Q. You are correct but here is another way of thinking of the answer that suits my logic better. Here is another explanation using doors 1 through 3:

Say you pick door 1. This means the host will show you what is behind door 2 or 3. He shows you what is behind door 3.

If the prize is behind door #1 than the host could show you either door 2 or 3. There is a 1 out of 2 chance that door 3 would be opened.

If the prize is not behind door #1 but behind door #2 than the host must show you door #3 and the odds are 1 to 1 that door three would be shown. Or twice as great as if the prize was behind door 1.

Since the odds of the prize being behind door #1 are 1/3 from the original guess and the odds of door 3 being shown if door #2 contains the prize is twice as great, than the odds that door #2 has the prize is 2/3

Hopefully I have this spoiler thing right....

If not then sorry.....

pjreid is correct. Your odds are never 33.3% or 66.6% Your odds remain constant at 50%. As pjreid pointed out the third box is a red herring. A joke box is always removed, so in reality you are only choosing between two boxes.

test of this spoiler thingy

right now i know what i am doing, i agree with PJ and the Cat Train. If you ever have a choice of 3 and only one is correct, then your odds are 1 in 3. Any idea of combinations and permuations are a waste of time 1 is right out of 3 choices - 33% probability of picked at randon the right one.

If Mr. Gameshow removes a bumsteer and gives you the opportunity to again make a choice, you could pick at random one correct box out of two therefore a 50% probability of getting it right.

The important thing in this whole riddle is that he gives you a choice again to re-pick. By doing this you reset the entire equation. Pick 1 box from these 2.... a 50% chance.

If he did not give you this choice, your chance was at the beginning of the game 33% and that's it - that was your chance at the beginning - and that was always your chance.

Maybe I am just a simple bloke and all, but i reckon them big city folk from them big city buildings, just put Member Mensa after their name and made sumpin simple all messed up.

Proteosome,

I don't quite follow but I have had several beers.

what i do know and I am 100% sure of is over a large sample you will be right 66.66666666666666666666666667% of the time by taking the door not chosen by you or monty.

Like I said take 10 doors instead of three. Do it with pieces of paper with someone else. YOu should clearly see the door you pick is of lower percentage and therefore better to go with the other option.

THE ONLY WAY IT IS 50-50 IS IF A DOOR IS ELIMINATED BEFORE YOU PICK.

Q and Proteosome are correct.

On the first pick, you are right or wrong:
- 1/3 times you are right and the only way to win is by staying.
- 2/3 times you are wrong and the only way to win is by switching.

Thus, the switch option results in 2/3 chance so I would switch every time.

I will admit that I mulled this over some as my first answer had 1/2 odds.


99gecko, I would be interested in hearing more about how or why Mensa got it wrong at first.

Happy Monday All,

JoeSoap,
I would be interested in hearing more about how or why Mensa got it wrong at first. Me too!.

Unfortunately I searched but found nothing regarding how the Mensa Journal got it wrong other than this:http://www.deanesmay.com/archives/000013.html.
Regardless, the notoriety of this one is due to an article in Parade magazine by famed Mensian Marilyn vos Savant (http://en.wikipedia.org/wiki/Marilyn_vos_Savant), highest IQ in the world (after DHC's not-so-famous Q;) )according to Guiness.

BTW, this is a very famous brain teaser. It even has it's own Wiki-hit:
The Monty Hall Problem (http://en.wikipedia.org/wiki/Monty_Hall_problem). Please read for an extended explanantion fo the solution.

cheers,
99gecko

Okay smarties, try this one:

You are detective instructed to determine the cause of death of 6 recreational swimmers, found 200 miles off-shore surrounding an unanchored pleasure craft/yacht. There is no one on board the vessel and the life jackets are all accounted for. The weather has been warm, calm, and still for at least a week in the entire ocean and they have been dead for less than a day. Other than the obvious drowning, there are no other signs of trauma, dispute/poisoning/foul play, which autopsies will confirm.

What caused their death?

I forgot to include this for the three door puzzle (http://edp.org/monty.htm). Everytime your browser loads the page it does a 3000 sample simulation, of the puzzle.

I have some more that I like.

Here is one.

You have a 3 gallon plastic jug and a 5 gallon plastic jug. Using only those 2 containers how do you get exactly 4 gallons of water in one of the containers?

the boat floated away. They swam to retrieve but got too tired and drown.

Thanks 99gecko,

After reading your links on the Monty Hall problem, I like (I may be biased though) my explaination the best for its simplicity.

I hadn't heard this one before and I'm glad you gave us the weekend otherwise I would've been in the 1/2 chance crowd and would've groaned when it got explained. There is satisfication in getting it right on your own.

I imagine there would be a few ways to do it but here is the way that comes to my mind:
fill the 5 gallon jug --> use to fill the three gallon jug (leaving 2 gallons in the 5 gallon jug) --> empty 3 gallon jug --> pour remainder (2 gallons) from 5 gallon jug into 3 gallon jug --> fill 5 gallon jug--> top-up 3 gallon jug (with 1 gallon) from 5 gallon jug --> 5 gallon jug now contains 1 gallon

Imagine you are in a room with 3 switches. In an adjacent room there are 3 bulbs (all are off at the moment), each switch belongs to one bulb. It is impossible to see from one room to another. How can you find out which switch's belong to which bulbs, if you may enter the room with the bulbs only once?

Q,
sorry but unfortunately the seas have been calm for a week.

Okay, the boat is up first and I am likely wrong but...

the boat caught fire and the people had to abandon the boat before they could get their lifejackets. Alternatively, space aliens plucked them from the boat and drowned them.

Now for an alternative way for the water problem:

fill the 3 gallon jug and transfer to the 5 gallon jug. Refill the 3 gallon jug and transfer again to the 5 gallon jug. You will now have 1 gallon of water in the 3 gallon jug. Empty the 5 gallon jug and transfer the 1 gallon into it. Fill and transfer 3 gallon and voila, you have 4 gallons.

And finally, the lightbulb:

Turn one light on for an extended period of time and mark the switch. Just before looking at the bulbs turn this light off and one of the other two on. The switch that is on corresponds to the bulb that is on. Touching the other two bulbs will determine which one is hot and will correspond to the switch that was left on for an extended period of time. The remaining bulb in the off position thus matches up with the last switch.

well done Proteosome.

After answering the above questions I decided to go back and look at the rest of this thread which I seemed to have been skipping over. I have one question on the plane puzzle.

If the magic conveyor belt increases its speed to keep the plane in place, than wouldn't it increase its own speed so that the friction generated would be enough to keep the plane in place?

I guess what I am wondering is whether the friction maxes out so that the plane can overcome it or whether the magic conveyor belt can continually increase its speed so that the wheels cannot move forward on the conveyor belt.

I understand the wheels are independent of the planes thrust I just question the hypothetical situation because friction must be overcome before the plane moves forward. And if the conveyor belt is able to continue to increase the friction than can the plane overcome it.

there will be a point where the planes friction needs to be overcome before forward movement. I think this would be almost right away though.

The wheels spinning faster and faster should not generate significantly more friction.

I felt like contributing a puzzler to this thread so I went back and dug up one I liked. Here it is:

The three wisest sages in the land were brought before the king to see which of them were worthy to become the king's advisor. After passing many tests of cunning and invention, they were pitted against each other in a final battle of the wits.

Led blind-folded into a small room, the sages were seated around a small wooden table as the king described the test for them.

"Upon each of your heads I have placed a hat. Now you are either wearing a blue hat or a white hat. All I will tell you is this- at least one of you is wearing a blue hat. There may be only one blue hat and two white hats, there may be two blue hats and one white hat, or there may be three blue hats. But you may be certain that there are not three white hats."

"I will shortly remove your blind folds, and the test will begin. The first to correctly announce the color of his hat shall be my advisor. Be warned however, he who guesses wrongly shall be beheaded. If not one of you answers within the hour, you will be sent home and I will seek elsewhere for wisdom."

With that, the king uncovered the sages' eyes and sat in the corner and waited. One sage looked around and saw that his competitors each were wearing blue hats. From the look in their eyes he could see their thoughts were the same as his, "What is the color of my hat?"

For what seemed like hours no one spoke. Finally he stood up and said, "The color of the hat I am wearing is..."

making a BIG assumption here


has to be blue hat.
1) if sage #1 hat was white and his competitors saw his hat they would have no idea what the colour of their own hat could be, and sage#1 could assume his hat was white. However, since (I assuming here) the king to be fair to all, since only one sage could wear the white hat in this scenario, the others would be at a disadvantage.
2) therefore his hat must be blue