the first time n/t

[Lucky Luciano]

16. see posts 12,13

[Ready4Change]

21. Well, I wrote a little JAVA program to PROVE HOW WRONG YOU ARE.

And dang if it didn't prove you're RIGHT! Heheh.

And very, reliably right as well. Pretty darn exactly 66% likely to pick the prize if you switch your choice in this situation.

I bow in acceptance of the facts.

[Lucky Luciano]

29. Hey, see my post 28 below...

I made a few calculations based on someone running a computer simulation. I am curious if you could test it out with your program and see how consistently your results match my expectations.

Thanks if you have the time!

[Ready4Change]

9. This is a case where I think odds are BS.

The theory as I've heard it is that when you make your first choice your 66.6% likely to be wrong, because any of the 3 choices has only a 33.3% chance of being right. Thus, one of the other 2 is most likely to winner.

Thus, when one is eliminated, the other is most likey the winner, and you should switch.

I think this is BS.

To me, whenever you have a choice, the odds are absolutely dependant on the unknowns at that time. Thus, when you have your second choice, your odds are 50-50, no matter what led up to that point.

Proponents of "always switch" claim that their technique always gives you a greater than 33.3% chance of winning. Of course it does. Never switching ALSO gives a greater than 33.3% as well.

Smoke and mirrors by people who over think things, imo.

[Lucky Luciano]

15. See posts 12,13 - odds play a very big role.

[Ready4Change]

22. Er, see post #21.

I meant to post that here. Oops.

[Commie Pinko Dirtbag]

10. Exactly 50/50 chances, but probability isn't everything.

Namely, you'll feel like a jackass if you switch and lose. Staying put and losing is slightly less embarassing. But that's just me.

In the coldness of numbers, you are essentially in a coin toss.

[Lucky Luciano]

14. see post #12,13

[BlueJazz]

11. Lucky Luciano ...where are you??

:)

[Lucky Luciano]

13. Petronius, below gave a nice succint and correct answer

I will elaborate a little bit:

We may always assume, by symmetry, that you always start out by selecting door one (ie the same argument works if you pick the other doors first). If the money is behind door one and you switch, then you lose. If the money is not behind door one, and they show you door 2, then the prize is definitely behind door 3. If the money is not behind door 1 and they show you door 3, then it is definitely behind door 2. There are two scenarios where switching gets you the money and one where switching loses the money. So switching gives you 2/3 odds and staying put gives you one third odds...DEFINITELY SWITCH!!!!

[petronius]

12. Definitely switch

Assuming you choose door A: there is a 1/3 chance you are right, and a 2/3 chance that the prize is behind B or C. If there is a 2/3 chance that the prize is behind B or C, and you are shown a skunk behind B, then there is a 2/3 chance that the prize is actually behind C (and still only a 1/3 chance that it is behind A). So you should switch.

[Lucky Luciano]

18. bingo!!

The only correctly reasoned answer.

[Thor_MN]

25. Let me put it another way, you are making TWO choices

In the first choice, you are picking one of three doors - you have a 33% chance of winning - which is made totally irrelevant because they open a door you didn't pick with a losing prize.

You are now faced with another choice - the only one that matters. If you switch, I think everyone agrees you have a 50% chance of winning. Where everyone seems to be getting lost is that now there is only one other choice - your original door - ALSO with a 50% chance of winning. I assume, of course, that one is alert enough to avoid choosing the open door with the skunk with your second choice.

Staying with your original choice is not staying with your original 33% chance, it's picking the same door again, but this time with a 50% chance of winning. It doesn't matter if you switch or stay - the odds in the second choice are 50-50.

[Lucky Luciano]

26. I'll try again...

If I fail to explain myself, then you can do what the other guy did and write a program to test it out. Assuming you pick door #1, there are exactly three scnearios. They are that the prize is behind door 1, 2, or 3.

If it is behind door 1 and you switch, then you lose.
If it is behind door 2, then they have no choice but to show you door 3 since they can't show the prize or the original door you chose. So, when you switch here, you are forced into door 2 and you win.
If it behind door 3, then they are forced to show door 2 (this is the conditional probability at work by the way)at which point switching brings you to door 3 and you win.

So, there were three scenarios. You win in two of them by switching and you lose in one of them by switching. Therefore switching gives you 2/3 odds and staying put gives you 1/3 odds.

[Thor_MN]

27. Wrong, wrong and wrong.

No matter where the prize is, you are making two choices and the first is irrelevent. It boils down to the second choice where you pick either your original door or switch to the other closed door and the odds are the same - 50-50.

There will always be at least one door that they can show you. At that point, you now have a 50-50 chance of being right.

In your examples, you leave out a couple of concepts which round out the possibilites. Consider that the prize is behind #1, they could show you either 2 OR 3. You switch, you lose. Your other two options are correct, you switch in those, you win. So we have four possibilities, in which you can lose 2 by staying and 2 by switching. To give it the full treatment, one should include both staying and switching, in which case there are 8 possibilities, but it still boils down to 50-50.

The odds are 50-50 in your second choice regardless if you switch or stay.

[Lucky Luciano]

28. You still miss the point

There are three possible locations...door 1, 2, 3.
As before , assume you picked door 1. We will discuss door 1 because that is where you disagree.
If it is behind door 1 = 1/3 chance....then given this, there is a 50% chance they will show you door 2 and a 50% chance that will show you door 3.

However, if the prize is behind door 2, then there is a 100% chance they show door 3 and similarly if the prize is behind door 3, then there is a 100% chance they show door 2.

The two cases given in the door one scenario together have the same likelihood of occurring as the one case door 2 scenario.

I must stand by my assertions.

Keep thinking it through.

Or try it a few thousand times with a computer program...you shall see that after 10,000 attempts at playing this game, that in over 99.7% (3 standard deviations) of the instances, you will be at 6667 wins + or - 141 by switching..ie you will have won between 6526 and 6807 times....If you run the game a million times, then 99.7% of the time 666,667 plus or minus 1,414, so you will have won by switching between 665253 and 668081 times...this is far from 50%!

My figures come from using 3 standard deviations from the mean which I am asserting to be 2/3.

Let us say that you get 0 for losing and 1 for winning to simplify things. If you have a two thirds chance of winning, then the mean is 2/3 * 1 + 1/3 * 0 = 2/3 = "The expected value of X." The expected value of X^2 is 2/3 * 1^2 + 1/3 * 0^2 = 2/3 as well. The variance is = expected value of X^2 - the square of the expected value. That is var = 2/3 - (2/3)^2 = 2/9. The standard deviation is the square root of this sqrt(2)/3. The Central Limit THeorem in probability says that for a large number of trials X_1,...X_n (these are the variables that get assigned a 0 for losing and a 1 for winning) that their sum S_n will be such that:

P = probability, | | = absolute value, s = standard deviation

P(|(S_n - n*mean)/(s*sqrt(n))| <= 3) is very closely approximated by using the standard normal distribution within 3 standard deviations of the mean - ie 99.7% of the time if you consult a chart.

So, for n= 1,000,000, we get:

P(|(S_n - 666,667)/Sqrt(2)*1000/3| <= 3) = 99.7%

so 99.7% of the time we get:

|S_n - 666,667| <= 1000*sqrt(2) = 1,414 which implies the S_n which is the number of wins by switching is in the range given above:

From 665253 to 668081. For it to be 500,000 it would be 354 standard deviations to the left - ie not happening unless your rndom number generator is fucked!

Now test this on a computer!

[Ready4Change]

30. Your math is beyond me, but the results match.

I made several runs of my program with iterations of 10,000, and the results were in within your range. I then changed the program to run with 1 million iterations, and while it took a bit longer to finish up, the results were still within the range you stated above.

The 10,000 runs did sometimes come CLOSE to the extremes of the range. I had some as low as about 6537, and at least one as high as 6803. I didn't run the million iterations enough to explore this.

[Thor_MN]

20. I disagree!!!

When a door is opened revealing a skunk, the only odds that really matter are that there is now a 100% chance that the money is behind the other two. Therefore, it doesn't matter if you switch or stay, your odds are now 50%.

[arwalden]

19. I Read About This...

... if the revealed prize isn't the big prize, and you're given a chance to trade your first choice for the remaining prize... if you DO swap, your chances for winning the big prize will increase.

When there are 3 choices, your chance of hitting the big prize is 33.3%. When one of the unselected prizes is shown NOT to be the big prize, your chances do not become 50/50... you still have a 33% chance of having chosen correctly the first time. BUT... if you choose again (from TWO possible prizes) your odds increase from 33% to 50%.

Something like that.

-- Allen

Now I'll go read what everyone else said.

[Ready4Change]

23. As I view it NOW.

I didn't believe this till I wrote a little program which proved me wrong. (See post #21) I think I understand how/why it works now:

Let me try to put it this way:

Imagine you dissolve 3 cups of sugar into 3 gallons of water. You then pour that water into 2 containers. The first contains 1 gallon, the second contains 2 gallons. (They would also have 1 cup and 2 cups of dissolved sugar, respectively.)

You now boil the water in the second container until there is only 1 gallon of fluid left there. However, while the water has boiled off, the sugar hasn't. Thus, the water in the second container is now twice as sweet as the water in the smaller container.

Does that get the idea across?

[Lucky Luciano]

24. Yes...that gets the point across in the spirit of Petronius's answer

I am glad you wrote that program...In mathematical communities, they call things like that Monte Carlo simulations....where randomness is used to verify with a very high probability that something is true.