This probability puzzle has been around a long time. Generally, it is referred to as the Monty Hall Problem because it is somewhat based on a show he hosted called "Let's Make A Deal". The problem is stated as follows. You are a contestant on "Let's Make A Deal" and you have been shown three doors. You are told that behind one of the doors, is a new car, and behind each of the other two doors is a goat. Naturally, you want to win the car and you are told you can pick any one of the three doors. This would mean your chance of winning the car based on this condition is one in three or one third. However, after you make your pick, Monty Hall opens a door you haven't chosen and shows that it's a goat. He then offers to let you change your selection. Would changing your selection decrease your odds of winning the car, increase your odds or make no difference whatsoever?
It might be a little easier to use an actual example.
Conditions in effect when playing this game:
ANSWER Car is behind Door Number One: You switch and always lose. Goats are behind the two doors you haven't chosen. You will be shown a goat and yet since you are always changing your choice, you always end up picking the other goat. This happens 1/3 of the time. Car is behind Door Number Two: You switch and always win. The car is behind a door you haven't chosen and Monty has no choice but to show you a "goat door" and the car is behind the remaining door. This happens 1/3 of the time. Car is behind Door Number Three: You switch and always win. (see above) (Happens 1/3 of the time). Adding up the probabilities, we see that by always switching, you win 2/3 of the time and loose only 1/3 of the time. Granted, this was based on your choosing door number one each time but if you look at the above 3 explanations, you will see that choosing door number 2 or door number 3 will also result in your winning the car two out of three times. Another Explanation: When you make your first pick, you will be wrong 2 out of 3 times or putting this another way, the odds are always 2 out of 3 that the car is definitely behind one of the two doors you did not choose. Since Monty Hall always eliminates one of the unchosen doors, two out of three times the car will be behind one of them and Monty has no choice but to show a "goat" door. And Another Explanation: Suppose there are 1,000 doors, 999 goats and 1 car. For your first choice, you select Door Number One. Then, (just as in the previous example), Monty Hall opens all the unchosen doors except one. So, Monty Hall opens 998 doors to reveal 998 goats and only door number one and door number 557 remain unopened. Do you think it would be worth taking a risk and switching from your original choice?
The above graphic shows a chessboard that has had its upper left square and its lower right square "sawed off". Underneath that is a domino that is exactly the size of 2 of the chessboard squares. If you had 31 of these dominoes, would you be able to arrange them so that ALL 62 squares were covered ?
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Let's assume that a $20 gold piece weighs twice as much as a $10 gold piece. So, which would you rather have, a half pound of $20 gold pieces or a pound of $10 gold pieces?
An airplane pilot overheard a woman say that she had missed her flight. The pilot said "I can give you a ride and I'll charge you nothing because you won't even be taking me out of my way." The woman was very suspicious of this offer especially when she said "But you don't even know where I'm going !!" The pilot said "It doesn't matter. It will not be out of my way." The pilot was not lying so just where the heck was he going?
How about an easy puzzle this month? A class consists of 58% girls and
42% boys. There are 6 more girls than boys in the class. How many
TOTAL students are there?
ANSWER So, there are 15 boys and 21 girls for a total of 36 students. Due to rounding of the percentages there are two more answers. 16 boys + 22 girls = 38 students and 17 boys + 23 girls = 40 students So, all three answers are correct !!!
You are offered an even-money bet involving the flipping of two coins.
You reason that since 2 Heads is 1 of the 4 combinations, then flipping 2 coins twice will give you a "50/50" chance of getting 2 heads at least once.
Is your reasoning correct ?
ANSWER 1st Outcome 2nd Outcome Successful ? 1) H H H H Y 2) H H H T Y 3) H H T H Y 4) H H T T Y 5) H T H H Y 6) T H H H Y 7) T T H H Y 8) H T H T N 9) H T T H N 10) T H T H N 11) T H H T N 12) T T H T N 13) T T T H N 14) T H T T N 15) H T T T N 16) T T T T N Notice that the successful outcomes are only 7 out of 16 and not 8 out of 16. (Note: Case #1 is considered only 1 success - not two). Therefore, you will only be successful 43.75% of the time and not 50%. If you still doubt the reasoning of this, try flipping 2 coins yourself OR write a computer program to simulate this. Even the RAND() function in MS Excel™ will allow you to run such a simulation.
When rolling 2 six-sided dice, there are 36 possible combinations. Out
of these 36 combinations, there is only one way that a twelve (2 sixes) can
appear. You are offered an even-money bet (let's say for $10) that if you can throw a twelve in 21 throws, you win $10. If not, you lose $10. Also, you may continue to make bets for as long as you'd like. You reason that a twelve is 1 out of the 36 dice combinations, therefore your chances are 50-50 after 18 rolls and with 21 tries you are going to come out a winner more often than not. Is this reasoning correct?
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