Odd Escape

[ZZII 527]

You are being held prisoner in a small room. Only you and a gaurd are inside the room. There are three identical doors that lead out of the room. The gaurd offers a choice: you may exit through one of the three doors. Two lead to your execution, one leads to escape. You have no knowledge of the doors and so you choose randomly, door #1. You indicate your choice to the gaurd and he says, "I cannot tell you how to escape, but I can tell you that door #2 will NOT lead to escape." Assuming that the gaurd is impartial and is telling the truth, what should you do?

-Stick with your original choice, door #1.
-Change to door #3.
-It doesn't matter, it's 50/50.

This is not an easy one, but it is also not a trick question by any means. Please justify (mathematically, if possible) any answer.

[Tom Bottiglieri]

i say go with 3

[Solace]

change to door #3.

when you chose door number 1 initially, you were chosing from out of three doors, and therefore had a 33% chance of getting the right one. with one of the wrong doors now eliminated, there are only 2 remaining. you are, however, still unaware as to wether or not your door leads to escape. if you stay with your door, you stick with the 33% chance you had originally. however, if you switch to door #3, you will have picked from 2, and your chances are now 50%.

i know it doesn't at first make sense, but I think that its right.

[mightywombat]

ah - ha

its the old Monty Hall problem from my statistics book.

Monty Hall had a game show "Let's Make a Deal" which dealt with the contestant choosing from 1 of 3 doors and then he'd reduce the number to 2 doors and you could choose again... or something like that. i don't really remember.

[Team311]

its easy ask the guard wich doors do not lead to escape

[Jones571]

change to door #3.

when you chose door number 1 initially, you were chosing from out of three doors, and therefore had a 33% chance of getting the right one. with one of the wrong doors now eliminated, there are only 2 remaining. you are, however, still unaware as to wether or not your door leads to escape. if you stay with your door, you stick with the 33% chance you had originally. however, if you switch to door #3, you will have picked from 2, and your chances are now 50%.

i know it doesn't at first make sense, but I think that its right.

u picked 1 your onyl option left is 3 so how would picking three improve your chance? you still picked one of the 2 proper choices by choseing over again you dont improve your odds

EDIT: yes jake that just proves you are the head wombat cuz ur the smarter one

[ZZII 527]

Solace, you've got it spot on. Your 33% never changes, but the other door "takes on" the odds from the one that is eliminated, becoming 67%. It's a hard one to grasp, but you got it.

Another way to look at it: Imagine 1,000 doors, one that is correct and 999 that are wrong. You pick one randomly. The gaurd eliminates 998 others. Which do you think is more likely the right one? The one you chose randomly or the one the gaurd left?

I've seen this riddle argued on another forum for several pages, but there is no denying the answer....I was very surprised someone got it so quickly. Goes to show you that this forum has some amazingly intelligent people.

How about this one:

Option A) Play the same set of lottery numbers every week for 52 weeks.
Option B) Play 52 different sets of lottery numbers in one week.

Of course your odds of winning are pretty slim either way, but is one better than the other?

[Solace]

does that mean i get a cookie? I like cookies.

[ZZII 527]

Score one cookie for Solace.

[Denman]

hmm.... i so do not agree with that answer.... its more phycological.


Anyway... both options are the same.
There is an equal chance that a set of numbers will turn up and there is a chance that the same set of numbers turn up twice in a row. Some people just believe that it will be more likely to turn up if the same numbers are used every time.
Did you say 52 numbers? is the 52 numbers + a bonus ball or what?
cos here in england there is choice of49, and 6 numbers are drawn plus a bonus ball.
So really ,,, it makes no difference.

[Alan Anderson]

This question causes more heated debate than any other popular probability-based problem I know.

The answer is simple if you start out with the assumption that your first pick has one chance in three of being correct. From that, it should be obvious that your first pick has two chances in three of being incorrect, right? So if you switch, you have a 2/3 chance of switching away from a losing door.

After the other losing door is revealed and removed from your options, you end up having a 2/3 chance of switching to a winning door. It's good to switch doors. 1/3 of the time you'll pick the winning door to begin with and switch to a loser, but 2/3 of the time you'll start with a losing door and switch to the winner.

[Steve W]

If I were the one asking the questions then I would throw in the free door to get the person to reconsider their first choice. By the way you are chosing then I would always have the upper hand. Therefore I always stay with my first choice as it is most likely right.

Remember that if you have a 67% chance of chosing the right door then you also have a 67% chance of chosing the wrong one. Optimist of pesimist your choice. :rolleyes:

[Frank Toussaint]

Regarding:
>How about this one:
>
>Option A) Play the same set of lottery numbers every week for 52 weeks.
>Option B) Play 52 different sets of lottery numbers in one week.
>
Here's a counter example:
Supose that there were seventy million possible sets of numbers to be played. Consider the following two options:
Option A') Play the same set of lottery numbers every week for seventy million weeks.
Option B') Play all seventy million different sets of lottery numbers in one week.
In option A' it is possible that your set of lottery numbers never comes up but in option B' you are certain to hit. Option B is better.

[ZZII 527]

Regarding:
>How about this one:
>
>Option A) Play the same set of lottery numbers every week for 52 weeks.
>Option B) Play 52 different sets of lottery numbers in one week.
>
Here's a counter example:
Supose that there were seventy million possible sets of numbers to be played. Consider the following two options:
Option A') Play the same set of lottery numbers every week for seventy million weeks.
Option B') Play all seventy million different sets of lottery numbers in one week.
In option A' it is possible that your set of lottery numbers never comes up but in option B' you are certain to hit. Option B is better.

Score the second cookie for Frank Toussaint. ;)

This question causes more heated debate than any other popular probability-based problem I know.

It's so true:

http://www.gtplanet.net/forum/showthread.php?s=&threadid=34178

[Jones571]

Regarding:
>How about this one:
>
>Option A) Play the same set of lottery numbers every week for 52 weeks.
>Option B) Play 52 different sets of lottery numbers in one week.
>
Here's a counter example:
Supose that there were seventy million possible sets of numbers to be played. Consider the following two options:
Option A') Play the same set of lottery numbers every week for seventy million weeks.
Option B') Play all seventy million different sets of lottery numbers in one week.
In option A' it is possible that your set of lottery numbers never comes up but in option B' you are certain to hit. Option B is better.

yea and after playing seventy millinon diffrent combos in one week you will need to win the lotto after spending 70 million dollars on tickets.

[ZZII 527]

yea and after playing seventy millinon diffrent combos in one week you will need to win the lotto after spending 70 million dollars on tickets.

Yea really! :) You gotta win big.

Which brings up another interesting point:

Option B guarantees the win in this scenario, but you only win on one set of #s.

Option A does not gaurantee the win, however you do have the possibility of winning multiple times.

So, in the end: if you are concerned with just the probability of winning once and only once, then Option B is better. If you are concerned total payoff probability, both options WOULD work out to be the same. An interesting twist to an already interesting riddle.

[Jones571]

But other tickets may not be "big" winners but they still have a smaller pay out. You would have to find out how much you would win form the jack pot plus all the other smaller pay outs for having a portion of the numbers correct. then you could find out if it would be worth it.
You also have the possibility were some one else ends up playing the same number so you have to split it so really your screwed no matter what

yea but that is alot of work and u need 70million to start with so really we have learned just save up your money and buy a segway lotto is bad

[Grommit]

This problem might cause heated debate, but it's hard to argue against mathematical fact. In fact, to prove disbelievers wrong, I wrote a program to do exactly this:

Three doors, randomly picked one to be correct.
Randomly pick a door.
Randomly choose a door that is incorrect and not picked to be opened.
Switch to the other door and see if it is correct.

The program won 66.7% of the time, after enough trials.

But I have better problems to solve ;):
http://www.chiefdelphi.com/forums/showthread.php?t=27647

Have fun!

[ZZII 527]

This problem might cause heated debate, but it's hard to argue against mathematical fact. In fact, to prove disbelievers wrong, I wrote a program to do exactly this:

Three doors, randomly picked one to be correct.
Randomly pick a door.
Randomly choose a door that is incorrect and not picked to be opened.
Switch to the other door and see if it is correct.

The program won 66.7% of the time, after enough trials.

But I have better problems to solve ;):
http://www.chiefdelphi.com/forums/showthread.php?t=27647

Have fun!

That's funny I did the same thing (and got the same results of course.) Score one cookie for Grommit.

Here's a new one:

Is it possible to place four lamp-posts of different colors on the inside of a circle so that a person walking all the way around the outside of the circle would see all 24 (4!) combinations of colored posts from left to right at some point along the circle? Obviously, a proof is required either way.

[Ian Mackenzie]

No, it isn't possible; here's my geometric proof. Take any four points representing the four lights. Draw lines connecting every pair of points, extending off to infinity in both directions; since the observed order of the lights only changes when one light appears to the observer to pass behind another, crossing a line represents a change in observed order. Since only six lines can be drawn (4 choose 2), and the observer can only cross each one twice in one pass around the circle, the observed order can only change 12 times (one of these being a change back to the originally observed order), so there is a maximum of 12 orders that can be observed (not the full 24).

[ZZII 527]

No, it isn't possible; here's my geometric proof. Take any four points representing the four lights. Draw lines connecting every pair of points, extending off to infinity in both directions; since the observed order of the lights only changes when one light appears to the observer to pass behind another, crossing a line represents a change in observed order. Since only six lines can be drawn (4 choose 2), and the observer can only cross each one twice in one pass around the circle, the observed order can only change 12 times (one of these being a change back to the originally observed order), so there is a maximum of 12 orders that can be observed (not the full 24).

Absolutely perfect! Score one cookie for Ian Mackenzie!
How long did that take you? For me, I drew some pictures and started rotating them around in front of me, keeping track of all the combinations for about a half hour. Then I realized that you could just draw lines like you said. Congrats, that's not an easy one to figure out.

[Ian Mackenzie]

Sweet, I love cookies. (Just ask anyone on my team...)

Actually, I got it pretty quickly, through a combination of laziness and luck. I thought about doing some actual rotations as you described, but I really didn't want to, and figured there should be a more elegant method.

Here's another really cool problem for you: You have two ropes, and the only thing you know about them is they take exactly 1 hour each to burn (i.e. if you lit either end, the flame would take exactly 1 hour to reach the other end). The ropes are not of constant cross-section, so half a rope will not necessarily take half an hour to burn, and the ropes are not the same (i.e. not only do they burn unevenly, they don't even burn equally). Given these two ropes and a lighter, how do you measure exactly 45 minutes?

[jonathan lall]

Well I think if you burn the first rope at both ends, their flames will meet and fizzle at 30 minutes, right? Because their uneven burning will still make them meet somewhere along that length. So if you also burned the second rope at the start (i.e. the same time as you lit the first rope), when the first rope's flames die out, it'll be 30 minutes, at which point you could light the other end of the second rope. When both ends of the second rope meet, it'll be another 15 minutes. This adds up to 45 minutes, I think. Unless I'm horribly missing something.

[ZZII 527]

Well I think if you burn the first rope at both ends, their flames will meet and fizzle at 30 minutes, right? Because their uneven burning will still make them meet somewhere along that length. So if you also burned the second rope at the start (i.e. the same time as you lit the first rope), when the first rope's flames die out, it'll be 30 minutes, at which point you could light the other end of the second rope. When both ends of the second rope meet, it'll be another 15 minutes. This adds up to 45 minutes, I think. Unless I'm horribly missing something.

That sounds right. I was on the right track of burning both ends of the first rope for 30 minutes but couldn't get the 15 second part...I ventured down the blind alley of trying to light the second rope in four places...but it only works when you light the ends I guess.

[Ian Mackenzie]

Well I think if you burn the first rope at both ends, their flames will meet and fizzle at 30 minutes, right? Because their uneven burning will still make them meet somewhere along that length. So if you also burned the second rope at the start (i.e. the same time as you lit the first rope), when the first rope's flames die out, it'll be 30 minutes, at which point you could light the other end of the second rope. When both ends of the second rope meet, it'll be another 15 minutes. This adds up to 45 minutes, I think. Unless I'm horribly missing something.

Exactly right. (Your solution, not the statement that you're horribly missing something :D).

[trev2023]

Well...I think I've got the paper solution to the first one...

Given that there are 3 doors and 1 of them is "escape", you can:

A) Choose door 1
B) Choose door 2
C) Choose door 3

Now given that 3 is the "correct" door, if you pick 1, 2 will be taken away and vice versa and if you pick 3, either 1 or 2 will be taken away.

A1) Stick with door 1
A2) Go with door 3
B1) Stick with door 2
B2) Go with door 3
C1) Stick with door 3
C2) Change to door 1 or 2 (whichever is not eliminated)

Now, as you can see, if you stick with your first choice (A1, B1, C1), you have a 1 in 3 chance (C1) of getting it correct while if you change it (A2, B2, C2) you have a 2 in 3 chance of getting it right (A2, B2).