Three men have dinner in a restaurant. The waiter comes with the bill and it is $75. The three men decide to pay for the bill together and they each pay $25. The waiter brings the money to the manager, who notices that the calculation was wrong. The dinner did not cost $75, but $70. The manager hands the waiter five one-dollar bills and tells him to give it back to the three men. The waiter for a moment wonders how to divide five bills among three people. Finally, he decides that they had not given him a tip yet. So he gives each of them one dollar back and keeps the two remaining.

The three men have now paid $24 each. $24 x 3 = $72. Then there are the two dollars that the waiter kept, making the total $74. But where did the last dollar of the original $75 go?"

------------------------------------
------------------------------------

Answer:
Jack bought 3 but ate 8/3, so he spent 1/3 of hot dog extra. John bought for 5 but also only ate 8/3, so he spent 7/3 extra. So John should get $7 and Jack gets $1.

While I'm thinking about Nebs' riddle, I'll post another one of mine...


John, Jack and Jeremy sat down to eat. John had 5 hot-dogs with him while Jack had 3. Jeremy, however, didn't have any, so he asked for a permission to eat with them. They ate all their hot-dogs together. ach of them ate the same amount of food (=hot-dogs). In the end Jeremy gave them 8 dollars and went away. Now there was a question how to share the money. John's and Jack's friends made different suggestions. Most said that John should get $5 and Jack $3, one even said that John should get $7 and Jack $1 while the rest said that each of them should get $4.

What would be the fairest solution?

Here's something I found...not sure if it exactly fits the thread.

There is a common English word that is nine letters long. Each time you remove a letter from it, it still remains an English word - from nine letters right down to a single letter. What is the original word, and what are the words that it becomes after removing one letter at a time?

Argh, don't you get that it WAS NOT in the movie? The riddle that was in the movie is a very well-known one and a very easy one, unlike the one I have posted.

??? It is the same riddle with 5 & 3 gallon jugs. ???

In a pub, the barkeeper has two containers for liquids. One fits 5 pints and the other one 3 pints. He also has a big cask of beer. What he needs is to end up with one pint of beer in EACH of the containers. However, he is not allowed to use any measurment instruments or any other containers or glasses etc. aside from what he has.

If it is a well known one, please post it cause I think it is the same.

Haha, everyone beat me to the answer.

But i was watching World Series

Anywho, this is the problem I posted in the BOM forums, and it seems like nobody had a chance to think about it.

It's called the Monty Hall problem:

Suppose you're on a game show, and you're given the choice of three doors: Behind one door is a car; behind the others, goats. You pick a door, say No. 1, and the host, who knows what's behind the doors, opens another door, say No. 3, which has a goat. He then says to you, "Do you want to pick door No. 2?" Is it to your advantage to switch your choice?

??? It is the same riddle with 5 & 3 gallon jugs. ???

In a pub, the barkeeper has two containers for liquids. One fits 5 pints and the other one 3 pints. He also has a big cask of beer. What he needs is to end up with one pint of beer in EACH of the containers. However, he is not allowed to use any measurment instruments or any other containers or glasses etc. aside from what he has.

If it is a well known one, please post it cause I think it is the same.

Mr. Rose had a flock of geese which he was selling. At first he sold half of his flock and and a half of a goose to Mr. Tyler. Then he sold a third of what he had left and a third of a goose to Mr. Foster. After that he sold one fourth of the rest and three-fourth of a goose to Mr. Jacobe. In the end he sold one fifth of the rest of his flock and a fifth of a goose above that to Mr. Coller.

After all that he had 19 geese left....

How many geese did he have in his flock at the beginning


PS: No geese were harmed in this riddle

I don't see any way it would help you. It's 50/50. The only thing you can assume is that the host does or doesn't want you to win. In that case, you would act based on the host's motives. Otherwise, I don't see how you can know anything about the odds beyond it being a 50/50 chance.

Oh come one, now we're down to the third-grade level

This isn't a lateral-thinking puzzle, it's about multiplication

Mr. Rose had a flock of geese which he was selling. At first he sold half of his flock and and a half of a goose to Mr. Tyler. Then he sold a third of what he had left and a third of a goose to Mr. Foster. After that he sold one fourth of the rest and three-fourth of a goose to Mr. Jacobe. In the end he sold one fifth of the rest of his flock and a fifth of a goose above that to Mr. Coller.

After all that he had 19 geese left....

How many geese did he have in his flock at the beginning


PS: No geese were harmed in this riddle

This is the first answer that comes to almost anyone's mind.

There is more to this problem than it seems like at first sight.

Haha, everyone beat me to the answer.

But i was watching World Series

Anywho, this is the problem I posted in the BOM forums, and it seems like nobody had a chance to think about it.

It's called the Monty Hall problem:

Suppose you're on a game show, and you're given the choice of three doors: Behind one door is a car; behind the others, goats. You pick a door, say No. 1, and the host, who knows what's behind the doors, opens another door, say No. 3, which has a goat. He then says to you, "Do you want to pick door No. 2?" Is it to your advantage to switch your choice?

I think I got it:
101 in the beginning
51 to Mr. Tyler
17 to Mr. Foster
9 to Mr. Jacobe
5 to Mr. Coller

Let's see...

In the beginning there is a chance of 3:1 to pick the car and a chance of 2:3 to pick the goats. Let's say you picked the goats which is rather likely than likely without knowing what is behind each door. There is a 2:3 chance to get the goats. That means there are two chances in three to get the goats and only one to get the car. If you get the goat (and there is a higher chance to) and you switch, then you win the car. The chance to win a car is 2:3 if you switch.
right?

More then you said? I don't see what that can be other then something to do with the Host's motives.

Simple fact:
If the host intended to open a door with a goat, he could have done that regardless of whether you picked one with a goat or the car.

If the host only decided to do so depending on which door you picked, that goes into the host's motives.

Absolutely correct.

Highlight for more

A better way to think about this is if you had 100 doors, with 1 car and 99 goats. You pick a door, then the host opens 98 doors with goats. Now you have 1 door with a goat, and one with a car. The chance for picking a car in the beginning was 1 in 100; therefore, the chance that the car is the same chance that the car was behind one of the doors that you didn't pick, which is 99%. If you want the car, you should switch ;-)

Regardless of host's motives, there is always a goat behind one of the doors you didn't pick. The host always opens a door with a goat from the two that you didn't pick.

Ok, I get the logic behind it, although I don't really like it that much. It's all based on the assumption that the host will open a door with a goat behind it. However, from the original problem it isn't stated thathe would always do so. In fact, in a real life situation I would think that the host might only open another door if you have already picked the door with the car. In any case, the solution to the problem all depends on the host's inability to open the door with the car.