can someone solve this math riddle for me - Page 3

On July 23 2007 09:09 Jathin wrote:


This is also known as the famous "Monty Hall" problem.

Hence the map name. There are 3 paths you can choose to expo to.

http://en.wikipedia.org/wiki/Monty_hall_problem

On July 23 2007 08:51 Cascade wrote:
You should switch.
It is 33% that you pick the right one at first, so 67% to get the prize if you switch.

You are falling in the trap of this riddle. One is tempted to say "two boxes, we do not know which one contains the prize, so 50%". It is however not two boxes of which we know nothing:
As you said, the host knows where the prize is, and as he always takes away the empty one, he is "moving" the probability to find the prize in both boxes into just one of the boxes. The 50% argument is only valid if the situation is completely symmetric, which it is not: the host has rejected to open the other box, thereby increasing the probability for it to contain the prize.

EDIT: also note that in two of the three possible situations you list, you will win if you switch.

On July 23 2007 11:43 Frits wrote:
The monty hall problem is a very simple riddle.

Choice 1 will always stay 33,3% no matter what they do, and there's sure to be a price. So the odds of the remaining doors together are 100%. So if you have 2 choices, and 1 is 33,3%, the other one has 66,6% chance of being the price. (100-33,3)

The case in which it wouldn't matter which door you took is when they take away one of the choices without knowing if that was the prize or not. (In which case it would be 33-33.)

Everyone here should be able to understand that.

On July 23 2007 10:09 evanthebouncy! wrote:


Hmmmmm after some wikipedia you are right! Amazing riddle I'm glad I heard it.

After re-reading your post just now you explained it perfectly.

So let me re-reason for abit...

[x] [1] [2]

3 boxes, when you pick one and stick to it, ignore w/e the host does, your chance is 1/3 and remains 1/3.

However, when you let the host remove 1 first then choose, your chance is 1/2

Now, you choose first, getting a 1/3 chance, the host removes an empty, leaving the chances improved.

Host cannot touch your door, and therefore the chances are only improved in the remaining 2 doors by the host identifying the false one, and therefore you should switch.

Brejkalh weird logic hahaha`

On July 23 2007 12:02 gameguard wrote:


just think of it like this.

2/3 chance of picking an empty box
1/3 chance to pick the Box containing prize
If you pick an empty box, the host will take away the other empty box. You switch now and win.
If you pick the prize box, you miss out when you switch.

The trick is that you are counting on the higher probability of picking the empty box at the begining.

On July 23 2007 12:08 Chill wrote:


This is the simplest way I've ever thought of it. Thanks.

On July 23 2007 12:16 evanthebouncy! wrote:


Cascade's idea was better for me, basically...

Pick one w/ .33 chance, the other 2 contains .66 chance
You take away empty box, thus condensing the .66 chance into 1 box, you pick that box.

On July 23 2007 12:40 geometryb wrote:
if anyone needs more convincing, you can just write a program to run this 1,000 times.

in the first group, you make the person not switch.

in the second group, you make the person switch.

and keep track of how many times each person wins and see who wins more. this avoids all the thinking that makes my head hurt.

On July 23 2007 12:45 Zelniq wrote:
oh i get it now

the only reason why you would do 9 * 3 calculation for this is to find out how much money the hotel ends up with at the end, which should equal 25.

each of the 3 tenants effectively ends up giving the hotel 9 dollars each, 9 * 3 = 27, but 2 of the dollars were also ended up with a greedy bellboy, so 27 - 2 = 25.


all the money is accounted for, lets see how much money each party ends up with at the end:

hotel : 25
bellboy : 2
tenants : 3

25 + 2 + 3 = 30