# Different ways of computing probability

There are twelve unique watches and five men. Each of the five men are asked to choose a watch for themselves. What is the probability that at-least two of them choose the same watch.

I could think of two different approaches of solving above problem.

1st approach: $$\text{answer} = 1 - \frac{12 \cdot 11 \cdot 10 \cdot 9 \cdot 8 }{ 12^5} \approx 0.618055$$

2nd approach: Consider $$x$$ such that \begin{aligned} x = &\hphantom{{}+{}} \text{no of ways exactly}~\textbf{two}~\text{men choose same watch} \\ &{}+ \text{no of ways exactly}~\textbf{three}~\text{men choose same watch} \\ &{}+\text{no of ways exactly}~\textbf{four}~\text{men choose same watch} \\ &{}+ \text{no of ways exactly}~\textbf{five}~\text{men choose same watch} \\ = &\hphantom{{}+{}} {5\choose 2} \cdot 12 \cdot {11\choose 3} \cdot 3! \\ &+ {5\choose 3} \cdot 12 \cdot {11\choose 2} \cdot 2! \\ &+ {5\choose 4} \cdot 12 \cdot {11\choose 1} \cdot 1! \\ &+ {5\choose 5} \cdot 12 \cdot {11\choose 0} \cdot 0! \\ ={} &118800 + 13200 + 660 + 12 = 132672 \end{aligned} Then $$\text{answer} = x\, / \,12^5 = 132672\, / \, 12^5 \approx 0.53317$$

The two approaches have different answers.
I believe the first approach has no issues, which means i am counting short in the second approach. But not able to figure out how?

• What about 2 choose watch A and 2 others choose Watch B, etc... – DJohnM Apr 20 at 10:25

Three men choose one model and two other men choose a second model: There are $$\binom{5}{3}$$ ways that three of the men can select the same model and twelve models for them to choose. There are eleven ways for the other two men to pick the same model. Hence, there are $$\binom{5}{3}\binom{12}{1}\binom{2}{2}\binom{11}{1} = 1320$$ such distributions.
Two men choose one model, two other men choose a second model, and the fifth man picks a third model: There are $$\binom{12}{2}$$ ways to select which two models are each picked by two men, $$\binom{5}{2}$$ ways for two of the men to pick the more expensive of those two models (or some other feature that distinguishes them if they are the same price), $$\binom{3}{2}$$ ways for two of the other three men to pick the other of those models, and ten ways for the remaining man to pick one of the remaining models. Hence, there are $$\binom{12}{2}\binom{5}{2}\binom{3}{2}\binom{1}{1}\binom{10}{1} = 19800$$ such distributions.
Adding these to the total gives $$118800 + 13200 + 660 + 12 + 1320 + 19800 = 153792$$ Dividing this amount by $$12^5$$ yields $$\approx 0.6180555556$$, which agrees with the answer obtained from your simpler first approach.