# Probability of 6 people in a lift getting of at the same floor in a building with 10 floors

My reasoning is that there are 10 different events that match our desires (because there are 10 floors), and that each person chooses one of the floors to get of on, and hence $$P=\frac{10}{6*{{10}\choose{1}}}=16.7\% .$$

But the solutions state that $$P=10^{-5}$$.

• I'd say $9^{-5}$ for it seems save to assume they will get off on another than the current floor – Hagen von Eitzen May 20 at 22:07
• What is your reasoning if there are $20$ floors and $2$ people? – Upstart May 20 at 22:20
• @Upstart I guess \frac{20}{2*(20 1)}=50%, or "they either come off at the same floor or they don't" – fazan May 20 at 22:39
• That means your probability just got increases by more than 3 times. You probability will be $0.5$ with $2$ persons and any amount of floor. Does that make sense that you and your friend get off the same floor with 50% chance in a $100$ storey building? – Upstart May 20 at 22:59

You are correct that there are $$10$$ favorable cases. However, your denominator does not make sense.

If each person has $$10$$ choices for exiting the lift, then the six of them have $$10^6$$ choices for exiting the lift by the Multiplication Principle since the choices they make are assumed to be independent.

Hence, the probability that each of the six people exit the lift on the same floor is $$\frac{10}{10^6} = \frac{1}{10^5} = 10^{-5}$$

It is $$10^{-5}$$. For this, associate a string $$(a_1,\dots,a_6)\in \{1,2,\dots,10\}^6$$, and note that, there is exactly $$10^6$$ such strings, which is the number of all different people/floor arrangements. The event that you are interested in is $$S=\{(k,k,\dots,k):1\leqslant k\leqslant 10\}$$, thus giving $$10^{-5}$$.

In your reasoning, your denominator is wrong. Since the persons are independent, you need to rely on multiplication rule, that is, $$10\times \cdots \times 10$$ (repeated $$6$$ times) is the number of arrangements.

• I'm sorry, but I don't understand your first paragraph. – fazan May 20 at 22:11
• For people-floor arrangements (of 6 people and 10 floors) consider a strings $(a_1,\dots,a_6)$ of length $6$, where each element can take one value between $\{1,2,\dots,10\}$ ($a_i=j$ means person $i$ get off at floor $j$). Number of all such strings is $10^6$, which is the total number of assignments. – TBTD May 20 at 22:18

The first person has a choice of getting off at any level.

But the $$2^{nd}$$, $$3^{rd}$$ $$4^{th}$$ $$5^{th}$$ $$6^{th}$$ have to get of at the same floor as the first one did, this leaves each of them with a choice of $$\dfrac{1}{10}$$.

Hence the final probability is $$\dfrac{1}{10^5}$$

• Your answer would be clearer if you said that each of the others has probability $1/10$ of exiting the elevator on the same floor as the first person. – N. F. Taussig May 21 at 23:34