3 Ants and Triangle Puzzle

Three ants are sitting at the three corners of an equilateral triangle. Each ant starts randomly picks a direction and starts to move along the edge of the triangle. What is the probability that none of the ants collide? Now find the same for n vertex polygon with n ants.

solution :

Every Ant can move in 2 possible directions , Backward  ( B )  and Forward ( F ).
for 3 ants we have 2^(3) combination of moves,so the whole sample space consists of 8 choices.

F F B

F F F

F B B

F B F

B F B

B F F

B B B

B B F

here out 8 only 2 , namely BBB and FFF are no collision choices. 

so probability of not colliding is 2 /8 = 1/4=0.25

think in this way:

how can you avoid collision ?

The ants can only avoid a collision if they all decide to move in the same direction (either clockwise or anti-clockwise). If the ants do not pick the same direction, there will definitely be a collision.

1) all move in clockwise direction

2) move in anti-clockwise direction

Therefore  we have only 2 ways to avoid collision irrespective of shape and no of ant. 

how to find probability for n vertex polygon with n ants ?

As we know Every Ant can move in 2 possible directions , backward and forward .Therefore we have 2^(n) combination of moves.but only two will avoid collision (when all move in either clockwise or ant-clockwise direction )
hence probability of not colliding is 2/ 2^(n)
and probability of colliding is 1-2/ 2^(n)
example:pentagon 
pentagon has five vertices so n is 5
probability of not colliding = 2/ 2^(n)=2/2^(5)=2/32=1/16
probability of colliding = 1-2/ 2^(n)=1-2/2^(5)=1-2/32=1-1/16=15/16

Here another way to watch same problem...

As we know Each ant has the option to either move clockwise or anti-clockwise. There is a one in two chance that an ant decides to pick a particular direction. Using simple probability calculations, we can determine the probability of no collision.

P(No collision) = P(All ants go in a clockwise direction) + P( All ants go in an anti-clockwise direction)

P(No collision)= 0.5 * 0.5 * 0.5 + 0.5 * 0.5 * 0.5 = 0.25

let's generalize this for  n vertex polygon with n ants

P(No collision) = P(All ants go in a clockwise direction) + P( All ants go in an anti-clockwise direction)
P(No collision)=(0.5)^n + (0.5)^n
P(No collision)=2*(0.5)^n

P(No collision)=1-2*(0.5)^n

example:pentagon 
pentagon has five vertices so n is 5
probability of not colliding = 2*(0.5)^5=0.0625
probability of colliding = 1-2*(0.5)^5=0.9375

i know both are same and after simplifying them final formula is 1/2^(n-1) &1-1/2^(n-1) for No collision & collision resp.

The Fox and The Duck

A duck, pursued by a fox, escapes to the center of a perfectly circular pond. The fox cannot swim, and the duck cannot take flight from the water (it’s a deficient duck). The fox is four times faster than the duck. Assuming the fox and duck pursue optimum strategies, is it possible for the duck to reach the edge of the pond and fly away without being eaten? If so, how?

This is not a simple mathematical puzzle to solve like normal common problems. Fox in this puzzle is too fast and clever for any normal and simple strategy.
From the speed of the fox it is obvious that duck cannot simply swim to the opposite side of the fox to escape.

idea:
step 1:
As we know fox is four times faster than the duck.Let the duck rotate around the pond in a circle of radius R/4. Now fox and duck will take exact same time to make a full circle.

                             

step 2:
Now reduce the radius the duck is circling by a very small amount (Delta). Now the Fox will lag behind, he cannot stay at a position as well.

step 3:
Now in due time duck will get to a position we wanted, 3/4*R distance away from shore where fox is on the exact opposite side of the pond. From there duck can swim safely to shore and fly away.