## Wednesday, October 22, 2008

### Boring Math Stuff

Mrs. Peel has a post on Monty Haul and how it doesn't apply to Deal or No Deal. It's boring math stuff, but is just the kind of thing I like. I understand and concur exactly with her conclusions.

Now, here's the situation on DoND: you have picked a case, eliminated the rest, and now you are down to two cases. You know one has the million, and the other has one dollar. Should you switch, or stick with the case you picked?

It makes no difference what you do. Here's my thinking:

If there are 30 cases, your chance of picking the million at the outset is 1 in 30. So the chance the million is in one of the other cases is 29/30.

You eliminate a case. The chance it was the million case is 1/29. There are now 28 cases left.

In fact, there are 28*27*26... ways to eliminate the remaining cases to get to having just two cases, the 1 and the 1 million. That's a lot of potential paths. The chance that you follow one of these paths is about 28!/29!, or 1/29.

So it seems to me that when you get to two cases, there's a 1/30 chance the million is in case 1, and a 29/30*1/29 = 1/30 in case 2.

Ergo, do what you want, since the odds are even now.

I'm not enough of a math stud to be sure this logic is correct, though. Anybody else out there who is?

Mrs. Peel said...

Your conclusion is definitely correct, heh. Yes, that's another way to think about it. That's like saying there are 26! different paths on the probability tree (you start with 26 cases, right? haven't seen that show in a while). Each one is equally likely, so yes, the path in which you select the million first and the path in which the million is in the last case are equally likely.

Bottom line: Wickedpinto is wrong. ;-)

(He was insisting over at IB that there is an advantage to switching in Deal or No Deal, which is why I wrote my post.)

By the way, I love your blog. Your stories crack me up.

Jeff_McAwesome said...

I have been wondering about this for a while. I always assumed that it was the exact same thing as the Monty Hall problem, but never checked it. You are correct but your logic is flawed. When you get down to two cases the odds that the million is in either case is .5, not 1/30.

The way to solve this problem is using Bayes Theorem (look it up on Wiki). The theorem says:

P(A/B)=
P(B/A)*P(A)/[P(B/A)P(A)+P(B/Ac)P(Ac)]

Sorry if that looks horrible, I need to say it to show where I'm going.

Then you define two events:

A= you pick the \$1 million case
B= you eliminate all other cases

The probability that you pick the million dollar case:
P(A)= (1/30)

The probability that you eliminate all the other cases GIVEN that you picked the million dollar case:
P(B/A)= (1) (you will always eliminate all the other cases if your case has the million)

The probability that you did NOT pick the million dollar case:
P(Ac)= (29/30)

The probability that you eliminate all the other cases GIVEN that you did NOT pick the million dollar case:
P(B/Ac)= (1/29) (as you said)

So, plug that into the equation to solve for the probability that you picked the million dollar case, given that you eliminated all the other cases:

P(A/B)=(1)*(1/30)/[(1)*(1/30)+(29/30)*(1/29)]

P(A/B) = .5

So the odds that the case you chose is the million dollar case *if you eliminate all but one case* is exactly 50%.

The reason this is different from the Monty Hall problem (even if it is expanded so that they open 30 doors) is the P(B/Ac) term. Since the odds are against you eliminating all but the million dollar case, then if you do, you have much better odds of your case being the million dollar case.

If you selected your case, and then Howie opened all but the million and one of the other cases, P(B/Ac) would be 1 and the odds that your case is the million dollar case are just 1/30, and we are back to the Monty Hall problem.

At any rate, I just wrote a massive comment saying that you are right and Wikedpinto is stupid.