# Stick Figure Puzzle

Wait But Why holds a work retreat for all of its stick figures. Everyone’s excited about it—it’s a chance to get to know each other better, which is rare because they’re typically confined to their own drawings and not able to socialize.

Soon, though, things take a dark turn, as it is announced that Wait But Why can no longer employ all of them and needs to downsize.

The decision is that all stick figures with perfect head circles will keep their jobs, and those whose heads Tim messed up by not connecting the circle will be erased.

The thing is, stick figures can’t look in the mirror and there’s no way for them to tell what their own head looks like. They can see all the other stick figures’ heads, but not their own.

To avoid the awkwardness of having to tell the messed up drawings that they’re going to be erased, Tim and Andrew make the following statements and then spinelessly leave the retreat:

Statement 1: From here forward, there will be no mention of anyone’s head quality whatsoever, including any facial expressions or body language that could give someone information about what their head looks like.

Statement 2: If at any point you discover that your head is messed up, you are required to erase yourself that night at midnight.

Statement 3: All stick figures must be perfectly obedient to the rules, perfect at logic, and aware that all the others are perfect at logic as well.

Statement 4: It is announced that there are 100 stick figures present at the retreat and at least one stick figure present has a messed up head.

Disturbed, the stick figures go on with Day 1 of the retreat and try to enjoy themselves as best they can. That night at 10pm, they all head to their bedrooms and go to sleep.

The next day (Day 2), the stick figures come out for the day, and everyone is still there. Same story on Day 3. In fact, it’s the same story through Day 8.

But when the stick figures emerge from their bedrooms on the morning of Day 9, the only ones present have perfect heads. It seems that all stick figures with messed up heads erased themselves on the night of Day 8.

The question is: How many stick figures erased themselves on Day 8 and why did things play out this way?

This is based on a famous logic puzzle. Think long and hard about it before you look at the solution. There are two hints below to help you along the way if you’re stuck.

HINT 1
How would things play out if only one of the stick figures present had a messed up head?

HINT 2
How about if there were two with a messed up head? On Day 1, what would each of those two think when they looked around at all the other stick figures?

SOLUTION
Eight stick figures erased themselves on Day 8. Here’s why:

Start by thinking about the case in which there were only one stick figure with a messed up head. He would look around the room on Day 1, see no one with a messed up head, and realize that he must have a messed up head (since Statement 4 states that at least one stick figure has a messed up head). Since he’s perfectly obedient (as per Statement 3), he would erase himself that night.

Now imagine that there’s not one, but two stick figures with a messed up head—let’s call them Guy A and Guy B. On Day 1, both Guy A and Guy B would look around the room and see one stick figure with a messed up head and the other 98 with a perfect head. Guy A assumes that Guy B will see no one else with a messed up head, know he himself must have one, and erase himself that night.

But when Day 2 rolls around and Guy A sees Guy B emerge that morning, fully intact, he knows that Guy B must have gone to bed without the knowledge that his head was messed up. The only way that could have happened, since Guy B knows that at least one person has a messed up head, is if Guy B saw someone else with a messed up head. Since Guy A saw perfect heads on the other 98 stick figures, he concludes that he must also have a messed up head.

Guy B goes through the same thought process, so on Day 2, both of them know that they have messed up heads and both erase themselves that night at midnight.

Taking it a step further, if there are three stick figures with messed up heads (Guy A, B, and C), on Day 1, everyone can see three messed up heads, except the three who have them, who only see two messed up heads. So Guy A looks at Guy B and Guy C and thinks, “Those two don’t realize they have messed up heads and only see the other’s messed up head—so they will expect each other to erase themselves on the night of Day 1. But when Day 2 rolls around, they’ll see each other and realize the truth, that they have a messed up head too—and they’ll both erase themselves on the night of Day 2.” But when Guy A wakes up on the morning of Day 3 and sees both Guy B and Guy C alive and well, it hits him that he himself must also have a messed up head—which is the only explanation for Guy B and Guy C going to bed on Day 2 still not knowing if they had a messed up head.

Guy B and Guy C have this exact same revelation on Day 3, and that night, all three of them erase themselves.

Continuing this pattern, if Day 8 is the day the erasings finally happen, there would have to be eight stick figures with messed up heads. Each of them would have spent the retreat seeing seven other messed up heads and each would have gone to bed on Day 7 assuming that that night would be the night of the erasings. When all the messed up heads emerged on Day 8, each one of them then realized that they must have a messed up head too.

Make sense? How’d you do?

If the hint and solution links won’t expand on your device, view them here.

___________

More puzzles on Wait But Why:

The Three Jellybean Problem

The Puzzle of the Pirate Booty

The Two Envelopes Problem

The Infinite Checkerboard Quandary

And one very big puzzle:

What Makes You You?

• 109

27 were erased. . . the 27 presidents after Lincoln. More presidents please?

• Wait But Why

Touche. Sometime soon.

• piper.

this is the best comment i’ve ever read.

• Barney

Okay well it took 30 minutes of thinking and jotting diagrams on my work report but I finally get it. Unbelievable puzzle.

• Grace

OH. MY. GOD.

• i dont like to brag but i nailed it.

nailed it.

• i dont like to brag but i nailed it.

#noHints

i’m kind of lame.

• #2 Ticonderoga

All that is left are little stick hands clutching pencils. So sad.

• Nic

It is a good puzzle, but makes me feel sad thinking of the poor little stick figures with messed up heads having to quietly erase themselves.

• Dude

Don’t forget your commitments. PRESIDENTS. DO IT.

• Sean

Using this logic, what is stopping the people with perfectly fine heads from erasing themselves?

• Frank

Or eight people erasing themselves at once? Couldn’t the number be 54 people or 12?

• Kyle

If there were 8 bad-heads, and morning 9 rolled around, then everyone would have erased themselves that night. It is like, “I see 2 bad people, and it has been 2 nights, so there must be 3 bad people. Okay I’ll kill myself tonight.”

• Tim Ryan

Agree with Kyle. The logic of the puzzle is sound.

• finn

how the hell do you erase yourself?? At some point a paradox occurs!!!

• PureLogic

At night after the 8th day it rained. The stick figures with an imperfect head filled up with rain water which made their heads explode.

• Strange but untrue

All of them erased themselves, because one of the seven figures with messed-up heads was secretly blind but shared a room with one of the non-messed up heads.

• Strange but untrue

No, wait, that leaves one messed-head guy alive. Okay, as above, except they all share a room and two of the sighted ones mention on Day 1 the possibility of a reunion next year.

• Robert

Being familiar with this kind of puzzle, as a stick figure I would immediately stop all interaction with other stick figures, i.e. I would never notice who erased themselves when. It is not prohibited by the way the puzzle is posed and will keep any dangerous input to my logic processor away from me. Come to think of it, even if I do not apply this strategy myself, I could always assume that other stick figures do – thus, logic alone is not enough to make me erase myself, even if I have a messed up head.
Given this possibility, I can only conclude that the simultaneous self-erasure of all stick figures with messed-up heads is a freakish coincidence and I have no way to determine how many there were.

• there_there

Cracked it! I just couldn’t let go. That was fun. Now back to workywork…

• there_there

By the way this puzzle would be a nice introduction for a Game Theory class

• boso

Ha, I got it. And that’s good for my selfesteem since I usually suck at riddles.

• Leo X

No way Tim, you messed up way more heads than 8 😉

• Shane

Got it! But not without the hints. Great puzzle!

• Andy

One of Tim’s stick figures has erased himself and been reincarnated as a real stick

http://www.ebay.com.au/itm/141259631308

• Alek

“Guy A assumes that Guy B will see no one else with a messed up head, know he himself must have one, and erase himself that night.”

Would it be more accurate to say, “Guy A assumes that as long he himself does not have a messed up head, Guy B will see no one else with a messed-up head, know that he (Guy B) has a messed-up head, and erase himself that night.”

Or, “Guy A knows that either Guy B will erase himself that night, or they both have messed up heads, and he will learn which one it is (and thus his own fate) the next morning.”

Guy A assuming something he can’t yet know would not be perfect logic.

Unless I’m totally missing something.

• Schuxu

The statements are equivalent. You just have to realize that “he” and “himself” refer to Guy B in the original wording and not to Guy A.

• Anonymous

The statements are equivalent, but the first has Guy A assuming that his own head is fine, which he has no reason to do. If they all have perfect logic, then they all know it’s possible they have messed-up heads. He shouldn’t be assuming anything.

• KCDC

I don’t understand how the nights passing correlates with the number of stick figures with imperfect heads. Can someone explain?

• Schuxu

“Statement 2: If at any point you discover that your head is messed up, you are required to erase yourself that night at midnight.”

• Laura

I hate story problems. My head is messed up.

• Gonz

Therefore, you must erase yourself tonight at midnight. 😛

• Jonathan

She’s still here, Gonz. Your turn… or am I next?

• Anna

Rationally, this makes sense, but thank goodness they are stick figures. No human being would ever be willing to admit to themselves that they personally must be the eighth if it meant their annihilation. They would just wander around thinking those seven guys were idiots for days and days.

• rohit

well.. i feel the logic is flawed. Since no one is able to see himself , the scenario will change altogether.

Let’s assume there are 2 guys with incomplete heads and we know that no one knows the exact count of incomplete head guys. So, after day 1 all the 99 will expect the 100th guy to erase himself but when he doesn’t do that, the next morning all 99 will feel that they have incomplete heads and thus, all will kill themselves.

Please tell me if I am wrong.

• Josh

You’re forgetting that 98 of them can see 2 messed up heads, and 2 that can see 1 messed up head. By the second day, they both know that they have messed up heads, because they can only see 1 other, the other 98 can see 2 and know that they are still safe.

• rohit

got it.. 😛

• d

@ rohit – could you explain it to me because I am still seeing it the way you did in your first message. There is no way that if the stick figures followed perfect logic, all 100 stick figures would not assume on day two of seeing 7 messed up heads in a row that there is a 50% chance they themselves have a messed up head.

As far as logic dictates, this means that on the day 8 all stick figures would erase themselves.

• Kev_a_Swing_Dancer

To me the logic only works up to 4 people. If the 3 don’t erase themselves, 4, not seeing any others, knows he must go (at least that’s what we’re thinking). If there are 5, each always sees imperfects. The logic of 3 can no longer exist if #5 sees 4 imperfects. The only reason 4 would erase himself is if he didn’t see a 5; and they ALL see the fifth and can’t distinguish a 5th from a 4th and from a 3rd. The 3rd is the last to “benefit” from the if-not-only-2-then-I-must-be-the-extra-one logic. The fallacy was only bringing the proof to 3 and letting the solver work out the 4th (maybe). If there are 5, …let’s make it easier, 6, …there’s no day 1 where someone wakes up and wonders why 5 imperfects are still walking around; there’s no incremental, additive information to get started with.

• Diane

I’m with you, Kev. I can’t get my head around why any more than four would work.

• Tobermory

Late to the party but… The key is the speed of propogation of knowledge. The act of other figures not erasing themselves tells you a little bit about what they can see.

Day 1: Everyone knows if there was 1 bad head, he would erase himself that night.
Day 2: Everyone knows what happened on day 1. So now everyone knows there are at least 2 bad heads. Anyone who only sees 1 bad head would know they were the second, and erase themselves.
Day 3: Everyone knows what happened on day 2. So now everyone knows there are at least 3 bad heads. Anyone who only sees 2 bad heads would know to erase themselves that night.
Day 4: Everyone knows what happened on day 3. So now everyone knows there are at least 4 bad heads. Anyone who only sees 3 bad heads would know to erase themselves that night.
Day 5: Everyone knows what happened on day 4. So now everyone knows there are at least 5 bad heads. Anyone who only sees 4 bad heads would know to erase themselves that night.
Day 6: Everyone knows what happened on day 5. So now everyone knows there are at least 6 bad heads. Anyone who only sees 5 bad heads would know to erase themselves that night.

Also, note that all the perfect-head people would always see one more bad head so wouldn’t get the full picture until the morning after Erasure Night.

• Failed

In this case the hints thew me off further.
Ah so if the figures with the messed up heads waited an extra day (due to having messed up heads) then all the figures will erase themselves.

• bwahaha

nailed it <3

• champ

“The question is: How many stick figures erased themselves on Day 8 and why did things play out this way?”
“The question is: How many stick figures erased themselves by Day 8 and why did things play out this way?”

• Baelnor

Why all the retarded presidents requests? Who cares about presidents? Unless of course you mean presidents of Chile. That might actually be remotely interesting.

• robeot

here’s my hopefully clear explanation:
(this is one of those problems where it really helps to look at the problem as an observer both inside and outside the system)

the rules state that these stick people perfectly logical and adherent to the rules. they’re like super spocks’ or some shit. as such, say they decide as a community to establish a rule that will tell them all, infallibly, if anyone one of them is required to erase themselves without the need of another telling them to do so.

first,: the workers realize they cannot see their own heads, therefore they -must- each achieve this knowledge from outside themselves somehow.

second: they note the only parameter that’s shared between them all is that they must erase themselves at midnight if they find out they have abnormal heads. therefore, they realize information can be exchanged through this midnight erasing event in the sense that they can observe how many co-workers are left from the original worker group with each passing of midnight.

third: being the perfectly logical and rule-adherent stick figures that they are, they use the two above ideas to realize they can use each passing day as a sort of counting mechanism.

four: on day 1 (before the workers go to bed), each worker will count the number of abnormal heads they see. they establish the rule that for however many abnormal heads each individual counts (Y), you wait that many days plus one before erasing yourself.

Y = number of abnormal heads amongst your co-workers that you count on day one.
D = day that you erase yourself on
D = Y + 1 (this is what rule #4 describes)

five: once anyone erases themselves based on these rules, no one else needs to erase themselves!

this rule works because anyone that has an abnormal head will always observe Y as one less than someone with a regular head (regular head folks see all of the abnormal heads and the abnormal head folks see all of the abnormal head folks except for his/herself). so if you count 7 abnormal heads and day 7 passes without anyone erasing themselves, you know you must erase yourself on day 8 because you now know you are one of the abnormal head folks since D = Y + 1 = 7 + 1 = 8 and no one had erased themselves yet. likewise, if you were in that same situation but had a regular head, you would count 8 abnormal heads so you would plan on erasing yourself on D = 8 + 1, but those 8 abnormal headed co-workers you counted would have already erased themselves on day 8.

hope that helps are least one person…

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• Stuart

Based on the Blue Forehead problem. I love the blue forehead problem.

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• Information Theory

Easier way to “guesstimate” the answer.
There are only 3 pieces of given information in the problem: 1, 8, 100.
The number of erased people is at least 1 and at most 100. The only other option possible is 8, because there are 8 days, and because no other information is given and no suggestions of any operations or permutations on the number 8.
For example, it can’t be the number 17, because there would be no way to get the number 17 from the given information given the rules.
So it had to be either 1, 8, or 100 people erased.
Using inductive reasoning and common sense, one chooses 8 because it would be silly for
A. the answer to be everyone, because that isn’t implied by the premise, or
B. the answer to be only 1, because more information is given, and we assume that we aren’t given useless information.

Therefore we can reasonably assume that 8 people were erased, because no other number makes sense.

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• Confused

I don’t understand why everyone is saying you would only realise there are 2 bad heads after 2 nights and 3 after 3 nights etc. The reality is there are 8 bad heads, therefore you would either see 7 bad heads (if you yourself had a bad head) or 8 (if you didn’t have a bad head) from the get go. Therefore how would you know if you yourself were one of those 8?

• Joseph

It is precisely because not everyone is seeing the same thing. (that’s the key point with a lot of those ‘trick’ logical puzzles. Everyone is presented with the same situation and rules, but not everyone sees the same things if you put yourself in their shoes due to different starting conditions).

If you followed the logic of the post. It is obvious that everyone will either see N or N+1 bad heads. They will all wait the same number of days that they saw bad heads before concluding that they also have bad head and remove themselves.

If you are in the N group and N days has passed, you know you are the bad head and erase yourself by the morning of N+1 day.

IF you are in the N+1 group, you wouldn’t need to worry because every bad heads would have deleted themselves the night before. If they hadn’t, it would be clear to you that the number of bad heads is actually N+2 and that you are one of them, and that no one has deleted themselves because some people are seeing N+3 bad heads (the actual total number). This revelation will dawn on you and everyone who sees N+2 heads simultaneously (because they are in the same situation as you) and so everyone with bad heads including you will delete themselves that night.

In other words, people who sees the number of bad heads minus 1 will always dawn on their situation one step ahead of the people who see the number of bad heads, and all of them simultaneously. The latter will only know they are the good heads once the former had erased themselves.

Take the simplest case of one bad head present, and work your way up by induction, and you’ll see what I mean.

• Joseph

Typos in the 4th paragraph of my previous post. The N+3 is actually supposed to be N+2 and the second N+2 is supposed to be N+1.

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