Riddle Me This, Batman!

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AuthorTopic: Riddle Me This, Batman!
Law Bringer
Member # 335
Profile Homepage #25
On the first day Joe is on the receiving end of a large number, n, of pitches and hits only one. Bob swings once and misses once.

On day two Joe hits the only ball to come his way, which gives him a lovely average. Bob gets m hits but misses once. His average is lower.

Joe's average is higher for each day, but (m-1)/(m+1) > 2/(n+1) because their limits as m and n approach infinity are 1 and 0, respectively.

—Alorael, who could have said that better. He blames travel.

[ Sunday, August 13, 2006 09:15: Message edited by: Forty Days and Nights of Solitude ]
Posts: 14579 | Registered: Saturday, December 1 2001 08:00
E Equals MC What!!!!
Member # 5491
Profile Homepage #26
Nothing is more entertaining than watching someone flat-out refuse to believe the door trick when you understand it. :)

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SupaNik: Aran, you're not big enough to threaten Ash. Dammit, even JV had to think twice.
Posts: 1861 | Registered: Friday, February 11 2005 08:00
? Man, ? Amazing
Member # 5755
Profile #27
quote:
Originally written by Ash Lael:

Nothing is more entertaining than watching someone flat-out refuse to believe the door trick when you understand it.
Two things. That applies to so much more than the door trick. You are highly entertaining. :P

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quote:
Originally written by Kelandon:

Well, I'm at least pretty sure that Salmon is losing.


Posts: 4114 | Registered: Monday, April 25 2005 07:00
Infiltrator
Member # 8
Profile #28
Way to leave the crucial premise out of the first post

As written, there's not enough information

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"Names is for tombstones, baby." -Mr. Big
Posts: 699 | Registered: Thursday, September 20 2001 07:00
...b10010b...
Member # 869
Profile Homepage #29
quote:
Originally written by WiKiSpidweb:

Joe and Bob go to two days of batting practice, Joe has the better batting average on both the first and second days. Bob has the better overall average after both days. How is this possible?
Alorael already answered this one, but I'd just like to point out that this is an example of a statistical phenomenon known as Simpson's paradox (http://en.wikipedia.org/wiki/Simpson's_paradox), and it's not uncommon in real data.

[ Monday, August 14, 2006 00:37: Message edited by: Thuryl ]

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The Empire Always Loses: This Time For Sure!
Posts: 9973 | Registered: Saturday, March 30 2002 08:00
Infiltrator
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It confused the hell out of me in 6th grade.

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"As our circle of knowledge expands, so does the circumference of darkness surrounding it." --Albert Einstein
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Posts: 536 | Registered: Sunday, September 7 2003 07:00
Warrior
Member # 7195
Profile #31
Hej, i don't know if i'm saying anything new, coz i didn't read every post.

The worst man ever was right in his own way, but only IF then the doors would be mixed again and you'd not know which was which.
But otherwise, here's the logic in common way of thinking:

THE HOST KNOWS THE ANSWER AND IS A ROBOT - IMPORTANT!!!
You've 1/3 chance the first time around and let's suppose it's the door number two, the one that's correct.
If you pick 1, the host opens 3(1/3), if you pick 3 he/she opens 1(1/3 again) and if you pick 2 he opens 1(1/2 of 1/3) or 3(again 1/2 times of 1/3).
So now for the upper problem:
If you choose door 1 and he/she opens door 3.
Look above. If your door's correct, you have 1/2 chance that the host'll open door 3 and if it's door 2, the correct one, there's 1 chance the 3'll be opened - get the conclusion??

Of course, that's just pure logic, which certainly does not work when we deal with human beings, especially if one was to implement psichology and stuff like that...

And also if there were 1000 doors and you picked door 1 and the otherone laft'd be, let's say, door 537, it'd be a much higher probability that it's that door, since probably humans make the show and they'd rarely hide the price behind door 1...

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I don't care what you say, I'm punk and Hardcore all the way!
Posts: 185 | Registered: Sunday, June 4 2006 07:00
The Establishment
Member # 6
Profile #32
quote:
since probably humans make the show and they'd rarely hide the price behind door 1
I'd estimate the probability to be about one in a thousand for the thousand door example. Most likely they would have a computer decide such that patterns don't unintentionally emerge.

Let's suppose we play an n door game. In each round, the player selects a door (he/she may stay with the door from the previous round or may switch). Then, one door that (1) is not the selected door, and (2) does not contains the prize is removed. This continues until 2 doors remain where the player chooses.

What is the optimal strategy for the player?

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Your flower power is no match for my glower power!
Posts: 3726 | Registered: Tuesday, September 18 2001 07:00
Nuke and Pave
Member # 24
Profile Homepage #33
I like these apparent paradoxes.
quote:
Originally written by *i:

...
Let's suppose we play an n door game. In each round, the player selects a door (he/she may stay with the door from the previous round or may switch). Then, one door that (1) is not the selected door, and (2) does not contains the prize is removed. This continues until 2 doors remain where the player chooses.

What is the optimal strategy for the player?

Warning: the rest of my post contains an answer to above question. If you don't want to see any spoilers, stop reading it now. Or just scroll down past my post. Although there is no guarantee that the next post doesn't also contain spoilers.

That was enough buffer text, so here is my answer: If your initial door doesn't contain the prize and you stay with it until there are only 2 doors left, the host will be forced to leave only the prize door unopened, giving you (n-1)/n chance of winning if you switch only in the last round. (There is 1/n chance that you've picked the prize door in the first round.)

Detailed explanation for this is that in a multi-round game the probability of your door containing a prize is determined when you initially pick it and doesn't change unless you pick a different door. (The host couldn't possibly open the door you've selected for current round, so which doors he doesn't open gives you information only about other doors.) This means that by sticking with your initial selection until final round, you keep its probability of containing the prize down at 1/n, while the host keeps giving you more and more information about the rest of the doors, until there is only 1 door left. The probability that this last door contains the prize will be 1 - 1/n = (n-1)/n.

[ Monday, August 14, 2006 10:32: Message edited by: Zeviz ]

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Be careful with a word, as you would with a sword,
For it too has the power to kill.
However well placed word, unlike a well placed sword,
Can also have the power to heal.
Posts: 2649 | Registered: Wednesday, October 3 2001 07:00
Board Administrator
Member # 1
Profile Homepage #34
The person should always switch. Period.

If you don't believe me, well, if you ever meet me in person, we can play it with cash, with 3 cards, two 2's and an Ace (instead of the awkward door arrangement). I'll be the host. You bet ten bucks. If you pick right, I give you $25. After you pick, I'll reveal a deuce from among the two cards you didn't pick. You won't switch. We'll play 10 times.

If you think that, after the reveal and not switching, you have a 50/50 chance of winning, omg! Great bet for you! 50/50 chance of getting $25 on a $10 bet.

Then, after you lose a bunch of money, you can still comfort yourself with how smart you are.

And if you ever get in this argument in real life, be sure to offer the wrong person a chance to put their money where their mouth is.

- Jeff Vogel

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Official Board Admin
spidweb@spiderwebsoftware.com
Posts: 960 | Registered: Tuesday, September 18 2001 07:00
Board Administrator
Member # 1
Profile Homepage #35
I'll add this instead of a stealth edit because I did make a mistake. The original statement of the problem is terrible.

A better statement is here:

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

To be specific, the problem must make clear that the host will ALWAYS reveal one of the other 2 doors to have nothing behind it. The problem as stated at the beginning of this thread leaves open the possibility that, say, the host only reveals a door when the player originally chooses correctly.

So, bad job all around.

And, if you don't understand the problem properly stated, well, you have no excuse. This is something you can verify for yourself in real life with 2 minutes, 3 cards, and a "friend".

- Jeff Vogel

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Official Board Admin
spidweb@spiderwebsoftware.com
Posts: 960 | Registered: Tuesday, September 18 2001 07:00
Electric Sheep One
Member # 3431
Profile #36
At a conference a few years ago, someone who seemed authoritative at the time propounded a paradox somewhat reminiscent of the Monty Hall problem, but with envelopes of cash, and doubling the money in between two choice stages, or something like that. Point was, it was supposed to be a truly unresolved problem, after a fair amount of study by professionals, at least at that time.

But that's as far as I remember. Ring any bells for anyone here?

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We're not doing cool. We're doing pretty.
Posts: 3335 | Registered: Thursday, September 4 2003 07:00
...b10010b...
Member # 869
Profile Homepage #37
quote:
Originally written by Student of Trinity:

At a conference a few years ago, someone who seemed authoritative at the time propounded a paradox somewhat reminiscent of the Monty Hall problem, but with envelopes of cash, and doubling the money in between two choice stages, or something like that. Point was, it was supposed to be a truly unresolved problem, after a fair amount of study by professionals, at least at that time.

But that's as far as I remember. Ring any bells for anyone here?

There are two situations you could be thinking about:

#1: In each round of the game, you have a 50% chance of doubling your current amount of money and a 50% chance of halving it. Since you have the potential to win an unbounded amount and only lose a finite amount (and can never lose all your money), if you have all the time in the world, you can keep playing until you have an arbitrarily large amount of money. This seems counterintuitive to some people, since on average after a large number of rounds your expected number of wins is 0 (which would put you back where you started), but there's nothing all that mysterious about it: you have as much chance of having a net total of 1 win as a net total of -1 win, and 1 win wins you more than -1 win loses you.

********

#2: You're given a choice between two envelopes, A and B. One has X amount of money, the other has 2X amount of money. You pick envelope A. It has $20,000 in it. Then you're given an option to switch to the other, unknown envelope, which contains either double or half the money of the one you picked. Since we established in situation #1 that you stand to gain by switching, you switch.

But wait! The logic applies to your current situation as well; the unknown envelope you're holding could just as well contain X and the envelope you originally picked could contain 2X, so you still stand to either double or halve your money by switching. This line of argument would seem to support switching back to your original envelope, but obviously switching between the two envelopes ad infinitum would be silly, so did you really stand to gain anything by switching in the first place?

The obvious trick is that the game in #2 isn't quite equivalent to #1, since you're really only playing one round of it -- you can't win an arbitrarily large amount of money by switching repeatedly. The second and more important trick is that there's no such thing as a uniform probability distribution extending to infinity, so in order to make a sensible decision about whether to switch, you have to have some idea about the actual probability of the game show putting a particular amount of money in the envelopes. In other words, the fact that the envelope you picked contains $20,000 is information that you have to use in making your decision. (If you're not told what's in either envelope, then obviously the two envelopes are equivalent and whether you switch is irrelevant.)

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The Empire Always Loses: This Time For Sure!
Posts: 9973 | Registered: Saturday, March 30 2002 08:00
The Establishment
Member # 6
Profile #38
quote:
The original statement of the problem is terrible.
For this specific case it does not matter as I specifically told you the outcomes until the final choice. Yes, in general the problem would need to be explained better revealing the choices of the host to get the problem in general.

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Your flower power is no match for my glower power!
Posts: 3726 | Registered: Tuesday, September 18 2001 07:00
Board Administrator
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"For this specific case it does not matter as I specifically told you the outcomes until the final choice."

Not correct. And a good example of why this problem is such a nightmare to discuss.

The way the problem is stated has an implicit assumption in it, namely: that the host was going to open a door no matter what you choose.

To see the problem, consider this alternate puzzle:

You're picking one of three doors. One has a bag of money, the other two have poo. You pick door 1. The host, knowing that door 1 has the money and not wanting contestant to win it, opens door 3, shows you poo, and gives you a chance to switch. Should you? (The answer is no. OMG.)

In other words, maybe the host opening door 3 was conditional on the contestant choosing door 1! The statement of the problem must make explicit that the host will reveal a bad door no matter what is chosen initially.

This is a comment mistake in stating the puzzle, leading to long, idiotic arguments like this one. Before you argue further, PLEASE read the wikipedia page, which discusses this error, and others.

In fact, the wikipedia link makes the whole discussion redundant. IBTL.

- Jeff Vogel

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Official Board Admin
spidweb@spiderwebsoftware.com
Posts: 960 | Registered: Tuesday, September 18 2001 07:00
Electric Sheep One
Member # 3431
Profile #40
quote:
Originally written by Thuryl:

#2: You're given a choice between two envelopes, A and B. One has X amount of money, the other has 2X amount of money. You pick envelope A. It has $20,000 in it. Then you're given an option to switch to the other, unknown envelope, which contains either double or half the money of the one you picked. Since we established in situation #1 that you stand to gain by switching, you switch.

But wait! The logic applies to your current situation as well; the unknown envelope you're holding could just as well contain X and the envelope you originally picked could contain 2X, so you still stand to either double or halve your money by switching. This line of argument would seem to support switching back to your original envelope, but obviously switching between the two envelopes ad infinitum would be silly, so did you really stand to gain anything by switching in the first place?

Thanks, Thuryl; I think that's the one. And the measure issue was discussed when I heard it; it did seem to be a weak point. I think I might remember some added complication, but probably I'm wrong or it wasn't important.

I am still wondering whether this is really resolved. For instance I could imagine playing the game with, say, Bill Gates. I would really only expect to be able to draw any conclusions based on the amount found in the first envelope, if I found a fair fraction of a billion dollars there. Anything less than that, and the relevant issue is simply how frivolous he's feeling that day, which I would have no way of knowing at all. (Okay, actually from what I've heard Bill rarely feels frivolous to any degree, but imagine a different Bill Gates.)

So in this case the measure issue doesn't really seem so relevant. If I find $20,000, or $200,000, I just don't see that there's any appreciable information to be extracted. Certainly not in comparison to the 200% vs. 50% payoff factor. So I think there must instead be a problem with minmax, probably related to the fact that this is a one-shot game.

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We're not doing cool. We're doing pretty.
Posts: 3335 | Registered: Thursday, September 4 2003 07:00
Electric Sheep One
Member # 3431
Profile #41
Another little statistical puzzle. I once heard the following theory attributed to George Bernard Shaw. Suppose the average probabilities of a baby being born male or female are equal (and to forestall pointless kibitzing, let both be 50% for the sake of this discussion). Nevertheless (the theory argues) there end up being more girls born than boys, because many families want to have at least one boy, so they keep having children until they get one, resulting in disproportionately many families with several sisters and one baby brother.

The easy version of the puzzle, which is the one I heard, is to show that this theory couldn't possibly work. The slightly harder version, which I believe I invented, is to show that it easily could.

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We're not doing cool. We're doing pretty.
Posts: 3335 | Registered: Thursday, September 4 2003 07:00
...b10010b...
Member # 869
Profile Homepage #42
quote:
Originally written by Student of Trinity:

So in this case the measure issue doesn't really seem so relevant. If I find $20,000, or $200,000, I just don't see that there's any appreciable information to be extracted. Certainly not in comparison to the 200% vs. 50% payoff factor. So I think there must instead be a problem with minmax, probably related to the fact that this is a one-shot game.
Yeah, it was my feeling that there was more to it than my analysis captured. After all, even in the repeated game, the amount of money you currently have gives you information -- it tells you when you have enough money and should stop playing. (Since the correct strategy in practice is obviously going to be "play until you have enough money", rather than "play forever".)

quote:
Another little statistical puzzle. I once heard the following theory attributed to George Bernard Shaw. Suppose the average probabilities of a baby being born male or female are equal (and to forestall pointless kibitzing, let both be 50% for the sake of this discussion). Nevertheless (the theory argues) there end up being more girls born than boys, because many families want to have at least one boy, so they keep having children until they get one, resulting in disproportionately many families with several sisters and one baby brother.

The easy version of the puzzle, which is the one I heard, is to show that this theory couldn't possibly work. The slightly harder version, which I believe I invented, is to show that it easily could.
The fact that the chance of a randomly-selected couple conceiving a boy as their next child is 50% does not imply that the chance of a boy is 50% for every particular couple. Am I on the right track? :P

[ Tuesday, August 15, 2006 01:00: Message edited by: Thuryl ]

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The Empire Always Loses: This Time For Sure!
Posts: 9973 | Registered: Saturday, March 30 2002 08:00
Warrior
Member # 6401
Profile #43
Perhaps this is what SoT meant by kibitzing, but the distinction between conception and birth is important. I'm not sure if this is true, but I heard somewhere that although more male embryos are conceived, (proportionally) more female babies are born, evening things out.

So in this scenario he stated that the chances of a male or a female baby being born were both 50%, but said nothing about conception. I don't suppose this is what he is getting at, but if lots more female babies are conceived, then inevitably more female babies will be born.

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I think this is really wonderful.
Posts: 147 | Registered: Tuesday, October 18 2005 07:00
...b10010b...
Member # 869
Profile Homepage #44
quote:
Originally written by Thin Air:

Perhaps this is what SoT meant by kibitzing, but the distinction between conception and birth is important. I'm not sure if this is true, but I heard somewhere that although more male embryos are conceived, (proportionally) more female babies are born, evening things out.

So in this scenario he stated that the chances of a male or a female baby being born were both 50%, but said nothing about conception. I don't suppose this is what he is getting at, but if lots more female babies are conceived, then inevitably more female babies will be born.

You just performed a reductio ad absurdum on your own argument. Think about it. :P

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The Empire Always Loses: This Time For Sure!
Posts: 9973 | Registered: Saturday, March 30 2002 08:00
The Establishment
Member # 6
Profile #45
quote:
This is a comment mistake in stating the puzzle, leading to long, idiotic arguments like this one. Before you argue further, PLEASE read the wikipedia page, which discusses this error, and others.
I've already read it. I see I forgot to include the sentence in my original statement, thought I had, oops. It's now fixed.

[ Tuesday, August 15, 2006 04:48: Message edited by: *i ]

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Your flower power is no match for my glower power!
Posts: 3726 | Registered: Tuesday, September 18 2001 07:00
Warrior
Member # 6401
Profile #46
Originally written by Thuryl:
quote:
You just performed a reductio ad absurdum on your own argument. Think about it. :P
That is entirely possible.

My main point was just that SoT was talking about the chances of a baby being born, while you were talking about the chances of a baby being conceived. They're not the same thing.

Maybe in reality more male babies are conceived and proportionally more female babies are born, but SoT didn't state this in his scenario. If for some reason more female babies are conceived in that case, that might explain why more female babies are born (even though both genders have the same chance of being born once they are conceived).

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I think this is really wonderful.
Posts: 147 | Registered: Tuesday, October 18 2005 07:00
Law Bringer
Member # 335
Profile Homepage #47
[Edit:
quote:
Suppose the average probabilities of a baby being born male or female are equal
Nothing about conception in there. There could be evil alien embryos too and it would have no effect.]

If we consider male babies the stop codon of reproduction, here are the choices:

50%: 1 male
25%: 1 female, 1 male
12.5%: 2 female, 1 male
6.25%: 3 female, 1 male

And so on. Anyone can drop dead at any time, so there will be a few families with females and no males.

You get a disproportionate number of families with several female sisters and one brother, but you get far more males than females overall. If you don't consider male babies a perfect form of birth control then you also have a larger (non-zero) chance of getting, for instance, 2 males and 1 female or whatever, which reduces the probability of girl-heavy families.

—Alorael, who again inflicted poor wording but probably correct thinking upon Spiderweb. It was more or less asked for.

[ Tuesday, August 15, 2006 07:17: Message edited by: My Soul For Guilt ]
Posts: 14579 | Registered: Saturday, December 1 2001 08:00
Post Navel Trauma ^_^
Member # 67
Profile Homepage #48
Another one to think about:
Keep tossing a coin until it comes up tails, then stop. You win $2^(number of heads)
ie your winnings in each case look like:
T : $1
HT: $2
HHT: $4
HHHT: $8
etc...

How much would you be willing to pay to play this game?

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Barcoorah: I even did it to a big dorset ram.

desperance.net - Don't follow this link
Posts: 1798 | Registered: Thursday, October 4 2001 07:00
Law Bringer
Member # 335
Profile Homepage #49
In theory, as much as I have. I think my theory is unworkable.

—Alorael, who would probably be willing to pay a dollar in practice. That way he can be sure of a profit.
Posts: 14579 | Registered: Saturday, December 1 2001 08:00

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