Probability problem

[Brad Johnson]

By context I mean that the problem, as presented, makes you think in the incorrect context.  The presentation makes you think in terms of solving for the second cup-choice probability when the actual question is one of overall probability based on the original three-cup configuration.  Attorneys use the same concept in courtrooms, adding secondary information to questions in order to confuse the context.

If you want to be painfully technical about it, there are actually three seperate probabilities to be calculated.  First is the probability you will choose the right cup from the original three on the first pick (1/3).  Second is the probability you will choose the right cup from the remining two on the second pick (1/2).  Third is the probability you will choose the right cup on the second pick, but from the original three choices (2/3). 

Brad

[CNYCacher]

First is the probability you will choose the right cup from the original three on the first pick (1/3).

Correct.

Second is the probability you will choose the right cup from the remining two on the second pick (1/2).

No.  Even though there are two cups, they are not evenly weighted because the person running the game knows where the $ is and they have imparted bias into the results by removing a sure loser.  The cup which you did not pick now represents the odds which were originally attributed to the two cups you did not pick.

Third is the probability you will choose the right cup on the second pick, but from the original three choices (2/3). 

[Brad Johnson]

No.  Even though there are two cups, they are not evenly weighted because the person running the game knows where the $ is and they have imparted bias into the results by removing a sure loser.  The cup which you did not pick now represents the odds which were originally attributed to the two cups you did not pick.

You're combining the first and second picks.  Again, I am breaking the problem into three exclusive calculations based on the choices available for that pick and that pick only - first pick probability (1/3), second pick probability (1/2), and an overall pick probability (2/3).

Brad

[CNYCacher]

You're combining the first and second picks.  Again, I am breaking the problem into three exclusive calculations based on the choices available for that pick and that pick only - first pick probability (1/3), second pick probability (1/2), and an overall pick probability (2/3).

Brad

You have to combine them.  There is a 2/3 probability the ball is under Cup B and a 1/3 probability that it is under Cup A.  You can't look at that situation and say "Two cups, 50-50!"

[AZRedhawk44]

Examination in my own little melon:

-The game-master can't take your cup, because you reserved it.
-The game-master can't take the money cup, because that violates the game.
-The game-master must take one of the incorrect cups.  He has a 33% chance of having 2 incorrect cups to choose from (since you chose the money cup), and a 66% chance of having 1 incorrect cup to choose from (since you chose a bad cup and he can't remove the money cup).

So, 33% of the time if you change... you lose.
66% of the time if you change, you win.

[CNYCacher]

I love this problem because it is so polarizing.  I'll admit that I argued against the correct answer quite vigorously when the problem was first presented to me.  The scenario which eventually got my brain to snap into the proper mindset was imagining the situation with 1000 cups.  You choose #1.  The game master then removes 998 losing cups leaving you with your original pick and #682

[AZRedhawk44]

A similar problem I heard in middle school:


You're walking down a road on a journey to a city, and it splits into two paths.  You don't know which path takes you to the city.  On each path, there's a guide.  One of the guides is incapable of telling a lie, and one of the guides is incapable of telling the truth... but you don't know which one.

You're allowed to ask 1 of the guides, 1 question.  That's it.

What question can you ask that guarantees you get a useful answer that takes you to the city?

[Scout26]

"What path will the other guide tell me to take ?"  And then take the other. 

[Brad Johnson]

What is your favorite color?

Brad

[Chuck Dye]

For the search inclined, try "price is right probability."

[Perd Hapley]

By context I mean that the problem, as presented, makes you think in the incorrect context.  The presentation makes you think in terms of solving for the second cup-choice probability when the actual question is one of overall probability based on the original three-cup configuration. 

As far as I can see, the presentation is about as clear and simple and straightforward as can be. How would you present it more transparently?

If you want to be painfully technical about it, there are actually three seperate probabilities to be calculated.  First is the probability you will choose the right cup from the original three on the first pick (1/3).  Second is the probability you will choose the right cup from the remining two on the second pick (1/2).  Third is the probability you will choose the right cup on the second pick, but from the original three choices (2/3). 

I think we can all agree on that first one. The third doesn't seem to make sense here, since the second pick isn't a choice between three cups; it's a choice between two. I think the second one is correct, if you are asking about the odds that a person will choose correctly on the second stage of the exercise. After all, there are six possible scenarios, and a correct choice will be made in three of those.* But Iain didn't ask, "What are the odds that you'll pick the right one?" He asked how holding or changing would affect your odds.


Finally, you seemed to disagree with my statement about the 2/3 probability that the 20 is in the other cup. Could you explain?
 


*I pick cup #1, it's in #1, I stay with #1, I win.
  I pick cup #2, it's in #1, I stay with #2, I lose.
  I pick cup #3, it's in #1, I stay with #3, I lose.
  I pick cup #1, it's in #1, I change cups, I lose.
  I pick cup #2, it's in #1, I change cups, I win.
  I pick cup #3, it's in #1, I change cups, I win.

[Perd Hapley]

If the above table is unsatisfactory, then let's do it this way:

  I pick the right cup, I stay with that cup, I win.
  I pick the wrong cup, I stay with that cup, I lose.
  I pick the wrong cup, I stay with that cup, I lose.
  I pick the right cup, I change cups, I lose.
  I pick the wrong cup, I change cups, I win.
  I pick the wrong cup, I change cups, I win.

[Iain]

I've got into an argument elsewhere now. Someone has proposed the following as an extreme statement of the problem, to make it clearer. However, I think it confuses, and isn't clear in its wording, which is everything:

Or ramp up the numbers, you are at ManU's next home game with 75,000 others. One has a disc in his / her pocket and it you can guess who you win a million quid.

You have no idea who it is, so you pick one person totally at random.

After the game 74,998 people leave. The remaining two being the person you picked and one other person. You are then assured that all the people who left DID NOT have the disc. So the disc remains.

The chances of your original choice having it are 75,000 to 1.

I reckon you should swop

What needs to be stated is whether or not someone is controlling those who leave. If not, then it's massively unlikely that one of the two left has the disc, so if that happens it makes no difference if you stick or not.

The Monty Hall problem is different if the host doesn't know where the car is, and randomly opens a door to reveal a goat (or lifts a cup to reveal nothing).

[GigaBuist]

Someone has proposed the following as an extreme statement of the problem, to make it clearer.

His explanation would be easier to follow if he didn't make up sports teams or borrow monetary values from Harry Potter.

[Hawkmoon]

What needs to be stated is whether or not someone is controlling those who leave. If not, then it's massively unlikely that one of the two left has the disc, so if that happens it makes no difference if you stick or not.

It doesn't make any difference if someone "controlled" who left. All that matters is that 74,998 people left, and none of them had the disk; two people remain and one of them has the disk. This is not a statement of probability, it is a statement of fact. That the disk is in the possession of one of the two people remaining.

I don't think the fact you randomly selected one of the two people who remain affects the ultimate probability. In the end, you have two people from whom to chose.

Same applies to the cups. In your first choice, you had a 1:3 chance. Then one sure loser was removed, so you now face two cups. Your chance is now 1:2. The question was: Are your odds improved by holding or changing? My view is that to hold or to change is now a choice between two cups, so now either decision has a 1:2 chance of being correct. Thus, your odds are neither improved nor reduced, regardless of which choice you make. Or, if you want a one word answer to the question as posed: "Yes."

[AZRedhawk44]

"What path will the other guide tell me to take ?"  And then take the other. 

Yup.  That was easy.


How about this one?


There are two envelopes, one contains twice as much money as the other. You pick one at random and find $100 inside. What is the "expected value" of the amount in the other envelope? Given the option, should you switch? Does it matter how much you found in that first envelope? Does it matter if you open the envelope?

[Viking]

I have a theory, but I won't post it, because I'm not sure of it, but I still can't shake it. And you'll laugh at me afterwards, explaining why I fail logic forever.

[GigaBuist]

It doesn't make any difference if someone "controlled" who left.

It matters very much.  They've eliminated 74,998 known false answers.  The odds of you pick it correctly the first time is 1:75000.  The odds that the remaining disc is in the other choice is 74999:75000

Same applies to the cups. In your first choice, you had a 1:3 chance. Then one sure loser was removed, so you now face two cups. Your chance is now 1:2. The question was: Are your odds improved by holding or changing? My view is that to hold or to change is now a choice between two cups, so now either decision has a 1:2 chance of being correct. Thus, your odds are neither improved nor reduced, regardless of which choice you make. Or, if you want a one word answer to the question as posed: "Yes."

That's not how it works.  Experiments have been done and simulations too.  Your odds increase when you pick the other choice once the known false answers are removed.

[AZRedhawk44]

Code for the OP problem if you keep your original choice... run this in SQL Server.

Code: [Select]

if object_id('results') is not null
begin
drop table results
end

create table results (test_iteration int, win_fail char(1))

declare @i int
set @i = 1

while @i < 100000
begin

declare @cup1 int
declare @cup2 int
declare @cup3 int
declare @rand_dealer decimal(7, 6)
declare @rand_player decimal(7, 6)
declare @player_choice int

set @rand_dealer = rand()
set @rand_player = rand()

if @rand_dealer between 0 and 0.333333
begin
set @cup1 = 1
set @cup2 = 0
set @cup3 = 0
end

if @rand_dealer between 0.333334 and 0.666666
begin
set @cup1 = 0
set @cup2 = 1
set @cup3 = 0
end

if @rand_dealer between 0.666667 and 1
begin
set @cup1 = 0
set @cup2 = 0
set @cup3 = 1
end

if @rand_player between 0 and 0.333333
begin
set @player_choice = 1
end

if @rand_player between 0.333334 and 0.666666
begin
set @player_choice = 2
end

if @rand_player between 0.666667 and 1
begin
set @player_choice = 3
end

insert into results
select @i, case when @cup1 = 1 and @player_choice = 1 then 'W'
when @cup2 = 1 and @player_choice = 2 then 'W'
when @cup3 = 1 and @player_choice = 3 then 'W'
else 'F' end as test_result

print @i

set @i = @i + 1
end
Then to analyze...

Code: [Select]

select win_fail, count(*)
from results
group by win_fail
I just ran it and got 33398 wins and 66601 fails.


Working on altering the code to always switch to a different option...

[AZRedhawk44]

With auto-switch turned on:

Code: [Select]

if object_id('results') is not null
begin
drop table results
end

create table results (test_iteration int, win_fail char(1))

declare @i int
set @i = 1

while @i < 100000
begin

declare @cup1 int
declare @cup2 int
declare @cup3 int
declare @rand_dealer decimal(7, 6)
declare @rand_dealer2 decimal(7, 6)
declare @rand_player decimal(7, 6)
declare @player_choice int

set @rand_dealer = rand()
set @rand_player = rand()

if @rand_dealer between 0 and 0.333333
begin
set @cup1 = 1
set @cup2 = 0
set @cup3 = 0
end

if @rand_dealer between 0.333334 and 0.666666
begin
set @cup1 = 0
set @cup2 = 1
set @cup3 = 0
end

if @rand_dealer between 0.666667 and 1
begin
set @cup1 = 0
set @cup2 = 0
set @cup3 = 1
end

if @rand_player between 0 and 0.333333
begin
set @player_choice = 1
end

if @rand_player between 0.333334 and 0.666666
begin
set @player_choice = 2
end

if @rand_player between 0.666667 and 1
begin
set @player_choice = 3
end

/*Now change the player's choice based upon the logic that:
1.  The dealer cannot eliminate a winning cup
2.  The player must change chosen cups

*/

if @player_choice = 1
begin
if @cup1 = 1
begin
set @rand_dealer2 = rand()
if @rand_dealer2 between 0 and 0.5
begin
set @player_choice = 2
end
if @rand_dealer2 between 0.500001 and 1
begin
set @player_choice = 3
end
end
if @cup1 = 0
begin
if @cup2 = 1
begin
set @player_choice = 2
end
if @cup3 = 1
begin
set @player_choice = 3
end
end
end

if @player_choice = 2
begin
if @cup2 = 1
begin
set @rand_dealer2 = rand()
if @rand_dealer2 between 0 and 0.5
begin
set @player_choice = 1
end
if @rand_dealer2 between 0.500001 and 1
begin
set @player_choice = 3
end
end
if @cup2 = 0
begin
if @cup1 = 1
begin
set @player_choice = 1
end
if @cup3 = 1
begin
set @player_choice = 3
end
end
end

if @player_choice = 3
begin
if @cup3 = 1
begin
set @rand_dealer2 = rand()
if @rand_dealer2 between 0 and 0.5
begin
set @player_choice = 1
end
if @rand_dealer2 between 0.500001 and 1
begin
set @player_choice = 2
end
end
if @cup3 = 0
begin
if @cup1 = 1
begin
set @player_choice = 1
end
if @cup2 = 1
begin
set @player_choice = 2
end
end
end

--End auto-switch logic

insert into results
select @i, case when @cup1 = 1 and @player_choice = 1 then 'W'
when @cup2 = 1 and @player_choice = 2 then 'W'
when @cup3 = 1 and @player_choice = 3 then 'W'
else 'F' end as test_result

print @i

set @i = @i + 1
end
First run of 100k gave me 55561 wins and 44438 loses.  Way ahead of the 33k/66k even split along probability lines, but not quite on par with the 66/33 win I was expecting.

So, I ran a second batch of 100k.

I wound up with 55374 wins and 44625 loses.

Still not in line with my expectations, so I increased the sample set to 1 meeeelion [/doctor_evil] tests.

That'll take a couple minutes.

[AZRedhawk44]

Even with a million tests....

I get 554246 wins and 445753 losses.

AJ or anyone else, you see any holes in my auto-switch logic?

[Perd Hapley]

Did you try "sudo switch cups"?

[AZRedhawk44]

Did you try "sudo switch cups"?

Not Linux.

SQL.

It's like Excel, but for adults.   

[GigaBuist]

AZ, I think I found the problem:

Code: [Select]

if @cup1 = 1
begin
set @rand_dealer2 = rand()
if @rand_dealer2 between 0 and 0.5
begin
set @player_choice = 2
end
if @rand_dealer2 between 0.500001 and 1
begin
set @player_choice = 3
end
end
if @cup1 = 0
begin
if @cup2 = 1
begin
set @player_choice = 2
end
if @cup3 = 1
begin
set @player_choice = 3
end
end

Something wrong there.  If the player has selected the correct door (your first block) there's no point in randomizing their next choice.  They're going to lose if they had the right pick the first time.  In the second block you're forcing them to win if their first selection is wrong.

Or something like that.  I'm gonna try and whip up a smaller/simpler Perl script.

[GigaBuist]

OK, not as short as I'd like but I kept it rather verbose and traditionally formatted.

Editing the post to include a version that respects a 'v' as the second argument to turn on verbosity, expanded the comments a bit, and put in the "work" to find a losing second guess when the situation presents itself.

Code: [Select]

#!/usr/bin/perl

use strict;
use warnings;

my $win_first_cnt = 0;
my $win_second_cnt = 0;

my $run_cnt = shift;
my $verbose = shift || "";
my $c;

for($c = 0; $c < $run_cnt; $c++) {
    my($winner, $orig_choice, $second_choice, $cup_elim);
    my($won_first, $won_second);

    # Randomly pick the winning cup
    $winner = int(rand(3)) + 1;

    # Randomly pick the original choice
    $orig_choice = int(rand(3)) + 1;

    # find a cup to eliminate, start with the last one.
    # move back until you've found something that's not the
    # winner and not the original choice.
    $cup_elim = 3;
    while($cup_elim == $winner || $cup_elim == $orig_choice) {
        $cup_elim--;
    }

    if ($orig_choice != $winner) {
        # If the first choice wasn't the winner then their
        # second choice will be unless they already had the winner
        $second_choice = $winner;
    } else {
        # doesn't matter what you pick, they're going to lose
        # on their second choice if they already had the winner
        # but we'll seek out the losing cup anyway
        $second_choice = 1;
        while($second_choice == $orig_choice || $second_choice == $cup_elim) {
            $second_choice++;
        }
    }

    if ($orig_choice == $winner) {
        $win_first_cnt++;
        $won_first = "Y";
    } else {
        $won_first = "N";
    }

    if ($second_choice == $winner) {
        $win_second_cnt++;
        $won_second = "Y";
    } else {
        $won_second = "N";
    }
    if ($verbose eq "v") {
        print("Winner: " . $winner . " ");
        print("First: " . $orig_choice . " ");
        print("Elim: " . $cup_elim . " ");
        print("Sec: " . $second_choice . " ");
        print("WF: " . $won_first . " ");
        print("WS: " . $won_second . "\n");
    }
}
print ("Won first " . $win_first_cnt . "\n");
print ("Won second " . $win_second_cnt . "\n");
print ("Runs: " . $run_cnt . "\n");

When you tell it to run 1,000,000 simulations I get this:

Won first 333770
Won second 666230
Runs: 1000000

Run it again:

Won first 332984
Won second 667016
Runs: 1000000

Putting it into code has gotten it a bit simpler for me to explain how it works.  If you always take your 2nd guess you'll win unless you won on the first guess.  Since you've only got a 33% chance of winning with that first guess you naturally have a 66% chance with the second.