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If 3 ants are on three different vertices of a triangle. what is the probability they will collide and they can move along the sides of the triangle - so in this case each ant can move along 2 sides of the triangle. So why is the probability that they will move in a clockwise direction (1/2)^3 ?? and probability in the counter clockwise direction is (1/2)^3 again. I understand that probability of moving itself is (1/2) but why is it cubed ?

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Because each ant moves independently with a probability $\frac{1}{2}$ to move anticlockwise or clockwise: $$\frac{1}{2}\times\frac{1}{2}\times\frac{1}{2}=\left(\frac{1}{2}\right)^{3}$$ – Shaktal Oct 8 '12 at 19:07
so what formula is this using to get this cubed result and why is it not addition of the probabilities ? – Phoenix Oct 8 '12 at 19:09
@Shaktal Sorry but, for independent events, P(A or B or C)=P(A)+P(B)+P(C) is as wrong as can be. – Did Oct 8 '12 at 19:15
just imagine it as follows: there are 8 different ways the ants can move. 000,001,010,011,100,101,110,111 where 0 is anticlockwise and 1 clockwise. $2^3=8$ – Alex Oct 8 '12 at 19:21

Nobody knows what probability really is; but in the last 250 years we have learnt to talk about it coherently. When there are $3$ ants choosing one of two "equally likely" directions "independently" then there are $2^3=8$ possible cases, which for "reasons of symmetry" are equiprobable. Among these $8$ cases there are exactly $2$ where all three ants start to move in the same direction.

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For independent events, $$ \Pr\left( \bigcap_{i=1}^n A_i \right) = \prod_{i = 1}^n \Pr\left( A_i \right). $$ In words, "the probability of the product is equal to the product of the probabilities".

In your example, each ant has a $1/2$ chance to move either left or right. Presumably, these events are independent, as we are given no information about the ants communicating with each other. If we let $A_i$ be the event "ant $i$ goes right", then we get $$ \Pr(A_1 \cap A_2 \cap A_3) = \Pr(A_1) \cdot \Pr(A_2) \cdot \Pr(A_3) = (1/2)^3, $$ which is the probability that all ants travel clockwise. A similar calculation gives $(1/2)^3$ as the probability that all ants travel counterclockwise.

To finish the problem, the probability that the ants do collide is equal to the probability that the do not all travel clockwise or all counterclockwise. This is given by $$ 1 - (1/2)^3 - (1/2)^3 = 3/4. $$

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