• Log InLog In
  • Register
Liquid`
Team Liquid Liquipedia
EDT 06:26
CEST 12:26
KST 19:26
  • Home
  • Forum
  • Calendar
  • Streams
  • Liquipedia
  • Features
  • Store
  • EPT
  • TL+
  • StarCraft 2
  • Brood War
  • Smash
  • Heroes
  • Counter-Strike
  • Overwatch
  • Liquibet
  • Fantasy StarCraft
  • TLPD
  • StarCraft 2
  • Brood War
  • Blogs
Forum Sidebar
Events/Features
News
Featured News
RSL Season 1 - Final Week6[ASL19] Finals Recap: Standing Tall12HomeStory Cup 27 - Info & Preview18Classic wins Code S Season 2 (2025)16Code S RO4 & Finals Preview: herO, Rogue, Classic, GuMiho0
Community News
Esports World Cup 2025 - Brackets Revealed14Weekly Cups (July 7-13): Classic continues to roll6Team TLMC #5 - Submission extension3Firefly given lifetime ban by ESIC following match-fixing investigation17$25,000 Streamerzone StarCraft Pro Series announced7
StarCraft 2
General
Who will win EWC 2025? Weekly Cups (July 7-13): Classic continues to roll Esports World Cup 2025 - Final Player Roster Esports World Cup 2025 - Brackets Revealed The GOAT ranking of GOAT rankings
Tourneys
FEL Cracov 2025 (July 27) - $8000 live event Sea Duckling Open (Global, Bronze-Diamond) RSL: Revival, a new crowdfunded tournament series $5,100+ SEL Season 2 Championship (SC: Evo) WardiTV Mondays
Strategy
How did i lose this ZvP, whats the proper response
Custom Maps
External Content
Mutation # 482 Wheel of Misfortune Mutation # 481 Fear and Lava Mutation # 480 Moths to the Flame Mutation # 479 Worn Out Welcome
Brood War
General
Flash Announces (and Retracts) Hiatus From ASL BW General Discussion Help: rep cant save ASL20 Preliminary Maps BGH Auto Balance -> http://bghmmr.eu/
Tourneys
Cosmonarchy Pro Showmatches CSL Xiamen International Invitational [Megathread] Daily Proleagues [BSL20] Non-Korean Championship 4x BSL + 4x China
Strategy
Simple Questions, Simple Answers I am doing this better than progamers do.
Other Games
General Games
Stormgate/Frost Giant Megathread Path of Exile Nintendo Switch Thread CCLP - Command & Conquer League Project The PlayStation 5
Dota 2
Official 'what is Dota anymore' discussion
League of Legends
Heroes of the Storm
Simple Questions, Simple Answers Heroes of the Storm 2.0
Hearthstone
Heroes of StarCraft mini-set
TL Mafia
TL Mafia Community Thread Vanilla Mini Mafia
Community
General
US Politics Mega-thread Things Aren’t Peaceful in Palestine Russo-Ukrainian War Thread Stop Killing Games - European Citizens Initiative Summer Games Done Quick 2025!
Fan Clubs
SKT1 Classic Fan Club! Maru Fan Club
Media & Entertainment
[Manga] One Piece Movie Discussion! Anime Discussion Thread [\m/] Heavy Metal Thread
Sports
Formula 1 Discussion TeamLiquid Health and Fitness Initiative For 2023 2024 - 2025 Football Thread NBA General Discussion NHL Playoffs 2024
World Cup 2022
Tech Support
Computer Build, Upgrade & Buying Resource Thread
TL Community
The Automated Ban List
Blogs
Men Take Risks, Women Win Ga…
TrAiDoS
momentary artworks from des…
tankgirl
from making sc maps to makin…
Husyelt
StarCraft improvement
iopq
Trip to the Zoo
micronesia
Customize Sidebar...

Website Feedback

Closed Threads



Active: 733 users

[SFW] Riddles / Puzzles / Brain Teasers - Page 13

Forum Index > General Forum
Post a Reply
Prev 1 11 12 13 14 15 38 Next All
kdgns
Profile Joined May 2009
United States2427 Posts
April 19 2012 18:41 GMT
#241
On April 20 2012 03:37 Crownlol wrote:
Show nested quote +
On April 20 2012 02:33 XiGua wrote:
I am betting that nobody will solve this without cheating.

I present to you,
THE WARDEN

The warden meets with 23 new prisoners when they arrive. He tells them, "You may meet today and plan a strategy. But after today, you will be in isolated cells and will have no communication with one another.

"In the prison is a switch room, which contains two light switches labeled 1 and 2, each of which can be in either up or the down position. I am not telling you their present positions. The switches are not connected to anything.

"After today, from time to time whenever I feel so inclined, I will select one prisoner at random and escort him to the switch room. This prisoner will select one of the two switches and reverse its position. He must flip one switch when he visits the switch room, and may only flip one of the switches. Then he'll be led back to his cell.

"No one else will be allowed to alter the switches until I lead the next prisoner into the switch room. I'm going to choose prisoners at random. I may choose the same guy three times in a row, or I may jump around and come back. I will not touch the switches, if I wanted you dead you would already be dead.

"Given enough time, everyone will eventually visit the switch room the same number of times as everyone else. At any time, anyone may declare to me, 'We have all visited the switch room.'

"If it is true, then you will all be set free. If it is false, and somebody has not yet visited the switch room, you will all die horribly. You will be carefully monitored, and any attempt to break any of these rules will result in instant death to all of you"

What is the strategy they come up with so that they can be free?
[image loading]



Ok, I'm not sure, but here ya go:


+ Show Spoiler +
The group designates a single person the "counter". This person is the only one who will reset switch A to the bottom position. A starts in the down position. If it is your first visit to the room, you move switch A to up. If switch A is in the up position, you don't touch it, you just move B (also, if you have been there before you move B). If the counter finds the switch in the up position, he moves it down and adds +1 to his count. When the counter's count reaches 22, he frees himself and all his co-prisoners!


what if the warden picks the counter to go 22 times in a row?
Nehsb
Profile Joined May 2009
United States380 Posts
Last Edited: 2012-04-19 18:45:38
April 19 2012 18:43 GMT
#242
On April 19 2012 08:44 Nehsb wrote:
Show nested quote +
On April 19 2012 07:54 keplersfolly wrote:
I'm not a regular on this forum so apologies if I'm quoting wrong :D

Solution discussion for Nehsb and Miicro:
+ Show Spoiler +
I'm impressed :D Was that off the top of your heads or had you heard it before? It took me and my colleagues a while to work it out from a position of total ignorance. Nehsb is more correct though as he doesn't assume the starting state of the levers and he hints towards fully understanding the issue (but falls a little short). Basically there are two related facets:-
You MUST flip a lever exactly once + There are 2 levers.
What this effectively means is that the prisoners decide that the left lever is the "noop" lever. So the counter only cares about the right lever and you follow the logic you both described, so if a non-counter enters the room and the right lever is flipped to indicate someone needs counting then they throw away their flip on the left lever. You can think of this as changing the puzzle conditions to be having just one lever but flipping it is optional. Still, congrats if you worked this out from scratch!


Ninja Edit:
Nehsb's solution accounts for the unknown starting state - good job!
Miicro's solution accounts for the fact there are two levers - good job!


I'd heard it before. There's also a really hard version that I'm curious if anyone has an easier solution to:

Show nested quote +
20 prisoners are in jail who are all gathered in the main hall and addressed by the warden who explains the following. "In 1 hour you will all be placed in solitary confinement and never see each other again. You will individually be escorted by the guards to the "Lever Room" in which the are a pair of levers that have no function except being able to be flipped either Up or Down. You can flip any number of levers, including all of them or none of them. We make no guarantees about the order you will be called to the Lever Room in, and we may take any of you there more than once. If any of you prisoners tell the guard "We have all been to the lever room" and this is true you will all be released, otherwise you will all be shot."You do not know the starting position of the levers. You are allowed to write down a strategy and give it to all of the prisoners, but the strategy each prisoner gets must be the same; you are not allowed to "pick a leader" or anything like that.


The changes are in bold:
- Still 2 levers
- You can now flip any number of levers
- Every prisoner must have the same strategy
- You do not know the starting states of the levers

The problem gets much easier if you do know the starting state of the levers.


A hint:

+ Show Spoiler +
Instead of choosing a leader as in the earlier puzzles, use the levers to split the people into two about equally-sized groups.
Crownlol
Profile Blog Joined October 2011
United States3726 Posts
April 19 2012 18:44 GMT
#243

what if the warden picks the counter to go 22 times in a row?


+ Show Spoiler +
His count is still at "one", and the A switch is still in the down position. He just diddles the B switch 21 times.
shaGuar :: elemeNt :: XeqtR :: naikon :: method
kdgns
Profile Joined May 2009
United States2427 Posts
April 19 2012 18:46 GMT
#244
On April 20 2012 03:36 Taekwon wrote:
A boy comes around and yells at four of his friends: "HEY! Who ate my cake?! I was saving it for later!"

Girl A said: "I didn't eat it!"
Boy B said: "Neither A nor B ate it."
Boy C said: "I swear on my momma's grave I didn't eat it!"
Girl D said: "C is telling the truth!"

One of these children is lying. Who is it?


+ Show Spoiler +
A can not be lying, if she did, B would also be lying
C can not be lying, if he did D would also be lying
D can not be lying, if she did, C would be lying
B could be lying, if he was the one to eat it, A C and D would still be telling the truth.

or it could be the boy that says he had his cake eaten that's lying, seriously what kid goes around saving cake for later, he probably ate his cake and have it too.
Orome
Profile Blog Joined June 2004
Switzerland11984 Posts
April 19 2012 18:58 GMT
#245
On April 20 2012 02:33 XiGua wrote:
I am betting that nobody will solve this without cheating.

I present to you,
THE WARDEN

The warden meets with 23 new prisoners when they arrive. He tells them, "You may meet today and plan a strategy. But after today, you will be in isolated cells and will have no communication with one another.

"In the prison is a switch room, which contains two light switches labeled 1 and 2, each of which can be in either up or the down position. I am not telling you their present positions. The switches are not connected to anything.

"After today, from time to time whenever I feel so inclined, I will select one prisoner at random and escort him to the switch room. This prisoner will select one of the two switches and reverse its position. He must flip one switch when he visits the switch room, and may only flip one of the switches. Then he'll be led back to his cell.

"No one else will be allowed to alter the switches until I lead the next prisoner into the switch room. I'm going to choose prisoners at random. I may choose the same guy three times in a row, or I may jump around and come back. I will not touch the switches, if I wanted you dead you would already be dead.

"Given enough time, everyone will eventually visit the switch room the same number of times as everyone else. At any time, anyone may declare to me, 'We have all visited the switch room.'

"If it is true, then you will all be set free. If it is false, and somebody has not yet visited the switch room, you will all die horribly. You will be carefully monitored, and any attempt to break any of these rules will result in instant death to all of you"

What is the strategy they come up with so that they can be free?
[image loading]


I didn't see it in the thread earlier, so here's my try:

+ Show Spoiler +

Since there's no way to store the information of how many people have been in the room with the levers themselves, one of the prisoners needs to act as counter. The procedure is simple:

Simplified solution assuming switch 1 is down to begin with: If one of the prisoners (except the counter) enters the room for the first time, they flip switch 1 up. If switch 1 is already flipped up, they instead flip switch 2. If any of them are brought in again after the first time, they also flip switch 2.

Whenever the counter enters the room and finds switch 1 flipped up, he flips it back down. Otherwise he flips switch 2.

Switch 2 is completely useless here, it's simply used because the prisoners have to flip one switch. All the information is stored in switch 1.

If the prisoners knew that switch 1 was flipped down to begin with, this would be the end of the puzzle. The counter would simply announce that everyone had been in the room after he sees switch 1 flipped up for the 22nd time.

Because he doesn't know however, we need to account for the fact that the switch might have been up to begin with. I've been thinking about an elegant solution to this, but I don't think there is one. The prisoners simply have to take the longer way around.

The idea is that you can account for the possible mistake of 1 by doubling the amount of counts. If each of the non-counting prisoners flips the switch up the first two times they enter the room and the counter waits for switch 1 being up 44 times, then one of those times being a potential error doesn't matter. There would still be 43 real counts remaining, which could only have come from all the 22 non-counters (21 * 2 + 1 * 1).

On a purely personal note, I'd like to show Yellow the beauty of infinitely repeating Starcraft 2 bunkers. -Boxer
Orome
Profile Blog Joined June 2004
Switzerland11984 Posts
April 19 2012 18:59 GMT
#246
On April 20 2012 03:37 Crownlol wrote:
Show nested quote +
On April 20 2012 02:33 XiGua wrote:
I am betting that nobody will solve this without cheating.

I present to you,
THE WARDEN

The warden meets with 23 new prisoners when they arrive. He tells them, "You may meet today and plan a strategy. But after today, you will be in isolated cells and will have no communication with one another.

"In the prison is a switch room, which contains two light switches labeled 1 and 2, each of which can be in either up or the down position. I am not telling you their present positions. The switches are not connected to anything.

"After today, from time to time whenever I feel so inclined, I will select one prisoner at random and escort him to the switch room. This prisoner will select one of the two switches and reverse its position. He must flip one switch when he visits the switch room, and may only flip one of the switches. Then he'll be led back to his cell.

"No one else will be allowed to alter the switches until I lead the next prisoner into the switch room. I'm going to choose prisoners at random. I may choose the same guy three times in a row, or I may jump around and come back. I will not touch the switches, if I wanted you dead you would already be dead.

"Given enough time, everyone will eventually visit the switch room the same number of times as everyone else. At any time, anyone may declare to me, 'We have all visited the switch room.'

"If it is true, then you will all be set free. If it is false, and somebody has not yet visited the switch room, you will all die horribly. You will be carefully monitored, and any attempt to break any of these rules will result in instant death to all of you"

What is the strategy they come up with so that they can be free?
[image loading]



Ok, I'm not sure, but here ya go:


+ Show Spoiler +
The group designates a single person the "counter". This person is the only one who will reset switch A to the bottom position. A starts in the down position. If it is your first visit to the room, you move switch A to up. If switch A is in the up position, you don't touch it, you just move B (also, if you have been there before you move B). If the counter finds the switch in the up position, he moves it down and adds +1 to his count. When the counter's count reaches 22, he frees himself and all his co-prisoners!


your version doesn't account for the fact that the switch could be up to begin with^^
On a purely personal note, I'd like to show Yellow the beauty of infinitely repeating Starcraft 2 bunkers. -Boxer
el_dawg
Profile Joined September 2011
United States164 Posts
Last Edited: 2012-04-20 15:52:31
April 19 2012 19:41 GMT
#247
Riddle:
You are meeting a friend you have not seen in a long time. You know that they now have 2 children, but do not know the genders of the children. When you knock on the door of your friend's house, a male child answers. What is the % chance that the other child is male? (hint: it is not 50%)

+ Show Spoiler +
33%

Before you knock on the door, there are 4 equally possible situations.
First child - Second child
Boy - Boy
Boy - Girl
Girl - Boy
Girl - Girl

Once you know that at least 1 child is male, the 4th possibility is no longer possible leaving only:
Boy - Boy
Boy - Girl
Girl - Boy

Since you don't know if the older or younger child opened the door, each of the 3 remaining possibilities are equally likely and the chance that both children are male is 33%.

For the math nerds out there, this is an example of conditional probability. http://en.wikipedia.org/wiki/Conditional_probability

Probability(A given B)=P(A and B)/P(B)
P("both are male" given "at least one is male")=P("both male" and "at least one male" are true)/P("at least one male" is true)=.25/.75=.33

Edit + Update:
This question turned out a lot more ambiguous than I had intended. The problem is in how one interprets "a male child answers." In case 1, a random child is chosen and happened to be male. In case 2, a male child is chosen if at all possible. Imagining a community of 100 houses with 2 children each, 25 BB, 50 BG, and 25 GG. In the first case, the 25 BB and 50% of the 50 BG houses are considered (since the chance of a boy answering the door in a BG house is 50%), leaving 25 BB and 25 BG with the probability of the unknown child being a boy equal to 25/50=50%. In the second case, a boy answers at all 50 of the BG houses which combined with the 25 BB houses makes 75 with 25/75=33% of the houses having the unknown child being a boy.

I have honestly always interpreted the question as the second case, but it seems like most people interpret it as the first case. The hint "The answer is not 50%" suggests that it is not case 1, but I should definitely have added some additional language to be more specific.
ishboh
Profile Joined October 2010
United States954 Posts
April 19 2012 19:52 GMT
#248
On April 20 2012 03:59 Orome wrote:
Show nested quote +
On April 20 2012 03:37 Crownlol wrote:
On April 20 2012 02:33 XiGua wrote:
I am betting that nobody will solve this without cheating.

I present to you,
THE WARDEN

The warden meets with 23 new prisoners when they arrive. He tells them, "You may meet today and plan a strategy. But after today, you will be in isolated cells and will have no communication with one another.

"In the prison is a switch room, which contains two light switches labeled 1 and 2, each of which can be in either up or the down position. I am not telling you their present positions. The switches are not connected to anything.

"After today, from time to time whenever I feel so inclined, I will select one prisoner at random and escort him to the switch room. This prisoner will select one of the two switches and reverse its position. He must flip one switch when he visits the switch room, and may only flip one of the switches. Then he'll be led back to his cell.

"No one else will be allowed to alter the switches until I lead the next prisoner into the switch room. I'm going to choose prisoners at random. I may choose the same guy three times in a row, or I may jump around and come back. I will not touch the switches, if I wanted you dead you would already be dead.

"Given enough time, everyone will eventually visit the switch room the same number of times as everyone else. At any time, anyone may declare to me, 'We have all visited the switch room.'

"If it is true, then you will all be set free. If it is false, and somebody has not yet visited the switch room, you will all die horribly. You will be carefully monitored, and any attempt to break any of these rules will result in instant death to all of you"

What is the strategy they come up with so that they can be free?
[image loading]



Ok, I'm not sure, but here ya go:


+ Show Spoiler +
The group designates a single person the "counter". This person is the only one who will reset switch A to the bottom position. A starts in the down position. If it is your first visit to the room, you move switch A to up. If switch A is in the up position, you don't touch it, you just move B (also, if you have been there before you move B). If the counter finds the switch in the up position, he moves it down and adds +1 to his count. When the counter's count reaches 22, he frees himself and all his co-prisoners!


your version doesn't account for the fact that the switch could be up to begin with^^

all he has to add to ensure that his solution is correct is add a couple conditions:
1) they decide who the counter is beforehand.
2) he doesn't start his count until after the first time he adjusts the switch
Slithe
Profile Blog Joined February 2007
United States985 Posts
April 19 2012 19:58 GMT
#249
On April 20 2012 04:52 ishboh wrote:
Show nested quote +
On April 20 2012 03:59 Orome wrote:
On April 20 2012 03:37 Crownlol wrote:
On April 20 2012 02:33 XiGua wrote:
I am betting that nobody will solve this without cheating.

I present to you,
THE WARDEN

The warden meets with 23 new prisoners when they arrive. He tells them, "You may meet today and plan a strategy. But after today, you will be in isolated cells and will have no communication with one another.

"In the prison is a switch room, which contains two light switches labeled 1 and 2, each of which can be in either up or the down position. I am not telling you their present positions. The switches are not connected to anything.

"After today, from time to time whenever I feel so inclined, I will select one prisoner at random and escort him to the switch room. This prisoner will select one of the two switches and reverse its position. He must flip one switch when he visits the switch room, and may only flip one of the switches. Then he'll be led back to his cell.

"No one else will be allowed to alter the switches until I lead the next prisoner into the switch room. I'm going to choose prisoners at random. I may choose the same guy three times in a row, or I may jump around and come back. I will not touch the switches, if I wanted you dead you would already be dead.

"Given enough time, everyone will eventually visit the switch room the same number of times as everyone else. At any time, anyone may declare to me, 'We have all visited the switch room.'

"If it is true, then you will all be set free. If it is false, and somebody has not yet visited the switch room, you will all die horribly. You will be carefully monitored, and any attempt to break any of these rules will result in instant death to all of you"

What is the strategy they come up with so that they can be free?
[image loading]



Ok, I'm not sure, but here ya go:


+ Show Spoiler +
The group designates a single person the "counter". This person is the only one who will reset switch A to the bottom position. A starts in the down position. If it is your first visit to the room, you move switch A to up. If switch A is in the up position, you don't touch it, you just move B (also, if you have been there before you move B). If the counter finds the switch in the up position, he moves it down and adds +1 to his count. When the counter's count reaches 22, he frees himself and all his co-prisoners!


your version doesn't account for the fact that the switch could be up to begin with^^

all he has to add to ensure that his solution is correct is add a couple conditions:
1) they decide who the counter is beforehand.
2) he doesn't start his count until after the first time he adjusts the switch


How does he know that he's the first one to adjust the switch? What if one of the other prisoners originally saw it in the down position and flipped it up?
Aim Here
Profile Blog Joined December 2009
Scotland672 Posts
April 19 2012 20:19 GMT
#250
On April 20 2012 04:41 el_dawg wrote:
Riddle:
You are meeting a friend you have not seen in a long time. You know that they now have 2 children, but do not know the genders of the children. When you knock on the door of your friend's house, a male child answers. What is the % chance that the other child is male? (hint: it is not 50%)

+ Show Spoiler +
33%

Before you knock on the door, there are 4 equally possible situations.
First child - Second child
Boy - Boy
Boy - Girl
Girl - Boy
Girl - Girl

Once you know that at least 1 child is male, the 4th possibility is no longer possible leaving only:
Boy - Boy
Boy - Girl
Girl - Boy

Since you don't know if the older or younger child opened the door, each of the 3 remaining possibilities are equally likely and the chance that both children are male is 33%.

For the math nerds out there, this is an example of conditional probability. http://en.wikipedia.org/wiki/Conditional_probability


I have to disagree with this one.
Probability(A given B)=P(A and B)/P(B)
P("both are male" given "at least one is male")=P("both male" and "at least one male" are true)/P("at least one male" is true)=.25/.75=.33

+ Show Spoiler +

You're wrong. The answer IS 50%. I know that there is a counterintuitive puzzle where conditional probability works like this, but you didn't give such a problem.

The puzzle only works like this if you know that one unspecified child is male, out of two unknown children. You correctly point out that if you specify one child by age, then the answer would be 50%, but you ruin the puzzle by finding a different method of singling out a child.

In this particular case, you know for sure that the child in front of you is male, therefore you automatically eliminate TWO of the possibilities (i.e. 'The child in front of you is female and the other is male' and 'Both children are female') and so there is a 50% probability that the child not in front of you is male. It also works for specifying that the younger child is male too, of course, or that the shorter child is male, or whatever.

In order to get the 33% chance, you have to make sure it's completely ambiguous which child is male.

This puzzle is absolutely great for causing arguments over probability, by the way, just like Monty Hall!

Vega62a
Profile Blog Joined December 2010
946 Posts
April 19 2012 20:29 GMT
#251
On April 20 2012 05:19 Aim Here wrote:
Show nested quote +
On April 20 2012 04:41 el_dawg wrote:
Riddle:
You are meeting a friend you have not seen in a long time. You know that they now have 2 children, but do not know the genders of the children. When you knock on the door of your friend's house, a male child answers. What is the % chance that the other child is male? (hint: it is not 50%)

+ Show Spoiler +
33%

Before you knock on the door, there are 4 equally possible situations.
First child - Second child
Boy - Boy
Boy - Girl
Girl - Boy
Girl - Girl

Once you know that at least 1 child is male, the 4th possibility is no longer possible leaving only:
Boy - Boy
Boy - Girl
Girl - Boy

Since you don't know if the older or younger child opened the door, each of the 3 remaining possibilities are equally likely and the chance that both children are male is 33%.

For the math nerds out there, this is an example of conditional probability. http://en.wikipedia.org/wiki/Conditional_probability


I have to disagree with this one.
Probability(A given B)=P(A and B)/P(B)
P("both are male" given "at least one is male")=P("both male" and "at least one male" are true)/P("at least one male" is true)=.25/.75=.33

+ Show Spoiler +

You're wrong. The answer IS 50%. I know that there is a counterintuitive puzzle where conditional probability works like this, but you didn't give such a problem.

The puzzle only works like this if you know that one unspecified child is male, out of two unknown children. You correctly point out that if you specify one child by age, then the answer would be 50%, but you ruin the puzzle by finding a different method of singling out a child.

In this particular case, you know for sure that the child in front of you is male, therefore you automatically eliminate TWO of the possibilities (i.e. 'The child in front of you is female and the other is male' and 'Both children are female') and so there is a 50% probability that the child not in front of you is male. It also works for specifying that the younger child is male too, of course, or that the shorter child is male, or whatever.

In order to get the 33% chance, you have to make sure it's completely ambiguous which child is male.

This puzzle is absolutely great for causing arguments over probability, by the way, just like Monty Hall!


+ Show Spoiler +
So essentially the way to make this puzzle work as 33% is to say, "There are two children. Of the two, at least one is male. What is the probability that the other is male?" Thereby eliminating the implicit ordering requirement.
Content of my posts reflects only my personal opinions, and not those of any employer or subsidiary
opsayo
Profile Blog Joined July 2008
591 Posts
April 19 2012 20:51 GMT
#252
On April 20 2012 05:29 Vega62a wrote:
Show nested quote +
On April 20 2012 05:19 Aim Here wrote:
On April 20 2012 04:41 el_dawg wrote:
Riddle:
You are meeting a friend you have not seen in a long time. You know that they now have 2 children, but do not know the genders of the children. When you knock on the door of your friend's house, a male child answers. What is the % chance that the other child is male? (hint: it is not 50%)

+ Show Spoiler +
33%

Before you knock on the door, there are 4 equally possible situations.
First child - Second child
Boy - Boy
Boy - Girl
Girl - Boy
Girl - Girl

Once you know that at least 1 child is male, the 4th possibility is no longer possible leaving only:
Boy - Boy
Boy - Girl
Girl - Boy

Since you don't know if the older or younger child opened the door, each of the 3 remaining possibilities are equally likely and the chance that both children are male is 33%.

For the math nerds out there, this is an example of conditional probability. http://en.wikipedia.org/wiki/Conditional_probability


I have to disagree with this one.
Probability(A given B)=P(A and B)/P(B)
P("both are male" given "at least one is male")=P("both male" and "at least one male" are true)/P("at least one male" is true)=.25/.75=.33

+ Show Spoiler +

You're wrong. The answer IS 50%. I know that there is a counterintuitive puzzle where conditional probability works like this, but you didn't give such a problem.

The puzzle only works like this if you know that one unspecified child is male, out of two unknown children. You correctly point out that if you specify one child by age, then the answer would be 50%, but you ruin the puzzle by finding a different method of singling out a child.

In this particular case, you know for sure that the child in front of you is male, therefore you automatically eliminate TWO of the possibilities (i.e. 'The child in front of you is female and the other is male' and 'Both children are female') and so there is a 50% probability that the child not in front of you is male. It also works for specifying that the younger child is male too, of course, or that the shorter child is male, or whatever.

In order to get the 33% chance, you have to make sure it's completely ambiguous which child is male.

This puzzle is absolutely great for causing arguments over probability, by the way, just like Monty Hall!


+ Show Spoiler +
So essentially the way to make this puzzle work as 33% is to say, "There are two children. Of the two, at least one is male. What is the probability that the other is male?" Thereby eliminating the implicit ordering requirement.

+ Show Spoiler +
Yeah the point is that you're supposed to introduce the idea that one is a male, what is the probability of the other being male?

Then you introduce the fact that you visited the house and opened the door - what is the probability now? (Since it changes to 50% upon opening the door).

A real good mind fuck.
el_dawg
Profile Joined September 2011
United States164 Posts
April 19 2012 20:54 GMT
#253
On April 20 2012 05:29 Vega62a wrote:
Show nested quote +
On April 20 2012 05:19 Aim Here wrote:
On April 20 2012 04:41 el_dawg wrote:
Riddle:
You are meeting a friend you have not seen in a long time. You know that they now have 2 children, but do not know the genders of the children. When you knock on the door of your friend's house, a male child answers. What is the % chance that the other child is male? (hint: it is not 50%)

+ Show Spoiler +
33%

Before you knock on the door, there are 4 equally possible situations.
First child - Second child
Boy - Boy
Boy - Girl
Girl - Boy
Girl - Girl

Once you know that at least 1 child is male, the 4th possibility is no longer possible leaving only:
Boy - Boy
Boy - Girl
Girl - Boy

Since you don't know if the older or younger child opened the door, each of the 3 remaining possibilities are equally likely and the chance that both children are male is 33%.

For the math nerds out there, this is an example of conditional probability. http://en.wikipedia.org/wiki/Conditional_probability


I have to disagree with this one.
Probability(A given B)=P(A and B)/P(B)
P("both are male" given "at least one is male")=P("both male" and "at least one male" are true)/P("at least one male" is true)=.25/.75=.33

+ Show Spoiler +

You're wrong. The answer IS 50%. I know that there is a counterintuitive puzzle where conditional probability works like this, but you didn't give such a problem.

The puzzle only works like this if you know that one unspecified child is male, out of two unknown children. You correctly point out that if you specify one child by age, then the answer would be 50%, but you ruin the puzzle by finding a different method of singling out a child.

In this particular case, you know for sure that the child in front of you is male, therefore you automatically eliminate TWO of the possibilities (i.e. 'The child in front of you is female and the other is male' and 'Both children are female') and so there is a 50% probability that the child not in front of you is male. It also works for specifying that the younger child is male too, of course, or that the shorter child is male, or whatever.

In order to get the 33% chance, you have to make sure it's completely ambiguous which child is male.

This puzzle is absolutely great for causing arguments over probability, by the way, just like Monty Hall!


+ Show Spoiler +
So essentially the way to make this puzzle work as 33% is to say, "There are two children. Of the two, at least one is male. What is the probability that the other is male?" Thereby eliminating the implicit ordering requirement.


+ Show Spoiler +
Hmm, I didn't think the child answering the door made a difference. I will have to rethink this a bit.

If we play a game where I flip 2 coins (re-flipping both if both are tails) and show you that one is heads while hiding the other, you would guess that the hidden coin has a 66% chance of being tails, right?

I really like this question, but the hardest part is finding a way that asks it correctly.
TheAngryZergling
Profile Joined January 2011
United States387 Posts
Last Edited: 2012-04-19 21:52:58
April 19 2012 21:51 GMT
#254
On April 20 2012 02:33 XiGua wrote:
I am betting that nobody will solve this without cheating.

I present to you,
THE WARDEN

The warden meets with 23 new prisoners when they arrive. He tells them, "You may meet today and plan a strategy. But after today, you will be in isolated cells and will have no communication with one another.

"In the prison is a switch room, which contains two light switches labeled 1 and 2, each of which can be in either up or the down position. I am not telling you their present positions. The switches are not connected to anything.

"After today, from time to time whenever I feel so inclined, I will select one prisoner at random and escort him to the switch room. This prisoner will select one of the two switches and reverse its position. He must flip one switch when he visits the switch room, and may only flip one of the switches. Then he'll be led back to his cell.

"No one else will be allowed to alter the switches until I lead the next prisoner into the switch room. I'm going to choose prisoners at random. I may choose the same guy three times in a row, or I may jump around and come back. I will not touch the switches, if I wanted you dead you would already be dead.

"Given enough time, everyone will eventually visit the switch room the same number of times as everyone else. At any time, anyone may declare to me, 'We have all visited the switch room.'

"If it is true, then you will all be set free. If it is false, and somebody has not yet visited the switch room, you will all die horribly. You will be carefully monitored, and any attempt to break any of these rules will result in instant death to all of you"

What is the strategy they come up with so that they can be free?


+ Show Spoiler +


there is a designated counter. if the left lever is up he adds 1 and flips it back down. if it is down he just toggles the right one. his count is 1 when he leaves the room for the first time (himself) regardless of the state of the levers.
for everyone else you flip the left switch up on your first opportunity to do so. otherwise toggle the right switch.

when the counter has flipped the left switch down 22 more times after his initial visit. he tells the warden that they all have flipped the switches.
Everything in life is most clearly explained through a Starcraft analogy.
Excludos
Profile Blog Joined April 2010
Norway8061 Posts
April 19 2012 22:06 GMT
#255
On April 20 2012 06:51 TheAngryZergling wrote:
Show nested quote +
On April 20 2012 02:33 XiGua wrote:
I am betting that nobody will solve this without cheating.

I present to you,
THE WARDEN

The warden meets with 23 new prisoners when they arrive. He tells them, "You may meet today and plan a strategy. But after today, you will be in isolated cells and will have no communication with one another.

"In the prison is a switch room, which contains two light switches labeled 1 and 2, each of which can be in either up or the down position. I am not telling you their present positions. The switches are not connected to anything.

"After today, from time to time whenever I feel so inclined, I will select one prisoner at random and escort him to the switch room. This prisoner will select one of the two switches and reverse its position. He must flip one switch when he visits the switch room, and may only flip one of the switches. Then he'll be led back to his cell.

"No one else will be allowed to alter the switches until I lead the next prisoner into the switch room. I'm going to choose prisoners at random. I may choose the same guy three times in a row, or I may jump around and come back. I will not touch the switches, if I wanted you dead you would already be dead.

"Given enough time, everyone will eventually visit the switch room the same number of times as everyone else. At any time, anyone may declare to me, 'We have all visited the switch room.'

"If it is true, then you will all be set free. If it is false, and somebody has not yet visited the switch room, you will all die horribly. You will be carefully monitored, and any attempt to break any of these rules will result in instant death to all of you"

What is the strategy they come up with so that they can be free?


+ Show Spoiler +


there is a designated counter. if the left lever is up he adds 1 and flips it back down. if it is down he just toggles the right one. his count is 1 when he leaves the room for the first time (himself) regardless of the state of the levers.
for everyone else you flip the left switch up on your first opportunity to do so. otherwise toggle the right switch.

when the counter has flipped the left switch down 22 more times after his initial visit. he tells the warden that they all have flipped the switches.


+ Show Spoiler +
Someone might come into the room first, not knowing if the counter has been there or not, and flips the switch up. If the counter then starts at 1 at his first visit, he will be stuck at 22 never reaching 23. The huge problem with this riddle is that you don't know the state of the switches before you enter the room.
el_dawg
Profile Joined September 2011
United States164 Posts
Last Edited: 2012-04-19 22:50:51
April 19 2012 22:38 GMT
#256
On April 20 2012 05:54 el_dawg wrote:
Show nested quote +
On April 20 2012 05:29 Vega62a wrote:
On April 20 2012 05:19 Aim Here wrote:
On April 20 2012 04:41 el_dawg wrote:
Riddle:
You are meeting a friend you have not seen in a long time. You know that they now have 2 children, but do not know the genders of the children. When you knock on the door of your friend's house, a male child answers. What is the % chance that the other child is male? (hint: it is not 50%)

+ Show Spoiler +
33%

Before you knock on the door, there are 4 equally possible situations.
First child - Second child
Boy - Boy
Boy - Girl
Girl - Boy
Girl - Girl

Once you know that at least 1 child is male, the 4th possibility is no longer possible leaving only:
Boy - Boy
Boy - Girl
Girl - Boy

Since you don't know if the older or younger child opened the door, each of the 3 remaining possibilities are equally likely and the chance that both children are male is 33%.

For the math nerds out there, this is an example of conditional probability. http://en.wikipedia.org/wiki/Conditional_probability


I have to disagree with this one.
Probability(A given B)=P(A and B)/P(B)
P("both are male" given "at least one is male")=P("both male" and "at least one male" are true)/P("at least one male" is true)=.25/.75=.33

+ Show Spoiler +

You're wrong. The answer IS 50%. I know that there is a counterintuitive puzzle where conditional probability works like this, but you didn't give such a problem.

The puzzle only works like this if you know that one unspecified child is male, out of two unknown children. You correctly point out that if you specify one child by age, then the answer would be 50%, but you ruin the puzzle by finding a different method of singling out a child.

In this particular case, you know for sure that the child in front of you is male, therefore you automatically eliminate TWO of the possibilities (i.e. 'The child in front of you is female and the other is male' and 'Both children are female') and so there is a 50% probability that the child not in front of you is male. It also works for specifying that the younger child is male too, of course, or that the shorter child is male, or whatever.

In order to get the 33% chance, you have to make sure it's completely ambiguous which child is male.

This puzzle is absolutely great for causing arguments over probability, by the way, just like Monty Hall!


+ Show Spoiler +
So essentially the way to make this puzzle work as 33% is to say, "There are two children. Of the two, at least one is male. What is the probability that the other is male?" Thereby eliminating the implicit ordering requirement.


+ Show Spoiler +
Hmm, I didn't think the child answering the door made a difference. I will have to rethink this a bit.

If we play a game where I flip 2 coins (re-flipping both if both are tails) and show you that one is heads while hiding the other, you would guess that the hidden coin has a 66% chance of being tails, right?

I really like this question, but the hardest part is finding a way that asks it correctly.

+ Show Spoiler +
I think what is missing from my question is whether or not the child opening the door was random. For example, in the Monty hall problem, it is understood that the host will never open the door with the car. So this question needs something to say that not only did a boy happen to answer the door, but a boy will always answer the door.

Edit: the monty hall problem still works if the reveal was random (assuming a goat was revealed), I'm just using it as an example where the revealed info is usually chosen by the asker who knows the status of all the doors/ coins/ children.
Diavlo
Profile Joined July 2011
Belgium2915 Posts
Last Edited: 2012-04-19 23:03:11
April 19 2012 23:01 GMT
#257
On April 20 2012 04:41 el_dawg wrote:
Riddle:
You are meeting a friend you have not seen in a long time. You know that they now have 2 children, but do not know the genders of the children. When you knock on the door of your friend's house, a male child answers. What is the % chance that the other child is male? (hint: it is not 50%)

+ Show Spoiler +
33%

Before you knock on the door, there are 4 equally possible situations.
First child - Second child
Boy - Boy
Boy - Girl
Girl - Boy
Girl - Girl

Once you know that at least 1 child is male, the 4th possibility is no longer possible leaving only:
Boy - Boy
Boy - Girl
Girl - Boy

Since you don't know if the older or younger child opened the door, each of the 3 remaining possibilities are equally likely and the chance that both children are male is 33%.

For the math nerds out there, this is an example of conditional probability. http://en.wikipedia.org/wiki/Conditional_probability

Probability(A given B)=P(A and B)/P(B)
P("both are male" given "at least one is male")=P("both male" and "at least one male" are true)/P("at least one male" is true)=.25/.75=.33


What's funny is that even if your answer is mathematically incorrect. The answer is not 50% even if you consider the chance of having a girl or a boy to be 50/50.

Can you guess why??




+ Show Spoiler +
The two child could be true twins, making the probably of having a second boy in the house slighty higher
"I don't know how many years on this Earth I got left. I'm gonna get real weird with it."
crate
Profile Blog Joined May 2009
United States2474 Posts
April 20 2012 00:03 GMT
#258
On April 20 2012 07:06 Excludos wrote:
Show nested quote +
On April 20 2012 06:51 TheAngryZergling wrote:
On April 20 2012 02:33 XiGua wrote:
I am betting that nobody will solve this without cheating.

I present to you,
THE WARDEN

The warden meets with 23 new prisoners when they arrive. He tells them, "You may meet today and plan a strategy. But after today, you will be in isolated cells and will have no communication with one another.

"In the prison is a switch room, which contains two light switches labeled 1 and 2, each of which can be in either up or the down position. I am not telling you their present positions. The switches are not connected to anything.

"After today, from time to time whenever I feel so inclined, I will select one prisoner at random and escort him to the switch room. This prisoner will select one of the two switches and reverse its position. He must flip one switch when he visits the switch room, and may only flip one of the switches. Then he'll be led back to his cell.

"No one else will be allowed to alter the switches until I lead the next prisoner into the switch room. I'm going to choose prisoners at random. I may choose the same guy three times in a row, or I may jump around and come back. I will not touch the switches, if I wanted you dead you would already be dead.

"Given enough time, everyone will eventually visit the switch room the same number of times as everyone else. At any time, anyone may declare to me, 'We have all visited the switch room.'

"If it is true, then you will all be set free. If it is false, and somebody has not yet visited the switch room, you will all die horribly. You will be carefully monitored, and any attempt to break any of these rules will result in instant death to all of you"

What is the strategy they come up with so that they can be free?


+ Show Spoiler +


there is a designated counter. if the left lever is up he adds 1 and flips it back down. if it is down he just toggles the right one. his count is 1 when he leaves the room for the first time (himself) regardless of the state of the levers.
for everyone else you flip the left switch up on your first opportunity to do so. otherwise toggle the right switch.

when the counter has flipped the left switch down 22 more times after his initial visit. he tells the warden that they all have flipped the switches.


+ Show Spoiler +
Someone might come into the room first, not knowing if the counter has been there or not, and flips the switch up. If the counter then starts at 1 at his first visit, he will be stuck at 22 never reaching 23. The huge problem with this riddle is that you don't know the state of the switches before you enter the room.


+ Show Spoiler +
The same solution works, you just count to 43 or whatever (I'm too lazy to figure out the exact number) and make sure everyone flips the counting lever twice
We did. You did. Yes we can. No. || http://crawl.akrasiac.org/scoring/players/crate.html || twitch.tv/crate3333
icystorage
Profile Blog Joined November 2008
Jollibee19346 Posts
April 20 2012 00:37 GMT
#259
On April 20 2012 09:03 crate wrote:
Show nested quote +
On April 20 2012 07:06 Excludos wrote:
On April 20 2012 06:51 TheAngryZergling wrote:
On April 20 2012 02:33 XiGua wrote:
I am betting that nobody will solve this without cheating.

I present to you,
THE WARDEN

The warden meets with 23 new prisoners when they arrive. He tells them, "You may meet today and plan a strategy. But after today, you will be in isolated cells and will have no communication with one another.

"In the prison is a switch room, which contains two light switches labeled 1 and 2, each of which can be in either up or the down position. I am not telling you their present positions. The switches are not connected to anything.

"After today, from time to time whenever I feel so inclined, I will select one prisoner at random and escort him to the switch room. This prisoner will select one of the two switches and reverse its position. He must flip one switch when he visits the switch room, and may only flip one of the switches. Then he'll be led back to his cell.

"No one else will be allowed to alter the switches until I lead the next prisoner into the switch room. I'm going to choose prisoners at random. I may choose the same guy three times in a row, or I may jump around and come back. I will not touch the switches, if I wanted you dead you would already be dead.

"Given enough time, everyone will eventually visit the switch room the same number of times as everyone else. At any time, anyone may declare to me, 'We have all visited the switch room.'

"If it is true, then you will all be set free. If it is false, and somebody has not yet visited the switch room, you will all die horribly. You will be carefully monitored, and any attempt to break any of these rules will result in instant death to all of you"

What is the strategy they come up with so that they can be free?


+ Show Spoiler +


there is a designated counter. if the left lever is up he adds 1 and flips it back down. if it is down he just toggles the right one. his count is 1 when he leaves the room for the first time (himself) regardless of the state of the levers.
for everyone else you flip the left switch up on your first opportunity to do so. otherwise toggle the right switch.

when the counter has flipped the left switch down 22 more times after his initial visit. he tells the warden that they all have flipped the switches.


+ Show Spoiler +
Someone might come into the room first, not knowing if the counter has been there or not, and flips the switch up. If the counter then starts at 1 at his first visit, he will be stuck at 22 never reaching 23. The huge problem with this riddle is that you don't know the state of the switches before you enter the room.


+ Show Spoiler +
The same solution works, you just count to 43 or whatever (I'm too lazy to figure out the exact number) and make sure everyone flips the counting lever twice


crate's solution is true, just to make sure


+ Show Spoiler +
just make sure everybody flips the switch down twice, so at some point (too lazy to figure the number too) it still insures everybody has AT LEAST flipped the switch once, which is the only requirement
LiquidDota StaffAre you ready for a Miracle-? We are! The International 2017 Champions!
opsayo
Profile Blog Joined July 2008
591 Posts
April 20 2012 01:03 GMT
#260
The prisoner light switch problem is extremely well known and there are very advanced solutions for it:

http://www.ocf.berkeley.edu/~wwu/papers/100prisonersLightBulb.pdf
Prev 1 11 12 13 14 15 38 Next All
Please log in or register to reply.
Live Events Refresh
The PondCast
10:00
Episode 55
CranKy Ducklings25
Liquipedia
[ Submit Event ]
Live Streams
Refresh
StarCraft 2
Nina 243
Lowko0
StarCraft: Brood War
firebathero 3328
Sea 2936
Mind 1068
Stork 730
Larva 452
BeSt 439
Light 251
Barracks 225
Zeus 222
PianO 159
[ Show more ]
Leta 132
TY 127
sorry 72
Last 71
ToSsGirL 68
GoRush 64
sSak 58
Backho 32
Sacsri 28
Rush 25
Sharp 23
JulyZerg 23
Noble 12
ajuk12(nOOB) 7
IntoTheRainbow 6
Bale 4
Hm[arnc] 4
Dota 2
Gorgc6227
singsing1521
canceldota513
XaKoH 335
League of Legends
JimRising 400
Counter-Strike
shoxiejesuss698
x6flipin365
allub232
sgares221
Other Games
Fuzer 233
DeMusliM153
SortOf108
Mew2King51
Trikslyr18
Organizations
Other Games
gamesdonequick2365
StarCraft 2
Blizzard YouTube
StarCraft: Brood War
BSLTrovo
sctven
[ Show 12 non-featured ]
StarCraft 2
• LUISG 22
• AfreecaTV YouTube
• intothetv
• Kozan
• IndyKCrew
• LaughNgamezSOOP
• Migwel
• sooper7s
StarCraft: Brood War
• BSLYoutube
• STPLYoutube
• ZZZeroYoutube
League of Legends
• Jankos1215
Upcoming Events
OSC
2h 34m
WardiTV European League
5h 34m
Fjant vs Babymarine
Mixu vs HiGhDrA
Gerald vs ArT
goblin vs MaNa
Jumy vs YoungYakov
Replay Cast
13h 34m
Epic.LAN
1d 1h
CranKy Ducklings
1d 23h
Epic.LAN
2 days
CSO Contender
2 days
BSL20 Non-Korean Champi…
2 days
Bonyth vs Sziky
Dewalt vs Hawk
Hawk vs QiaoGege
Sziky vs Dewalt
Mihu vs Bonyth
Zhanhun vs QiaoGege
QiaoGege vs Fengzi
Sparkling Tuna Cup
2 days
Online Event
3 days
[ Show More ]
BSL20 Non-Korean Champi…
3 days
Bonyth vs Zhanhun
Dewalt vs Mihu
Hawk vs Sziky
Sziky vs QiaoGege
Mihu vs Hawk
Zhanhun vs Dewalt
Fengzi vs Bonyth
Esports World Cup
4 days
ByuN vs Astrea
Lambo vs HeRoMaRinE
Clem vs TBD
Solar vs Zoun
SHIN vs Reynor
Maru vs TriGGeR
herO vs Lancer
Cure vs ShoWTimE
Esports World Cup
5 days
Esports World Cup
6 days
Liquipedia Results

Completed

JPL Season 2
RSL Revival: Season 1
Murky Cup #2

Ongoing

BSL 2v2 Season 3
Copa Latinoamericana 4
Jiahua Invitational
BSL20 Non-Korean Championship
Championship of Russia 2025
FISSURE Playground #1
BLAST.tv Austin Major 2025
ESL Impact League Season 7
IEM Dallas 2025
PGL Astana 2025
Asian Champions League '25
BLAST Rivals Spring 2025
MESA Nomadic Masters

Upcoming

CSL Xiamen Invitational
CSL Xiamen Invitational: ShowMatche
2025 ACS Season 2
CSLPRO Last Chance 2025
CSLPRO Chat StarLAN 3
BSL Season 21
K-Championship
RSL Revival: Season 2
SEL Season 2 Championship
uThermal 2v2 Main Event
FEL Cracov 2025
Esports World Cup 2025
Underdog Cup #2
ESL Pro League S22
StarSeries Fall 2025
FISSURE Playground #2
BLAST Open Fall 2025
BLAST Open Fall Qual
Esports World Cup 2025
BLAST Bounty Fall 2025
BLAST Bounty Fall Qual
IEM Cologne 2025
TLPD

1. ByuN
2. TY
3. Dark
4. Solar
5. Stats
6. Nerchio
7. sOs
8. soO
9. INnoVation
10. Elazer
1. Rain
2. Flash
3. EffOrt
4. Last
5. Bisu
6. Soulkey
7. Mini
8. Sharp
Sidebar Settings...

Advertising | Privacy Policy | Terms Of Use | Contact Us

Original banner artwork: Jim Warren
The contents of this webpage are copyright © 2025 TLnet. All Rights Reserved.