5. Escaped Prisoner Riddle
Take a look at the image above. Police hire a detective to find an escaped prisoner. The detective narrows his search down to the prisoner hiding in one of three different houses. After taking a closer look at an aerial photo of the three houses, the detective instantly knew which house the prisoner was hiding in. In which house is the prisoner hiding in and how did the detective know this?
Answer: The prisoner is hiding in house A as this is the only house where the car is pointing towards the street for a quick escape.
4. Manhunt Riddle
A prisoner escaped from a prison and began to make a dash on foot. He ran for about 3 miles with intermittent stops to catch his breath until he saw a police car coming towards him. Instead of turning in the opposite direction and making a run for it, the man ran towards the police car briefly before turning and running into a nearby woods to hide. Why did the man run towards the police car?
Answer: The man was more than halfway across a bridge when he spotted the police car, so the quickest way to leave the bridge was to run towards the police car and then turn and run into the woods to hide.
3. Prisoner and Marbles Riddle
A King decides to give a prisoner a chance of being set free. The King placed two marbles in a non-transparent jar that was glued to a table. One marble was supposed to be black and the other was supposed to be blue. If the prisoner picks the blue marble, he would be set free, however, if he picks the black marble he would be executed. The King was very cruel as the prisoner witnessed him placing two black marbles in the jar and no blue marbles. The prisoner was not allowed to say anything. How did he escape with his life?
Answer: The prisoner grabbed one of the marbles in the jar, concealed it in his hand and then quickly swallowed it. As he was not allowed to speak, he showed everyone that the marble remaining in the jar would be black. Since the other marble was swallowed, it was assumed to be a blue marble. Therefore, the King had no choice but to free the prisoner.
2. Prisoner Light Switch Riddle
There is a jail with 50 prisoners in it, each of them in solitary confinement. One day, the jailer decides to give the prisoners a chance to be set free providing they can solve his puzzle. If not, the prisoners will stay in the prison for life with no chance of parole whatsoever. In the jail, there is a room with nothing but a light switch, however, the light switch does not do anything if switched on or off. At regular intervals, the jailer will bring a prisoner at random to the room where the prisoner can choose to flip the light switch up or down as many times as he/she likes before being escorted back to his/her cell. This will continue until one of the prisoners tells the jailer that all 50 of them have been in the room. If the assertion is correct, all the prisoners will be released. If the assertion is incorrect, they will not be set free. The light switch is currently in the down position. The jailer gives the prisoners some time to meet and come up with a plan before escorting all the prisoners back to solitary confinement. How do the prisoners know when everyone has been in the room at least once?
Answer: During the meet-up, the prisoners should nominate one person to be the “counter”. The “counter” will always flick the light switch up, whereas the rest will flick the light switch down. That way, the prisoners who flick the light switch down will flick the switch down if it is in the up position and they will each do this once apart from the “counter”. If the same prisoner is in the room and he/she has flicked the switch already, he/she will leave the switch and won’t do anything. Therefore, once the “counter” flicks the switch up 49 times, he/she will know that everyone would have been in the room at least once and would tell the jailer this, thus freeing everyone.
1. Prisoner Hats Riddle
Four prisoners are sentenced to death for committing horrible crimes. However, the jailer decides to play a game of life and death with them in the form of a puzzle. If the prisoners answer the puzzle correctly, they can go free. But if they answer incorrectly, they will be executed. Each prisoner is given either a black hat or white hat. The prisoners are aware that there are two black hats and two white hats but each do not know what colour hat they are wearing. Three of the prisoners are lined up in a room and can only see the hats in front of them but cannot see behind them. The fourth prisoner is placed in a separate room where he cannot see nor be seen by any of the other prisoners. Communication between the prisoners, verbal or non-verbal, is strictly prohibited and if one prisoner decides to look behind, they will all die. If one of the prisoners can determine what colour hat he is wearing with 100% certainty, all the prisoners will go free but if not, all will be executed. All four prisoners are set free. How could this be?
Answer: Prisoner 4 is of no use in this puzzle as he was placed in a separate room, isolated from the rest of the prisoners meaning he is not 100% certain of what colour hat he is wearing. Prisoner 3 is also not fully certain of what colour hat he is wearing as he cannot see any of the other prisoners’ hats in front of him, just a lovely view of a wall. Prisoner 1 can see the hats of prisoners 2 and 3 and notices that they are different colours. If they were the same colour, then prisoner 1 would have been certain that he would have had a hat of the opposite colour. However, this is not the case in this scenario. As prisoner 1 has observed that the hats of the prisoners in front of him are of opposing colours, he cannot be fully certain of whether he is wearing a black or a white hat. As a result, he stays silent. As time passes by, prisoner 2 picks up on the silence as a sign that prisoner 1 does not know what colour hat he is wearing meaning the hats in front of prisoner 1 must be of opposing colours. Therefore, by looking at the colour of prisoner 3’s hat, prisoner 2 can deduce with 100% certainty that his hat must be the opposite colour to that of prisoner 3’s.