# Pause and Try This Prison Riddle (Spoiler: You’ll Give Up!)

Predict the right number of switches and go free but if you guess wrong, you’ll never leave this prison puzzle!

Get 15% off new domain registrations at http://www.domain.com with promocode SCAMSCHOOL.

Find Brian on social media:

http://instagram.com/scamschoolbrian

This episode of Scam School was made by:
Brian Brushwood
Brandt Hughes
Bryce Castillo
Roberto Villegas
Jeff Schuessler

Filmed on location at The Rusty Mule in Austin, TX.

source

Fahad Hashmi is one of the known Software Engineer and blogger likes to blog about design resources. He is passionate about collecting the awe-inspiring design tools, to help designers.He blogs only for Designers & Photographers.

### 45 thoughts on “Pause and Try This Prison Riddle (Spoiler: You’ll Give Up!)”

• September 24, 2017 at 9:40 am

I got so close, but I got hung up on the starting position – I didn't think to make each prisoner flip the counter twice instead of just once.

• September 24, 2017 at 9:40 am

Q) If Brian has been in prison for the past 8 years after a episode of scam school, How is he still doing this show?

A) A magician never reveals his secrets.

• September 24, 2017 at 9:40 am

I thought the guy that was the last of the 21 to be called in had to proclaim it then and there or everyone would be killed. I mean otherwise why even bother and just keep keep going there once in a while and ignore the entire thing…

• September 24, 2017 at 9:40 am

what if the counter is taken first?

• September 24, 2017 at 9:40 am

Can't you make it faster by just designating the first guy to be the "counter"? On the first day, he sets his count to 1, and makes sure switch A is off. (If it's on, turn it off, if it's not on, play with switch B). Everyone else leaves switch A alone if it is off, but if it is on, and they have never flipped it before, they flip it to off. The counter then increments his count whenever switch A is on, and switches it off, switching B and leaving the count alone if it's already off. Eventually, his count will reach 21, and he can announce the end.

• September 24, 2017 at 9:40 am

brian brushwood i watched scam school since i was 7 and im 14 and a half now, dude im so glad you uploaded

• September 24, 2017 at 9:40 am

that guy rocks that moustache

• September 24, 2017 at 9:40 am

I said for sure this was completely impossible because I thought that if you were the last person and didn't say so, that would also get you executed.

• September 24, 2017 at 9:40 am

Too tough – there's stuff coming out of my ears – my brain is boiling.

• September 24, 2017 at 9:40 am

that just confused tf out of me

• September 24, 2017 at 9:40 am

Is this that guy from haking the sistem on Netflix

• September 24, 2017 at 9:40 am

Small problem, the guards could have withheld one prisoner allowing only 20 prisoners so prisoner 21 could have made a mistake. The riddle only shows that prisoner 21 has been in the room twice in a row.

• September 24, 2017 at 9:40 am

my answer they used the prison ups aka trash mail to send a peice of paper with a number on it to all prisoners and if the number you received in the prison ups is 20 say all where here if you are just being called also the prison ups is a real thing

• September 24, 2017 at 9:40 am

Cool puzzle but how do the prisoners know how to pick a counter when no communication is allowed?

• September 24, 2017 at 9:40 am

Piss on the floor once there is 21 per spots you will know

• September 24, 2017 at 9:40 am

Nice, very nice👏👏👏

• September 24, 2017 at 9:40 am

• September 24, 2017 at 9:40 am

You don't have to make everyone do it twice. The counter just need to flip it 21 times in total to make sure everyone has flipped the switch their one time.

• September 24, 2017 at 9:40 am

And I still dont get it!

• September 24, 2017 at 9:40 am

I thought of the answer and I got it right but I still don't know how it work

• September 24, 2017 at 9:40 am

The answer confused me even more than the original riddle did.

• September 24, 2017 at 9:40 am

just a poorly posed problem,  try again.

• September 24, 2017 at 9:40 am

well i got it but then again i saw this before

• September 24, 2017 at 9:40 am

I’m so confuzzled by this

• September 24, 2017 at 9:40 am

didn't say you couldn't leave a peice of clothing behind

• September 24, 2017 at 9:40 am

Wow. That was annoyingly convoluted.

• September 24, 2017 at 9:40 am

man sry!!🙏 can't take it anymore😣😣

• September 24, 2017 at 9:40 am

It's easier then this. just nobody say everyone has been in there

• September 24, 2017 at 9:40 am

I wish it was made clear that the switches can be given names A and B. Could also be that you get put into the room blindfolded and can't tell which switch is the right or left one if you don't know from which side to look at the two switches. Without that info my solution was to just never proclaim anything and technically survive until you die of natural causes. The guards or whatever even state that the inmates will never see each other again, implying the problem is unsolvable and that they will never be released.

The problem of watching these at home is that you can't ask for clarification on the rules so the wording has to be perfect the first time.

• September 24, 2017 at 9:40 am

The problem is that I thought that if you were the last one to enter, you'd have to immediately state that everyone has been in there. That info was missing.

• September 24, 2017 at 9:40 am

The prisoners better hope that they get taken to the switches often, because they have to go in there almost 1000 times (45 each) with this strategy, mostly just waiting on the counter. If it only happens about once per day, then it would take almost 3 years for them to escape. I'm sure that one of the inmates would crack long before the strategy is over.

• September 24, 2017 at 9:40 am

They should leave a piece of clothing only for the first time they were selected except one. Therefore when the 21st prisoners sees 20 pieces of clothing he will proclaim he's the last one to go.

• September 24, 2017 at 9:40 am

Not possible.. 1 or 2 people may never go in and the count could be greater then the number.. especially if there is NO communication between them.. who cares if you flipped it twice who would ever know… i call shenanigans!

• September 24, 2017 at 9:40 am

I'm going to see if I can solve it before the ad ends. Spoilers!

You set a basic setting for the switches, and one counter. Let's say you choose two down.

The counter is the only man who can guess.

The counter will reset the switches to down. If a prisoner sees the down switch, they aren't the counter, and they haven't changed it before or have only changed it once, they go to one up. If the previous three conditions are met, but the switches are one up and one down, (ad ends here, I paused video and this is where edit begins) the prisoner should make it two up. If the former conditions are met but the switches are both up, only the counter can move them, and to the down position. The first counter visit should be used moving both switches to down to be a sort of reset.

He counts the people moving by how many switches he flips down. If two are up, at least two other people have been in the room. If only one is up, one new person has been in the room. Excluding the first visit, when the counter has flipped 40 switches down, he knows everybody has been in the room.

• September 24, 2017 at 9:40 am

shouldnt have the music playing while reading out the riddle

• September 24, 2017 at 9:40 am