On the game show et´s Make a Deal, Monty Hall shows you three doors. Behind one of the doors is a new car, the other two hide goats. You choose one door, perhaps #1. Now Monty shows you what´s behind door #2 and it´s a goat.He gives you the chance to stay with original pick or select door #3. What do you do?
You should always abandon your original choice in favor of the remaining door (#3). When you make your first choice the chance of winning is 1 in 3 or 33%. When you switch doors, you turn a 2 in 3 chance of losing in the first round into a 2 in 3 chance of winning in the second round.
Search: Monty Hall problem
What 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.
You are killed in a plane crash and find yourself in front of 2 doors: one leads to heaven and one will lead you to hell for eternity. There is an identical troll at each door. You find instructions posted on the wall behind you. You can ask only one question and you can only direct it to only one of the trolls. One troll will always lie to you - regardless of your question - and the other will always tell you the truth. And only the trolls themselves know which one will lie and which one will be truthful. That is all that you are told.... What is the one and only question that will ensure you passage to heaven, and why?
Ask any of the tolls this question. "If I were to ask the other troll which is the door to Heaven, which door would he point to?" Now when the troll answers by pointing to one of the doors you simply take the other door.
You are standing next to three switches. You know these switches belong to three bulbs in a room behind a closed door – the door is tight closed, and heavy which means that it's absolutely impossible to see if any bulb is on or not. All three switches are now in position off.
You can do whatever you want with the switches and when you are finished you open the door and go into the room. While in there you have to tell which switch belongs to which bulb.
How will you do that?
Turn on the first switch and wait for a while.
Turn off the first one and turn on the second.
Go into the room.
One bulb is shining, the second bulb is hot and the third one nothing.
There are 4 big houses in my home town. They are made from these materials: red marbles, green marbles, white marbles and blue marbles.
Mrs Jennifer's house is somewhere to the left of the green marbles one and the third one along is white marbles.
Mrs Sharon owns a red marbles house and Mr Cruz does not live at either end, but lives somewhere to the right of the blue marbles house.
Mr Danny lives in the fourth house, while the first house is not made from red marbles.
Who lives where, and what is their house made from ?
From, left to right:
#1 Mrs Jennifer - blue marbles
#2 Mrs Sharon - red marbles
#3 Mr Cruz - white marbles
#4 Mr Danny - green marbles
If we separate and label the clues, and label the houses #1, #2, #3, #4 from left to right we can see that:
a. Mrs Jennifer's house is somewhere to the left of the green marbles one.
b. The third one along is white marbles.
c. Mrs Sharon owns a red marbles house
d. Mr Cruz does not live at either end.
e. Mr Cruz lives somewhere to the right of the blue marbles house.
f. Mr Danny lives in the fourth house
g. The first house is not made from red marbles.
By (g) #1 isn't made from red marbles, and by (b) nor is #3. By (f) Mr Danny lives in #4 therefore by (c) #2 must be red marbles, and Mrs Sharon lives there.
Therefore by (d) Mr Cruz must live in #3, which, by (b) is the white marbles house. By (a) #4 must be green marbles (otherwise Mrs Jennifer couldn't be to its left) and by (f) Mr Danny lives there.
Which leaves Mrs Jennifer, living in #1, the blue marbles house.