logicBruce is an inmate at a large prison, and like most of the other prisoners, he smokes cigarettes. During his time in the prison, Bruce finds that if he has 3 cigarette butts, he can cram them together and turn them into 1 full cigarette. Whenever he smokes a cigarette, it turns into a cigarette butt.
One day, Bruce is in his cell talking to one of his cellmates, Steve.
"I really want to smoke 5 cigarettes today, but all I have are these 10 cigarette butts," Bruce tells Steve. "I'm not sure that will be enough."
"Why don't you borrow some of Tom's cigarette butts?" asks Steve, pointing over to a small pile of cigarette butts on the bed of their third cellmate, Tom, who is out for the day on a community service project.
"I can't," Bruce says. "Tom always counts exactly how many cigarette butts are in his pile, and he'd probably kill me if he noticed that I had taken any."
However, after thinking for a while, Bruce figures out a way that he can smoke 5 cigarettes without angering Tom. What is his plan?

Bruce takes 9 of his 10 cigarette butts and turns them into 3 cigarettes total (remember, 3 cigarette butts can be turned into 1 cigarette). He smokes all three of these, and now he has 4 cigarette butts.
He then turns 3 of the 4 cigarette butts into another cigarette and smokes it. He has now smoked 4 cigarettes and has 2 cigarette butts.
For the final step, he goes and borrows one of Tom's cigarette butts. With this cigarette butt plus the 2 he already has, he is able to make his 5th cigarette to smoke. After smoking it, he is left with 1 cigarette butt, which he puts back in Tom's pile so that Tom won't find anything missing.

## Similar riddles

See also best riddles or new riddles.

logicmathshortI know a number which is spelled in an alphabetical order. Do you?

Forty.

logicshortThis is a newspaper headline:
Workers Strike - Want to Make Less Money!
What is going on?

The workers work at the mint and are tired of being overworked. They want to work less, which is making less money, since money is made at the mind!

logicshortDuring the Summer Olympics, a fellow competed in the long jump and out-jumped everybody. He didn't just win the event, he actually broke the world record held for that event. Nobody broke his record for the remainder of the Olympics, and still today his name is in the record books.
However, even though he holds the world record, he never received a medal in the long jump. How did he manage to do so well, but not receive a medal?

He was competing in the decathlon. He won the long jump event, but didn't perform very well in the other events. He lost the decathlon, so he didn't receive any medals (even though he hold the world record for long jump).

cleanfunnylogicshortCan you name three consecutive days without using the words Monday, Tuesday, Wednesday, Thursday, Friday, Saturday, or Sunday

Yesterday, Today, and Tomorrow.

logicmathshort

Think of a number. Double it. Add ten. Half it. Take away the number you started with. What is your number?

Your number is 5.

cleanlogicshortMr. Smith has 4 daughters. Each of his daughters has a brother. How many children does Mr. Smith have?

He has 5 children, all of the daughters have the same 1 brother.

logicshortwhat am II am a five letter word. If you remove all letters one by one from the last, I sound the same! What am I?

QUEUE! QUEU! QUE! QU! Q!

logicYou're standing in front of a room with one lightbulb inside of it. You cannot see if it is on or off. Outside the room, there are 3 switches in the off positions. You may turn the switches any way you want to. You stop turning the switches, enter the room and know which switch controls the lightbulb. How?

You turn 2 switches "on" and leave 1 switch "off" and wait about a minute. Then enter the room, but just before you enter, turn one switch from "on" to "off". Once in the room, feel the lightbulb - if it is warm, but off, it has to be the last switch you turned off. If it is on, it has to be the switch left on. If it is cold and is off, it has to be the switch you left in the off position.

animallogicpoemsA man stands on one side of a river, his dog on the other. The man calls his dog, who immediately crosses the river without getting wet and without using a bridge or a boat. How did the dog do it?

The river was frozen.

logicmathprobabilityWhat is the least number of people that need to be in a room such that there is greater than a 50% chance that at least two of the people have the same birthday?

Only 23 people need to be in the room.
Our first observation in solving this problem is the following:
(the probability that at least 2 people have the same birthday + the probability that nobody has the same birthday) = 1.0
What this means is that there is a 100% chance that EITHER everybody in the room has a different birthday, OR at least two people in the room have the same birthday (and these probabilities don't add up to more than 1.0 because they cover mutually exclusive situations).
With some simple re-arranging of the formula, we get:
the probability that at least 2 people have the same birthday = (1.0 - the probability that nobody has the same birthday)
So now if we can find the probability that nobody in the room has the same birthday, we just subtract this value from 1.0 and we'll have our answer.
The probability that nobody in the room has the same birthday is fairly straightforward to calculate. We can think of this as a "selection without replacement" problem, where each person "selects" a birthday at random, and we then have to figure out the probability that no two people select the same birthday. The first selection has a 365/365 chance of being different than the other birthdays (since none have been selected yet). The next selection has a 364/365 chance of being different than the 1 birthday that has been selected so far. The next selection has a 363/365 chance of being different than the 2 birthdays that have been selected so far.
These probabilities are multiplied together since each is conditional on the previous. So for example, the probability that nobody in a room of 3 people have the same birthday is (365/365 * 364/365 * 363/365) =~ 0.9918
More generally, if there are n people in a room, then the probability that nobody has the same birthday is (365/365 * 364/365 * ... * (365-n+2)/365 * (365-n+1)/365)
We can plug in values for n. For n=22, we get that the probability that nobody has the same birthday is 0.524, and thus the probabilty that at least two people have the same birthday is (1.0 - 0.524) = 0.476 = 47.6%.
Then for n=23, we get that the probability that nobody has the same birthday is 0.493, and thus the probabilty that at least two people have the same birthday is 1.0 - 0.493) = 0.507 = 50.7%. Thus, once we get to 23 people we have reached the 50% threshold.