FIRED FROM CANINES CANINE IM BULLET AIMED AT THW BRAIN SLOWLY AND SURELY I FIND MY TRACK COME SUNSHINE OR RAGING RAIN ISWIM AND WRIGGLE MAKING MY WAY THREE INCHES EVERY PING DAY TARGET I REACH IN FEW WEEKS SAY AND MAKE SURE OF ITS DEADLY DECAY WHO AM I RIDDLES WITH ANSWERS TO SOLVE - PUZZLES & BRAIN TEASERS

1/10

For a group of n friends, there are n! (n factorial) ways to draw the names out of the hat. Since a circle does not have a beginning and end, choose one person as the beginning and end of the circle. There are now (n-1)! ways to distribute the remaining people around the circle. Thus the probability of forming a single circle is

(n-1)! / n!

Since n! = (n-1)! * n (for n > 1), this can be rewritten as

(n-1)! / (n*(n-1)!)

Factoring out the (n-1)! from the numerator and denominator leaves

1/n

as the probability. Did you answer this riddle correctly?
YES  NO  

Russian Roulette
Probability puzzles require you to weigh all the possibilities and pick the most likely outcome.


Puzzle ID: #17681
Fun: *** (2.59)
Difficulty: ** (2.07)
Category: Probability
Submitted By: JMCLEOD****
Corrected By: cnmne








You are in a game of Russian Roulette with a revolver that has 3 bullets placed in three consecutive chambers. The cylinder of the gun will be spun once at the beginning of the game. Then, the gun will be passed between two players until it fires. Would you prefer to go first or second?


Answer
Label the chambers 1 through 6. Chambers 1 through 3 have bullets and chambers 4 through 6 are empty. After you spin the cylinder there are six possible outcomes:

1. Chamber 1 is fired first: Player 1 loses
2. Chamber 2 is fired first: Player 1 loses
3. Chamber 3 is fired first: Player 1 loses
4. Chamber 4 is fired first: Player 2 loses (First shot, player 1, chamber 4 empty. Second shot player 2, chamber 5, empty. Third shot player 1, chamber 6 empty. Fourth shot player 2, chamber 1 not empty.)
5. Chamber 5 is fired first: Player 1 loses (First shot, player 1, chamber 5 empty. Second shot player 2, chamber 6, empty. Third shot player 1, chamber 1 not empty.)
6. Chamber 6 is fired first: Player 2 loses (First shot, player 1, chamber 6 empty. Second shot, player 2, chamber 1, not empty)

Therefore player 2 has an 4/6 or 2/3 chance of winning. Did you answer this riddle correctly?
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