you are playing deal or no deal and you are down to 1 case. 1 case has the million in it. Statistically should you keep the case you choose in the beg. or switch cases?

I would say it is evens.

You're better off to change it, as you're 2/3 to get the $1M to 1/3 losing it.

Dg Dave is right. If you have ever seen the film "21" it might help. I am not quoting directly from the film, but in a probability class, the teacher gives the following problem. You are in a game show and there are 3 doors. Behind one is a car and behind the other 2 is a goat. The contestant picks a door, lets say 1 and the host opens one of the other doors, lets say 2 to reveal a goat (this is deliberate as s/he knows where the car is.) The host then gives the contestant the option to switch. When the contestant first chose, the probability of choosing the car was 33.3%. This means that the probability of it not being behind door 1 is 66.6%. We know that the car is not behind door 2, so there is a 66.6% chance that it is behind door 3. Hence the switch is best, as the probability is twice as much. Some people keep their case out of a belief in fate or sentimental value. I have felt true love, I feel it every day, so I know that maths and science are nothing, they cannot explain such a phenomenon as love. Therefore, I do not put a lot of emphasis on it. But, from a purely mathematical point of view, you should switch. I hope I explained that well, otherwise, google "21 film probability class" and you may find some of the links interesting.

I don't follow that. As the car wasn't behind door 2 and if the person has selected door 3 and not 1 then your probabilities reverse. If it isn't behind door 2 then it is a 50/50 chance of it being behind either of the other doors.

Try:

http://uk.youtube.com/watch?v=mhlc7peGlGg&feature=related

or google "variable change" and see what you come up with.

Fascinating. So Sue would be right, using variable change, ONLY if the host didn't know where the car was and randomly picked the wrong door. BUT, in the scenario that tstrip asked about Deal or No Deal, the host may know where the million dollar case is, but the game player is the one eliminating the cases. Wouldn't then the odds be 50/50 since it didn't make a difference if the host knew where it was or not since he wasn't picking any cases? Whew! This is confusing.

A different type of question abour Deal or No Deal.

A contestant must choose 1 case out of 30 that has the $1M. Aren't the odds of choosing the correct case 1 in 30?
After eliminating 28 cases you now have 2 cases, the one you originally picked and one more to go. According to variable change, or what you're saying here, is the case you originally picked has only a 33.3% chance of it holding the money. So if the case left has the money, what were the odds of NOT choosing the $1M case for elimination during game play?

Boxes without the $1M have already been eliminated, except one. The problem with the doors is a highly simplified example. As the odds of choosing $1M were so small at the beginning, the odds of the other box containing it are so large. That's because the chances of you picking $1M are 1 in 30 and not are 29 in 30. You will pick the $1M 1 time in 30, on average. However, you won't 29 in 30. So that means the chances of it being in the other box are much greater, 29 times in 30, it will be. You already know it's not in the 28 other boxes as YOU have already eliminated them. The original problem is just an example. It doesn't work if the host doesn't know where the car is, because he may choose the one with the car. Which brings us on to Q2. The odds of not chosing it during game play are complicated to work out. I HATE probability, but I think it would work something like this. (It might be worth asking a mathematician, which I am NOT.)

The probability of not choosing the $1M box to start with is 29 in 30. The probability of not choosing it in gameplay is 1 in 28 (there are only 28 boxes to pick, because you have already selected one to be your own and you must leave one at the end.) So I think that it's (29/30)x(1/28) = 29/840

That's 29 in 840 but PLEASE don't take my word for it.

In the original question there are just two cases left so I still think it is 50/50.

I know it's a wikipedia article and I haven't read the whole thing, so I can't vouch for its accuracy, but you could try this:

http://en.wikipedia.org/wiki/Monty_Hall_Problem

That still talks about the choice of three, we are talking about the choice of two.

Quote:

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you are playing deal or no deal and you are down to 1 case. 1 case has the million in it. Statistically should you keep the case you choose in the beg. or switch cases?

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That is a choice of just two boxes, not three.

Yes, the chance would be 1 in 2 for each if there were only two boxes to start with, i.e. you had the choice of two boxes, 1 containing $1M and the other containing, well, not $1M. However, in Deal or no Deal, you start with 30 (I think - it's 22 in the British version) and work through them all to 2. Then you get a chance to switch and so the laws of variable change apply, for which the Monty Hall Problem is a simplified example in order to demonstrate and explain it (variable change.) If it was demonstrated with thirty it would just be SO compicated, too complicated. The important thing is that you pick a box at the start and the chances of it containing $1M are 1 in 30. You then eliminate 28, leaving one and your box. Therefore variable change applies and there is a greater chance of it being in the box which you have not chosen.

I've got a college (16-18) level of maths education but it is in maths and maths mechanics not stats, but, I can't get my head around this! Maybe someone further educated in maths can come and explain to Jon the Simple how this works

I'm with Sue on this at the moment, surely the odds must change as you eliminate the boxes? Oh how something that looks so simple becomes utterly complex

1 in 22 to start, but, when you take away one box, surely it is then 1 in 21 and so down to 1 in 2?

I've just had a look around and it is saying the only way that this works is if the quizmaster KNOWS what is behind each door. Oh it is too early in the morning

I have heard something very similar before, but am no closer to understanding the reasoning.

I see it the same way as Sue, too. I think it is irrelevant that the quizmaster knows, or how many boxes you start with, and the only choice is between Box A, and Box B, ie. 50%.

Ozzz The Simple will now join Jon The Simple, and wait for a simple explanation.

For a really simple decision try "eeny meeny miny mo" which works out the same as 50/50 si I'm with Sue !