Chances of being picked last in a hat

Question

A couple of friends and I are struggling with coming up with an answer to this question. It's seemingly simple but I need a little help.

6 people put their name into a hat. One by one, names are pulled without replacement. What are the odds that your name will be picked last?

My thoughts are: in each pick you're looking at the odds that your name is not getting picked. So first it's $\frac{5}{6}$, then $\frac{4}{5}$, then $\frac{3}{4}$, $\frac{2}{3}$, $\frac{1}{2}$. And it's the chance you don't get picked first AND you don't get picked second, third ... so

$$\frac{5}{6} \cdot \frac{4}{5} \cdot \frac{3}{4} \cdot \frac{2}{3} \cdot \frac{1}{2} = \frac{1}{6}$$

Agree/Disagree?

Thank you in advance for your help.

a similar problem: Probability: 10th ball is blue – miracle173 May 17 '16 at 15:57 — miracle173, May 17 '16 at 15:57
The answer is 1/6 by symmetry. – Timothy Shields May 18 '16 at 02:57 — Timothy Shields, May 18 '16 at 02:57

score 18 · Answer 1 · answered May 17 '16 at 15:55

18

Correct. The odds that your name is picked last is the same as the odds for any other person's name being picked last. So for each person, the probability is $\dfrac{1}{6}$.

answered May 17 '16 at 15:55

Aravind

6,150

That argument extends to sampling with replacement as well (what are the chances I'm the last one to have my name appear at least once). – Clement C. May 18 '16 at 01:00

score 8 · Answer 2 · answered May 17 '16 at 15:55

8

It is true, and your calculation is fine. An easier way is to imagine pulling all the names and moving the first to the end. Clearly the first is yours with probability $1/6$

answered May 17 '16 at 15:55

Ross Millikan

374,822

score 4 · Answer 3 · answered May 17 '16 at 17:38

The solution can also be expressed using factorials. $6!=720$ is the total number of combinations in which the names can be drawn out of the hat. Favourable outcomes are $5!=120$, as the first five names may be drawn in any sequence. The probability that your name is picked last is $$\frac {5!}{6!} = \frac 16$$

score 4 · Answer 4 · answered May 17 '16 at 22:06

Here's another intuitive way to think of it: The order the names are picked is a sequence. Your name is just as likely to be at the end of that sequence as at the beginning of that sequence (or anywhere else in that sequence), and the chance of yours being picked first is $\dfrac{1}{6}$, so the chance of yours being picked last is $\dfrac{1}{6}$.

score 2 · Answer 5 · answered May 17 '16 at 22:16

Since all you care about in this question is the last name picked, this whole scheme is just a fancy mechanism to pick that one name out of the hat. Because that mechanism does not at any time discriminate between the names, your probability of being picked is exactly the same as anyone else's. So the probability is $1/6$.

score 2 · Answer 6 · answered May 17 '16 at 22:20

Here might be simplest way to understand the problem:

The list of people is shuffled randomly. What is the probability of your assignment to a given position? There are six positions, so 1 in 6.

The fact that the position in question just happens to be last fades from view.

Chances of being picked last in a hat

6 Answers6