Brain Teasers

[Pauldohert]

August.

The only way to achieve this is for the monday to be the 31st and the Thursday to be the 7th.

So the current month starts on a Tuesday and must end on a thursday so that the first thursday can be the 7th. This means it must have 31 days also. Only July and August are consecutive months, in the same year, which both have 31 days therefore it must be August.

Spot on - a nice uncontraversial one.

 

[Adam Caramon]

yes but you don't know what door will be eliminated therefore you must list all possible combinations from the starting position. Above you have excluded door 3 but you don't know that door will be excluded therefore it must be included in possible outcomes of the game.

Technically, you're correct. But what you're trying to determine has no value.

Imagine the same scenario, but the host opens every door. Technically your exact same table still holds. You can look at your original choice, all the possible outcomes of switching or not switching and determine the possibilities.

But to then draw a conclusions such as "You win 2/3's of the time when you switch" is incredibly misleading because its quite obvious that the person selecting will pick the winning door.

The point in time that you can draw meaningful conclusions from the data is when the question is actually posed: "Do you want to switch?" If a door has been eliminated, probability to choosing the correct door changes.

 

[Pauldohert]

But that's just it, at the point that this question becomes viable, you do know that door three is excluded, so it doesn't have to be included in the possible outcomes.

A door can always be excluded - you know this before you start. So before you start your odds are 1/3 - and since you know nothing more about your original pick now - its odds of being coorect are still 1/3- and the remaining odds 2/3 must be on the other door.


Right now we know who the marks are - lets set the venue - to play for money.

The best way for the unconvinced to be convinced is to play the host in the real world with a contestant. Then its more obvious nothing has really changed since the original pick. And they are better of swithing - the more doors the more obvious it is.

The choice is do you think - you were right first pick. The answer is probably not, I think we are all agreed on that.

1/3 chance
1/4 chance
.....
.....
1/n chances of being right first pick.

 

[chergh]

Right now we know who the marks are - lets set the venue - to play for money.

Yeah I'll happily play this with anyone arguing about it, the only stipulation is they are not allowed to switch.

 

[ChipperT]

A local entrepeneur bought 100 pounds of strawberries for $2.00 per pound and expected to double his investment by selling the strawberries for $4.00 per pound on a convenient street corner. The seller only managed to sell 50 pounds of strawberries the first day and he sold the remainder on the second day. The fresh strawberries had a content of 99% water, but because of the hot weather, the strawberries dehydrated and contained only 98% water on the second day. How much profit did the seller make?

 

[Pauldohert]

Here's another.

At the Infinity Hotel there are an infinite number of rooms. A man walks upto reception and asks for a room. He is told there are already an infinite number of guests staying at the hotel so all the rooms are occupied.

How do they get him a room to stay in?

I dunno -

Since Infinity + 1 is a valid number - So infinity hotel gets one bigger and the guest gets to stay?

As an aside -
Parallel lines meet at infinity - I never understood that one!

 

[Pauldohert]

Yeah I'll happily play this with anyone arguing about it, the only stipulation is they are not allowed to switch.

If the stakes are high enough we can have a joint venue midway between Newbury and Liverpool.

You could play it with the witnesses next time they knock! It'll convince you they are illogical, but at the same time there is a God. I'm joking, Just joking.

 

[chergh]

The point in time that you can draw meaningful conclusions from the data is when the question is actually posed: "Do you want to switch?" If a door has been eliminated, probability to choosing the correct door changes.

If the person eliminating the door knows where the prize is and you know the history of the game the odds don't change.

If the host eliminates a door and then shuffles the doors so you don't know which one you originally picked the odds are 50/50 but as long as you know what you're original choice is then the odds don't change.

 

[Adam Caramon]

and since you know nothing more about your original pick now - its odds of being coorect are still 1/3

Wrong. Once a door is eliminated, the odds change. You know one of the choices is no longer valid. Your using what's called junk science.

Let me give you an example using your logic. I say "Pick a number between 1 and 100,000", you pick 1, then I say "the number is not greater than 2, would you like to change your guess?"

Your original odds of winning were 1 in 100,000. Since you still don't know if your original guess was correct or not, the odds are still 1 in 100,000. The other possible choice, 2, must contain all the rest of the odds. So 2 will win 99,999/100,000 times.

 

[namliam]

seeing as only losing doors are "lost" in the equation, the odds turn out to be 50%-50%

Winning	Chosen	Rmvd	Win?
1	1	2	TRUE
1	1	3	TRUE
1	2	3	FALSE
1	3	2	FALSE
2	1	3	FALSE
2	2	1	TRUE
2	2	3	TRUE
2	3	1	FALSE
3	1	2	FALSE
3	2	1	FALSE
3	3	1	TRUE
3	3	2	TRUE

6 wins - 6 loses


If you make it 4 doors, removing two duds.... 12 wins, 11 loses (52% win)

Winning	Chosen	Rmvd	Rmvd 2	Win?
1	1	2	3	TRUE
1	1	3	4	TRUE
1	1	2	4	TRUE
1	2	3	4	FALSE
1	3	2	4	FALSE
1	4	2	3	FALSE
2	1	3	4	FALSE
2	2	1	3	TRUE
2	2	3	4	TRUE
2	2	1	4	TRUE
2	3	1	4	FALSE
2	4	1	3	FALSE
3	1	2	4	FALSE
3	2	1	4	FALSE
3	3	1	2	TRUE
3	3	2	4	TRUE
3	3	1	4	TRUE
4	1	2	3	FALSE
4	2	1	3	FALSE
4	3	1	2	FALSE
4	4	1	2	TRUE
4	4	2	3	TRUE
4	4	1	3	TRUE

However 5 Doors, 15 wins - 20 loses (43% win)

Winning	Chosen	Rmvd	Rmvd 2	Removed 3	Win?
1	1	2	3	4	TRUE
1	1	3	4	5	TRUE
1	1	2	4	5	TRUE
1	2	3	4	5	FALSE
1	3	2	4	5	FALSE
1	4	2	3	5	FALSE
1	5	2	3	4	FALSE
2	2	2	3	4	TRUE
2	2	3	4	5	TRUE
2	2	2	4	5	TRUE
2	1	3	4	5	FALSE
2	3	2	4	5	FALSE
2	4	2	3	5	FALSE
2	5	2	3	4	FALSE
3	3	2	3	4	TRUE
3	3	3	4	5	TRUE
3	3	2	4	5	TRUE
3	2	3	4	5	FALSE
3	1	2	4	5	FALSE
3	4	2	3	5	FALSE
3	5	2	3	4	FALSE
4	4	2	3	4	TRUE
4	4	3	4	5	TRUE
4	4	2	4	5	TRUE
4	2	3	4	5	FALSE
4	3	2	4	5	FALSE
4	1	2	3	5	FALSE
4	5	2	3	4	FALSE
5	5	2	3	4	TRUE
5	5	3	4	5	TRUE
5	5	2	4	5	TRUE
5	2	3	4	5	FALSE
5	3	2	4	5	FALSE
5	4	2	3	5	FALSE
5	1	2	3	4	FALSE

6 doors, 24 wins, 30 loses (44%)
At the logic of your first choice is 5/6 wrong, that should mean that the odds of it being correct/winning is 1/6 or 16%

If above is correctly worked out, it does seem to be slightly favorable when there are multiple doors (> 4)? Though I can offcourse have errors here
BUT the big differences vs the 50-50 are diffinatly not there I dont think

 

[chergh]

Wrong. Once a door is eliminated, the odds change. You know one of the choices is no longer valid. Your using what's called junk science.

Let me give you an example using your logic. I say "Pick a number between 1 and 100,000", you pick 1, then I say "the number is not greater than 2, would you like to change your guess?"

Your original odds of winning were 1 in 100,000. Since you still don't know if your original guess was correct or not, the odds are still 1 in 100,000. The other possible choice, 2, must contain all the rest of the odds. So 2 will win 99,999/100,000 times.

Sorry adam that's junk science.

You're example should be the answer is 1 or X where X is any number in your range except Y. 99,999/100,000 times the answer will be X.

 

[chergh]

seeing as only losing doors are "lost" in the equation, the odds turn out to be 50%-50%

Winning	Chosen	Rmvd	Win?
1	1	2	TRUE
1	1	3	TRUE
1	2	3	FALSE
1	3	2	FALSE
2	1	3	FALSE
2	2	1	TRUE
2	2	3	TRUE
2	3	1	FALSE
3	1	2	FALSE
3	2	1	FALSE
3	3	1	TRUE
3	3	2	TRUE

6 wins - 6 loses

The error here is that which door he removes is irrelevant, as he will never remove the winning door. What matters is if you switch or not.

If the winning door is 1 and you have chosen 1 it doesn't matter if he removes door 2 or 3 as if you switch when given the option you will lose if you do.

 

[Pauldohert]

A local entrepeneur bought 100 pounds of strawberries for $2.00 per pound and expected to double his investment by selling the strawberries for $4.00 per pound on a convenient street corner. The seller only managed to sell 50 pounds of strawberries the first day and he sold the remainder on the second day. The fresh strawberries had a content of 99% water, but because of the hot weather, the strawberries dehydrated and contained only 98% water on the second day. How much profit did the seller make?

So 50 * 2 profit for day one assuming they haven't dehydrated.
= 100

Day two -

it was 99 water and 1 strawberry.

Its now 2% strawberry an 98% water.

So has the weight halved (99/100 is 99% 49/50 is 98%) and the profit is 25 * 2 = 50

So he makes 50.

?

 

[Adam Caramon]

Sorry adam that's junk science.

Well, now we agree

You're example should be the answer is 1 or X where X is any number in your range except Y. 99,999/100,000 times the answer will be X.

At this stage you're correct. You're still talking hypothetical. The odds of you successfully guessing the correct number are 1/100,000. The odds of you guessing an incorrect number are 99,999/100,000.

As soon as you add in the "Would you like to switch" question, the above no longer applies.

 

[Access_guy49]

After reading all this, I have to say, the simple explination for the door problem seem to be that your simply Betting that you were wrong on your first guess.
Because your odds are 1/3 initially, and you made your pick with those odds, given that you have the option to switch to another door.. they are basically asking you.. "Do you think you were right when you choose that door with odds of 1/3"
Your best bet is to assume you were wrong and choose the other door.
Which gives you the 66% odds...yes?

Oh, and the farmer made $198

 

[Pauldohert]

Nam

1 1 2 TRUE
1 1 3 TRUE

Choosing 1 when the winner is one is only one example, because inevitably the host can remove 2 or 3.
Choosing 1 is one possible choice - somehow you are counting it as 2.

The point is - its irrelevent to the odd that the host can reveal a wrong answer - he can reveal a wrong answer for whatever you first pick as you illustrate. He never as you suggest though gives you two wrong answers.

If you choose 1 - he reveals either, 2 or 3 - just one revaeal. Even if he has two choices of what the one reveal would be.

So you have twice the number of correct choices as there really are.

so 3 out of 9 are correct, and 6 out of nine are wrong.

 

[chergh]

As soon as you add in the "Would you like to switch" question, the above no longer applies.

yes it does. Really think about it. Why does giving you an option to switch change the odds?

 

[Pauldohert]

Let me give you an example using your logic. I say "Pick a number between 1 and 100,000", you pick 1, then I say "the number is not greater than 2, would you like to change your guess?"

Your original odds of winning were 1 in 100,000. Since you still don't know if your original guess was correct or not, the odds are still 1 in 100,000. The other possible choice, 2, must contain all the rest of the odds. So 2 will win 99,999/100,000 times.

Thats correct. 99,999 times - the right answer is differant from the one I originally picked. All you have done is clarify that 99, 998 of the ones i didn't pick were wrong - which I already knew.

Its still 99,999 times more likely i was originally wrong, and since 2 is the only answer possible left I didnt go for - its 99,999 times more likely that 2 is correct.



Look at it the other way - play the game over a few times in your head - you are saying my chances of getting the right answer out of 100,000 is 50/50 each time.


So 2 will be right 99,999 out of 100,000
or 3 will correct 99,999 out of 100,00
.......

Or 100,000 will be correct 99,999 out of 100,000.

I doubt thats clarified it but still.

 

[dan-cat]

In effect, for the first choice to be correct it has to perform two criteria.
Be the 1/1000 and now the 1/2 - long shot - against the 2nd door which only has to beat the 1/2 odds.

This is the crux of it for me. The showing of the third door allows the first door to conform to the first criterion you described.

There is now an equal chance that door 1 and 2 will confirm to the second criterion unless the first door is somewhat fatigued from the first event.

The first door doesn't have to conform to the first criterion on both occasions.

 

[ChipperT]

So 50 * 2 profit for day one assuming they haven't dehydrated.
= 100

Day two -

it was 99 water and 1 strawberry.

Its now 2% strawberry an 98% water.

So has the weight halved (99/100 is 99% 49/50 is 98%) and the profit is 25 * 2 = 50

So he makes 50.

?

Sorry, your answer is not correct.