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[jonts26]

Since this is getting nowhere, I suggest people entertain themselves with a particularly mind wrinkling stats puzzle.

http://scienceblogs.com/evolutionblog/2011/11/the_tuesday_birthday_problem.php

[DStu]

Maybe we can discuss the card problem once this is solved, because it's a different problem, and it does not really help.

So back to the Estates. So we have 2 Estates and 3 Coppers left. Do you agree that each of the
EECCC
ECECC
ECCEC
ECCCE
CEECC
CECEC
CECCE
CCEEC
CCECE
CCCEE

is equaly likely for the last 5 cards? If not why not? And there happens to be just 4 out of 10 setups where the Estate is in the last position...

[timchen]

BUT - You know that it will be 0% of the time, since you have 3 of 3 in hand. This only holds true, however, because you've revealed 100% of the remaining.

This. I thought this is about the only argument you can make. But seriously:

Suppose we have this deck with 7 coppers and 3 estates. We draw the first hand to contain 1 estate.
 
According to you, the chance for the next card in the remaining 5 to be an estate is 30%. At least for us.

Suppose there is a guy who just came along. We told him that there are 3 coppers and 2 estates in the remaining 5- card deck. What would he say about the chance for the first card among the 5 to be an estate? 40% without question.

Why and how can our answers be different?

[Fabian]

Galzria,

Just want to make sure I'm getting your argument, please answer the following!

1. I shuffle my deck and draw my starting hand. My starting hand is 3 Estate and 2 Copper. Is the probability of me having an Estate on the bottom of my library a) 0% b) 30% c) 40% d) other?

The odds that the bottom card is an Estate IS still 30% - BUT - You know that it will be 0% of the time, since you have 3 of 3 in hand. This only holds true, however, because you've revealed 100% of the remaining. Thusly:

2. I shuffle my deck and draw my starting hand. My starting hand is 2 Estate and 3 Copper. Is the probability of me having an Estate on the bottom of my library a) 0% b) 30% c) 40% d) other?

The odds that the bottom card is an Estate IS still 30% - BUT - In this case, intuitively, 1 in 5 remain to be an Estate, so it would SEEM to be 20%, but it isn't, because it's initial probability hasn't changed. Until you reach 0% or 100%, it remains 30% because that was the information the system was given to start. EVERY card is 30% until all are revealed. That is, even when you can know 100% of the time, if it IS or ISN'T, IT'S odds are still 30%.

3. I shuffle my deck and draw my starting hand. My starting hand is 1 Estate and 4 Copper. Is the probability of me having an Estate on the bottom of my library a) 0% b) 30% c) 40% d) other?

As above, The odds of the bottom card is an Estate IS still 30% - BUT - In this case, intuitively, 2/5 remain to be an Estate, so it would SEEM to be 40%, but it isn't.

-- Either way! -- I can't provide the proof right now, so I'm happy to abstain and let it stand with the intuitive answer until I can.

This is more than I could have hoped for, thanks!

[GendoIkari]

The odds that the bottom card is an Estate IS still 30% - BUT - You know that it will be 0% of the time, since you have 3 of 3 in hand. This only holds true, however, because you've revealed 100% of the remaining. Thusly:

Wait, what? I'm pretty sure you're using the word "odds" here differently than everyone else. If something happens 0% of the time, then the odds of it happening are 0%. You can't say that the odds are 30%, but it will happen 0% of the time. That's not a paradox; that's just defining the word "odds" differently.

[GendoIkari]

Since this is getting nowhere, I suggest people entertain themselves with a particularly mind wrinkling stats puzzle.

http://scienceblogs.com/evolutionblog/2011/11/the_tuesday_birthday_problem.php

Nooooooo, not the Tuesday Birthday Problem!!!

[Galzria]

For the record, I am now +1ing all of the exceptional trolling in this thread for it's exceptionalness.

To be fair, I wasn't intentionally trolling, and if I could find this stupid proof, it wouldn't have gotten so out of hand. I've also tried to back down until I CAN, as it's as obvious to me as it is to you and anybody else that the right answer APPEARS to be 40%. It's the nature of the problem.

Without the math and proof behind it, I'm more than happy to let it go. I thought I knew where the material was, and I was wrong, so now am stuck in an awkward position of having done this twice academically, and yet not being able to provide more information.

If I can uncover where I put it, I'll be happy to come back with that information.

-- Either way! -- I can't provide the proof right now, so I'm happy to abstain and let it stand with the intuitive answer until I can.

... which will be, if you want to provide a correct proof, ONCE AND FOR ALL!

That's rather uncalled for. Fair enough that I havn't been able to provide a proof for my reasoning yet, but I've done my best to explain it in Lamens terms without it. I've also acknowledged that at this time, I'm willing to let it go with 40% until I can. It took me many weeks of staring at the solution to believe it, because yes, it is VERY unintuitive. Still, I havn't been rude about my reasons for my suggestion.

[GendoIkari]

Galzria, the thing is you have already stated that "your odds of winning are > 50%." You've stated that certain things will happen a certain percentage of the time. (Like having an Estate on the bottom if you draw 3 Estates in your hand). These percentages are what everyone here is talking about when we are talking about "odds." The "odds of an Estate being on the bottom, given that you have 3 Estates in hand" is the same as saying "if you were to play 1000 opening hands, and consider only the ones where you start with 3 Estate in hand, how many of them will have an Estate on the bottom?" The answer is, as you would agree, 0. None of those games would have an Estate on the bottom.

Now, there are a lot of other games in there (about 30%) where you would have an Estate on the bottom, but those games aren't being considered. They are irrelevant and have no bearing on the question at hand. It's a different question completely. No one here is asking about the odds of an Estate being on the bottom. We are asking "if you start with 3 Estates in hand, and do this a whole bunch of times, how many of those times will you have an Estate on the bottom?"

Read my post about the Monty Hall problem. This is different than that simply because there was a chance that you would have drawn less than 3 Estates in hand, but you didn't. With Monty Hall, there was NOT a chance that the door revealed would have been the car. Monty knows where the car is, and opens a non-car door every time. Basically, Monty is not giving ANY new information to you at all!  You already knew that at least 1 of the 2 doors you didn't choose had a goat, and, well, that's still all you know. No new information was given. If Monty hadn't know where the car was, and just picked a door at random, then it's a different story. Now new information has been given; because a door that originally had a chance to have a car now doesn't have a car.

*Edit* Just saw that Jont's link addresses this with Monty Hall:

The classic example of this is the Monty Hall problem. (I shall assume you are familiar with this problem. If you are not, I know a good book you should read.) The common fallacy is to ignore what we know about how Monty makes decisions. Thus, when he opens an empty door we tend to think, mistakenly, that we have only learned that that door is empty. In reality we have learned that Monty, who makes his decisions in a rigidly controlled way, chose to take a certain action.

*Changed 1 Estate in hand to 3 Estates in hand because it better illustrates the point, without changing the discussion.

[DG]

If you've got a bag of red and black dogs, and draw out a blue dog, should you give it to Monty Hall?

[Kuildeous]

If you've got a bag of red and black dogs, and draw out a blue dog, should you give it to Monty Hall?

I can't answer that. I need to know if the blue dog's birthday is on a Tuesday.

[WanderingWinder]

If you've got a bag of red and black dogs, and draw out a blue dog, should you give it to Monty Hall?

Do you have an orange goat in another bag, or is it only a pink elephant?

[DStu]

To be fair, I wasn't intentionally trolling, and if I could find this stupid proof, it wouldn't have gotten so out of hand. I've also tried to back down until I CAN, as it's as obvious to me as it is to you and anybody else that the right answer APPEARS to be 40%. It's the nature of the problem.

If you are talking about the proof for your card game, and the best strategy gives you 50% winchance:


Obviously there exists a strategy with p=0.5, namely just guess "black" at the first turn.
Now every other strategy that does not guess "black" on the first turn does not have higher winchance:
Every strategy that don't guess "black" on the first turn, loses with prob. 50% already after the first step, namely if the card is black. So no matter what you do on later turns, you can't do better than 50%.
qed

[Kuildeous]

If I can find the information, I'll bring it out, but for now, follow the intuitive answer of %40 - However, it's still really %30. 

Sadly, my own math is rusty. I recognize that it's a conditional probability question. I recognize that the answer is not 30%. Like you, I do not personally have the math handy to show this (though others have provided excellent dissertations).

I will say that hard numbers show that it's 40%. I created a little simulation in Excel. Out of a batch of 30k draws, the simulation picked out how many draws had exactly 4 Coppers and had an Estate at the bottom. The percentage was indeed 40%. Granted, while my original sample size was 30k, the sample size of first hands that contain exactly 1 Estate ranged between 12k and 13k. Still, pretty good sample size. Every time I generated new data based on the randomizer, the final number was about 40%.

In fact, I expanded it to consider other assumptions.

If the first hand contains 5 Copper: 60% occurrence of bottom card being Estate.
If the first hand contains 4 Copper: 40% occurrence of bottom card being Estate.
If the first hand contains 3 Copper: 20% occurrence of bottom card being Estate.
If the first hand contains 2 Copper: 0% occurrence of bottom card being Estate (seemed silly to test this, but I like to double-check my code).

So, I'm sorry to say that your claim of 30% does not hold up to actual card draws.

I don't doubt that you thought you had the material to back this up. As Gendo pointed out, this is not exactly like the Monty Hall problem. It sounds similar: You may have read something that sounds similar, but it really isn't, and conditional probability is the culprit.

[Fabian]

Kuildeous, here's the actual numbers though, proof here.

If the first hand contains 5 Copper: 30% occurrence of bottom card being Estate.
If the first hand contains 4 Copper: 30% occurrence of bottom card being Estate.
If the first hand contains 3 Copper: 30% occurrence of bottom card being Estate.
If the first hand contains 2 Copper: 0% occurrence of bottom card being Estate (odds are still 30% though)

[Kuildeous]

Not to belabor the point (too late), but I think this illustration might help shed some light.

I have a simulation with 30,057 random card draws. Of those 30,057, there are 9,110 results of the bottom card being an Estate, yielding 30.3%. This is to be expected. So far, so good.

A) Of those 30,057, there are 2,570 results where the first hand has 0 Estates. Of those 2,570, there 1,536 results of the bottom card being an Estate, yielding 59.8%. This coincides with an earlier claim that the probability is 60%.

B) Likewise, there are 12,464 results where the first hand has 1 Estate. Of those 12,464, there are 5,102 results of the bottom card being an Estate, yielding 40.9%. Again, this coincides with the claim that the probability is 40%.

C) There are 12,484 results where the first hand has 2 Estates. Of that, there are 2,472 results of the bottom card being an Estate, yielding 19.8%.

D) Naturally, 0% of the 2,539 results where the first hand has 3 Estates had an Estate on the bottom.

This is where the condition is important. If you add all these up (probability of A times the sample size of A and so forth), then you get the final number of 30%. It is absolutely true that the last card (indeed any one card) being an Estate is 30% when you don't know anything about the deck. Once you know the first hand, then you know that you are either in case A, B, C, or D. The probability that the final card of a fresh shuffle being 30% is still there, but you are now no longer looking at a fresh shuffle. You are now looking at one of four cases. Knowledge of the first hand gives you better information.

[sjelkjd]

if I could find this stupid proof, it wouldn't have gotten so out of hand

You can't find the proof because it doesn't exist.  If you won't believe logic, maybe you'll believe a simulation:

Code: [Select]

#include <algorithm>
#include <vector>
#include <iostream>
using namespace std;

int main()
{
vector<char> cards;
for(int i=0;i<7;i++)
cards.push_back('C');
for(int i=0;i<3;i++)
cards.push_back('E');
const int TRIALS = 1000000;
int success = 0;
int total = 0;
for(int j=0;j<TRIALS;j++)
{
random_shuffle(cards.begin(), cards.end());
int eCount = 0;
for(int i=0;i<5;i++)
{
if(cards[i] == 'E') eCount++;
}
if(eCount == 1)
{
int money = 2;
for(int i=0;i<4;i++)
{
if(cards[5+i] == 'C')
{
money++;
}
}
if(money == 5)
{
success++;
}
total++;
}
}
double ratio = ((double)success)/total;
cout << ratio << endl;
}Guess what?  Answer is 40%

[eHalcyon]

Perhaps Galzria is thinking of some form of the Gambler's Fallacy.  Suppose that I flip a fair coin.  What is the chance that I flip heads?  The answer is, of course, 50%.

Now let's reword the problem: Suppose that I am flipping a fair coin.  What is the chance that the next flip will be heads?  The answer is still, of course, 50%.

Now here's the rewording that matches it up to our initial problem: Suppose that I flip a fair coin 10 times.  The first 5 flips come up TAILS.  What is the chance that the last flip will be heads?  The answer is STILL 50%.

The thing is, this doesn't apply to cards at all.  Each individual flip of the coin has a 50% chance of coming up heads.  You don't know what it is until it happens.  The same is not true of cards -- they are what they are.  If you have 5 red cards and 5 black cards, revealing 5 red guarantees that the next 5 are black.

I'm guessing that Galzria is remembering a proof for some concept that doesn't actually apply to the question at hand, though at first glance it might seem relevant.

[jonts26]

Perhaps Galzria is thinking of some form of the Gambler's Fallacy.  Suppose that I flip a fair coin.  What is the chance that I flip heads?  The answer is, of course, 50%.

Now let's reword the problem: Suppose that I am flipping a fair coin.  What is the chance that the next flip will be heads?  The answer is still, of course, 50%.

Now here's the rewording that matches it up to our initial problem: Suppose that I flip a fair coin 10 times.  The first 5 flips come up TAILS.  What is the chance that the last flip will be heads?  The answer is STILL 50%.

The thing is, this doesn't apply to cards at all.  Each individual flip of the coin has a 50% chance of coming up heads.  You don't know what it is until it happens.  The same is not true of cards -- they are what they are.  If you have 5 red cards and 5 black cards, revealing 5 red guarantees that the next 5 are black.

I'm guessing that Galzria is remembering a proof for some concept that doesn't actually apply to the question at hand, though at first glance it might seem relevant.

The difference between the cards and the coin is that the coin flip results are independent of each other, whereas the card flips are not. Past results will influence future ones. Which is what you said, I just wanted to make sure we had proper terminology.

[Robz888]

This reminds me of the Lightning Town problem (I don't know what other people call it--I call it the Lightning Town problem).

You have a town that experiences weirdly frequent lightning storms. The storms occur randomly, but on average, once a week. There is a lightning storm today (Tuesday). What day is the most likely day of the next storm?

Tomorrow (Wednesday). Most people incorrectly say either a week from today, or every day is equally likely. They are equally likely to have a storm, true, but they are not equally likely to be the next day with a storm. For instance, for the day after tomorrow (Thursday) to be the next day with a storm, there would have to be no storm on Wednesday. For Friday to be the next day with a storm, there would have to be no storm on Thursday or Wednesday, which is even less likely. So we discover that tomorrow is the next most likely day for a lightning storm, because there is the least amount of time for a storm to interrupt what you could call the next-day chain.

[Galzria]

This reminds me of the Lightning Town problem (I don't know what other people call it--I call it the Lightning Town problem).

You have a town that experiences weirdly frequent lightning storms. The storms occur randomly, but on average, once a week. There is a lightning storm today (Tuesday). What day is the most likely day of the next storm?

Tomorrow (Wednesday). Most people incorrectly say either a week from today, or every day is equally likely. They are equally likely to have a storm, true, but they are not equally likely to be the next day with a storm. For instance, for the day after tomorrow (Thursday) to be the next day with a storm, there would have to be no storm on Wednesday. For Friday to be the next day with a storm, there would have to be no storm on Thursday or Wednesday, which is even less likely. So we discover that tomorrow is the next most likely day for a lightning storm, because there is the least amount of time for a storm to interrupt what you could call the next-day chain.

We always did that as Popquiz:


Teacher:Sometime next week (Monday-Friday) will be a popquiz, but I'm not going to tell you when, because I don't want you to put off studying until the night before.
Student 1: Well, it can't be Friday then, because if it hadn't happened on Monday through Thursday, we would KNOW it was Friday, and we would study Thursday night.
Student 2: And if it can't be Friday, then we KNOW it can't be Thursday, because if we havn't had it by Wednesday afternoon, we would KNOW to study that night.
Student 3: ...
Class: Our popquiz therefor MUST be on Monday, so we should study this weekend.

[Axxle]

This reminds me of the Lightning Town problem (I don't know what other people call it--I call it the Lightning Town problem).

You have a town that experiences weirdly frequent lightning storms. The storms occur randomly, but on average, once a week. There is a lightning storm today (Tuesday). What day is the most likely day of the next storm?

Tomorrow (Wednesday). Most people incorrectly say either a week from today, or every day is equally likely. They are equally likely to have a storm, true, but they are not equally likely to be the next day with a storm. For instance, for the day after tomorrow (Thursday) to be the next day with a storm, there would have to be no storm on Wednesday. For Friday to be the next day with a storm, there would have to be no storm on Thursday or Wednesday, which is even less likely. So we discover that tomorrow is the next most likely day for a lightning storm, because there is the least amount of time for a storm to interrupt what you could call the next-day chain.

We always did that as Popquiz:


Teacher:Sometime next week (Monday-Friday) will be a popquiz, but I'm not going to tell you when, because I don't want you to put off studying until the night before.
Student 1: Well, it can't be Friday then, because if it hadn't happened on Monday through Thursday, we would KNOW it was Friday, and we would study Thursday night.
Student 2: And if it can't be Friday, then we KNOW it can't be Thursday, because if we havn't had it by Wednesday afternoon, we would KNOW to study that night.
Student 3: ...
Class: Our popquiz therefor MUST be on Monday, so we should study this weekend.

And then she has it Friday after you guys study all weekend.  There's some fallacy here.

[Galzria]

This reminds me of the Lightning Town problem (I don't know what other people call it--I call it the Lightning Town problem).

You have a town that experiences weirdly frequent lightning storms. The storms occur randomly, but on average, once a week. There is a lightning storm today (Tuesday). What day is the most likely day of the next storm?

Tomorrow (Wednesday). Most people incorrectly say either a week from today, or every day is equally likely. They are equally likely to have a storm, true, but they are not equally likely to be the next day with a storm. For instance, for the day after tomorrow (Thursday) to be the next day with a storm, there would have to be no storm on Wednesday. For Friday to be the next day with a storm, there would have to be no storm on Thursday or Wednesday, which is even less likely. So we discover that tomorrow is the next most likely day for a lightning storm, because there is the least amount of time for a storm to interrupt what you could call the next-day chain.

We always did that as Popquiz:


Teacher:Sometime next week (Monday-Friday) will be a popquiz, but I'm not going to tell you when, because I don't want you to put off studying until the night before.
Student 1: Well, it can't be Friday then, because if it hadn't happened on Monday through Thursday, we would KNOW it was Friday, and we would study Thursday night.
Student 2: And if it can't be Friday, then we KNOW it can't be Thursday, because if we havn't had it by Wednesday afternoon, we would KNOW to study that night.
Student 3: ...
Class: Our popquiz therefor MUST be on Monday, so we should study this weekend.

And then she has it Friday after you guys study all weekend.  There's some fallacy here.

Of course. That's a bit of the point, but as with the Lightning Tower, the thought process that brings you to the answer of Tomorrow/Monday is the same. In truth, the teacher is NOT randomly choosing a day once per week. But it's the same logic. The longer you go, the more likely it is to occur the following day. Since she's looking for a surprise, she's looking for the smallest chance of knowing when it'll be.

[toaster]

The pop quiz is actually a different problem...it's about game theory or logic more than probability, and the reasoning behind it is quite different.  The most common set up for it is as the Paradox of the Unexpected Hanging.

[Donald X.]

The difference between the cards and the coin is that the coin flip results are independent of each other, whereas the card flips are not. Past results will influence future ones. Which is what you said, I just wanted to make sure we had proper terminology.

That terminology is "hypergeometric distribution." There's a wikipedia article on it and everything.

Also: You are standing on one side of a river, with a fox, a goat, and a basket of cabbages. One of them always lies, one always tells the truth, and one is a basket of cabbages. You have to determine which is which using only one weighing on a scale.

[Galzria]

The pop quiz is actually a different problem...it's about game theory or logic more than probability, and the reasoning behind it is quite different.  The most common set up for it is as the Paradox of the Unexpected Hanging.

Yes, that would be it almost exactly. And like I said, they aren't the same problem. But the idea behind the prisoners hanging, or popquiz, was to illustrate a point, disregarding the fallacies of the logic. The key being "surprise", rather than semi-controlled randomness. It was still designed to get you to "the most likely day is tomorrow".

But following that logic, the Lightning Tower is a cleaner problem, because it erases the unknown factor. It still brings you to the most likely day, because anything after would be less probable. And that's, at the basic level, what the Prisoner's Hanging / Popquiz was trying to show. It just leaves too many invariables, and a loose definition "surprise". It draws out to conclusions that can't be made with certainty.