probability question

I made up the following probability question in hopes of explaining another probability question to someone, and now it’s got me confused:

I explained how the answer is no and that you would still have a 1/1,000,000 chance of being a winner.

Of course the other 5 people have an awesome chance of winning. My first instinct was that each one has a 1/2 chance of winning. Then it occurred to me that I must be wrong. Since one person has an only 1/1,000,000 chance of winning, the remaining five must each have a better than 1/2 chance of winning. How do I determine exactly what that probability is?

What you’ve got is a variation on the Monty Hall problem. This is a favorite on this board. Check out these threads:
http://boards.straightdope.com/sdmb/showthread.php?t=426065&highlight=monty+hall
http://boards.straightdope.com/sdmb/showthread.php?t=426065&highlight=monty+hall
http://boards.straightdope.com/sdmb/showthread.php?t=411943&highlight=monty+hall
Short answer: Yes, your chances are still one in a million. But if you randomly trade with another of the remaining six ticket holders, your chance improves to 1/2. So counter-intuitively, everyone’s chances improve by trading tickets with someone else. (It doesn’t make sense at first glance, but trust me: submit your will to it before it drives you mad!)

That’s what I originally thought. But other five (there are five remaining ticket holders by the way) each have a slightly less than 3/5 chance of winning, not 1/2. I’m trying to figure out exactly how much of a chance the other five have.

No, the only one who’s chances will improve after trading improve is the one who is currently at 1 in a million.

I don’t usually participate in the Monty Hall threads, and upon reflection, I realized why not. I’m screwing it up.

Anyway.

Either your conceptual problem is as follows, or I’m misunderstanding the question: Your odds do not change simply because you are acquainted with the cheater. Being buddies with the inside man does not change the fact that you are in the same position as the other five “surviving” ticket holders.

Each of the six survivors have an equal chance. That is, 3/1,000,000. You would insist that this does not add up to a probability of 1. You would be correct. This shocks the intuition, I know, because to create your whole probability of one, you need to re-introduce the “eliminated” tickets.

If you want the individual probabilities to add up to 1 post-elimination, everyone needs to trade tickets. Then, everyone will have a 3/6 chance, which does, in fact, add up to one.

For an interesting perspective on counterintuitive probability puzzles, read The Curious Incident of the Dog in the Nighttime, a novel which discusses the Monty Hall problem briefly.

I mean 3/1,000,000

Each person DOES NOT have a 3/1,000,000 chance of winning. Five of the men have an almost 3/5 chance of winning. The only one who has a 3/1,000,000 chance of winning is the one you’re calling the cheater.

I think you’re misunderstanding the question. D_White has set up the problem so that “Monty” always picks “you” to be in the final six. The other two (or occasionally three) losers in the final six are randomly picked, and the three winners are guaranteed to be present in the final six. That means “you” really are different from the other people.

Probability is based on what you do and don’t have knowledge of. In this case, you know that “you” are guaranteed a spot in the final six, so your odds of winning are still 3/1,000,000, because appearing in the final six tells you nothing about your chance of winning. However, you also know that all three winners are guaranteed to be in the final six, so appearing in the final six does affect the chance that any of the other five people are winners. Their chance of winning is now about 3/5.

The exact odds for the other five people are 3/5 times the probablity that “you” aren’t a winner plus 2/5 times the probablity that “you” are a winner. That’s (999,997/1,000,000)(3/5) + (3/1,000,000)(2/5) = 2,999,997/5,000,000.

Also, remember that probability is based on what you do and don’t have knowledge of. That means this calculation is from your perspective, taking into account what you know. The odds would be different from the perspective of “Monty” (since he already knows who the winners are) or from the perspective of one of the other five finalists (since, presumably, they don’t know Monty has rigged the final).

Thanks, zut. I just got done calculating and I came up with a slightly different answer than yours.

You came up with 2,999,997/5,000,000 which equals .5999994

I came up with 5,999,998/10,000,000 which equals .5999998

.5999998 x 5 = 2.999999

The guy who has a 1/1,000,000 chance = .000001

You’re mixing up your odds. The guy with the 1/1,000,000 chance doesn’t really have a 1/1,000,000 chance; he has a 3/1,000,000 chance, since there are three prizes. If you factor that into your calculations, yours should match mine.

Why are you guys overthinking this?

If the guy with inside inside info rules out 999,994 losing tickets, there’s only 6 possible tickets left, of which I have one, that can be a winner. If three of the six remaining tickets are winners, I have a 1 in 2 chance of being a winner.

Though the pool started with 1,000,000 tickets. It’s not down to 6. The original number is irrelevant. Half of the remaining “unknown” tickets are winners. That’s all that matters.

The guy eliminating the losers is not Monty Hall. He’s just making sure that you are in the final 6.

Period! Question over.

The reason this is being “overthought” is because your answer is completely incorrect.

As you say, the guy eliminating the losers makes sure that you are in the final 6. That means that “being in the final six” tells you nothing at all about your chances of winning, and thus your chances of winning stay the same as they were at the beginning: 3/1,000,000.

zut,

You’re right. I’m wrong. The ace of spades analogy in Cecil’s original column helped me see the error of my ways.

My apologies,
milquetoast

Doh! I don’t know why I keep doing that. Thanks again

I don’t know about anybody else, but it would probably be simpler for me to think about if the original question was rephrased so that there was only one prize, and you eliminated all but two of the lottery ticket holders, yourself and some random guy. Then the question can still be asked “Is your probability of winning 1/2?”

Cecil tackled a problem very similar to this one during the whole Monty Hall debacle, where he said:

To which reader Jim Balter of Los Angeles replied:

And Cecil replied:

I think the only difference is that in D_White’s question, you cheat and find out that you’re one of the possible winners ahead of time. Of course there’s probably some other subtlety I’m missing.

This seems to get a little fogged because of some ambiguity about what the “inside man” is up to. This could be one of two things:

a) he’s only talking to you in the first place because he’s got a list of 999,994 losers and you’re not on it, in which case yours odds really are at 3 in 6, or

b) he’s got a list of 5 potential winners and he adds you just for the hell of it to make you feel special till the drawing, in which case, yeah; it’s still 3/1,000,000.

I feel like the question is phrased so you could take it either way. Whether the inside guy is a Monty Hall-type figure depends on why he made you one of his special 6.

I’m sure I missed something obvious, though. Go ahead and rip me a new one, I deserve it.

Yup. I did; I missed the phrase “other than yourself if you should be one.” Although why any sane human being would torture him or herself in this fashion is beyond me. You’d have to have some kind of disappointment fetish.

What thetruewheel said.

I’m reasonably confident that the intention of D_White’s scenario is that the “inside man” is Monty Hall. In other words, “Monty” sets it up ahead of time that you will be part of the final six, whether or not you’re a winner. Since that is set up from the start, your chances of winning never change.

If you were to alter the scenario, so that there is no precondition that you will be part of the final six, then learning you are part of the final six does alter your chances. This scenario is more akin to the one you quoted from Cecil.

Okay, I see what you and thetruewheel were getting at – so in the first case, “Monty” essentially just happens to find almost all the losers while not checking on you at all, which tells you nothing about whether or not you are also a loser.

Whereas if you were to go in to Monty and say “Look up all the losing tickets with your cheatin’ computer,” and you didn’t happen to be among them, then you would have a much better chance of having a winning ticket.

Because in the first scenario, you’ve basically been artificially excluded from the analysis, and the overwhelming likelihood is that if Monty had included you in the analysis, you would have been among the losers, so you’re still overwhelmingly likely to have a losing ticket.