Do I help my chances in picking the bucket with gold, if I re-select after an empty bucket is removed? [duplicate]

Possible Duplicate:
The Monty Hall problem

Let's say there are ten covered buckets. One of the buckets holds gold, while the rest are empty. I can't lift or touch the buckets, but I can randomly choose one from the set. Moreover, after my first selection, an empty bucket is revealed and I am asked if I want to re-select.

Would random re-selection help my chances of winning? And if don't re-select are my chances to win 1/10 or 1/9?

-

marked as duplicate by joriki, Dilip Sarwate, Salman Paracha, Asaf Karagila, Ross MillikanApr 16 '12 at 17:29

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

–  Martin Wanvik Apr 16 '12 at 15:11
I have seen it...but that doesn't answer the second part of my question. –  Salman Paracha Apr 16 '12 at 15:12
Because the question is if there are three doors, and one of them is removed. Are my chances to win improve to 50% (if I did not change my selection?) –  Salman Paracha Apr 16 '12 at 15:13
I don't understand where you see the difference. It seems the difference is just in the different numbers? All the conceptual tools you need for this sort of question are in that article. –  joriki Apr 16 '12 at 15:15
The Monty Hall problem is the subject of this question. Unless I'm missing a subtle difference between this question and that one (other than the numbers), this should be closed as a duplicate of that. In case you agree that these are conceptually the same question with different numbers and you're having trouble transferring the reasoning to the different set of numbers, please point out more specifically what you're finding difficult. –  joriki Apr 16 '12 at 15:20

2 Answers

The Wikipedia article linked in the comments gives a nice discussion of this kind of problem; I especially like the "Other host behaviors" section. If you understand that section, I think you've mastered the Monty Hall problem. Since you are particularly interested in your second question ("And if don't re-select are my chances to win 1/10 or 1/9?"), I'll address that question.

Notice that no matter what you pick (empty or gold), there are at least 8 empty buckets out there that you didn't select. So, it's no trouble for me to peek in all the buckets, pick one of the unselected empty buckets at random, and show you that it's empty. In fact, I can pick 8 of the unselected empty buckets at random and show them to you without giving you any new information about your choice. You still have the 1/10 odds you started with.

However, if I start showing you unselected buckets without peeking inside, and they turn out to be empty, that's a different situation. Notice that there's no suspense in the first version (you know I'm going to show you empty buckets), but there is suspense in the second version (neither of us know what's in the bucket). In this version, for every bucket I show you, you gain new information and can adjust your probability estimate-- after I reveal one empty bucket, you are excited that it was empty and can adjust your odds of winning to 1/9.

-
Well, presumably the host knows which is the empty bucket ("after my first selection, an empty bucket is revealed"). If so, the odds change $-$ see my answer. –  TonyK Apr 16 '12 at 15:43

The key question is whether you know in advance that an empty bucket will definitely be revealed to you. If not, then the motive of the revealer comes into the equation -- perhaps an empty bucket is only revealed when the chooser chooses the golden bucket. This point is the crux of the Monty Hall problem, and a failure to take it into account is the reason for the endless debate.

But taking your question at face value, it seems that you do know this in advance. So your odds definitely improve (from 1/10 to 9/80) by changing your selection. 9/80 is 9/10 (the probability that you got it wrong first time) multiplied by 1/8 (the probability that you get it right second time).

To elaborate on your second question: if you don't change your selection, your chance is obviously 1/10 $-$ you have to get it right first time. I suspect that your uncertainty over this simple matter is a result of the confusion deliberately sown in the minds of the betting public by professional bookmakers: a chance of 1/10 is described as "9 to 1" in betting circles. This practice should be against the law.

-