[SOLVED] Question for Daniel0

[seventheyejosh]

I have this math related question I want to ask, and figured you'd be the one who'd know stuff

 

So I'm in an intermediate stats class, and there is a 'case study' or whatever you want to call it on the side of the page. I'll write what it says, but basically it states an interesting situation, but instead of elaborating or telling you the answer, it merely says it is way beyond the scope of this textbook. Well, that is fine and all, but I'd like to know about it. Mind you, this isn't homework, and I won't die if i don't figure it out, I'm just really curious.

 

Thanks for any thought you give it.

 

Blurb:

 

(entitled) The Random Secretary:

 

One classic problem of probability goes like this: A secretary addresses 50 different letters and envelopes to 50 different people, but the letters are randomly mixed before being put into the envelopes. What is the probability that at least one letter gets into the correct envelope? Although the probability might seem like it should be small, its actually 0.632. Even with a million letters and a million envelopes, the probability is 0.632. The solution is beyond the scope of this text-way beyond.

 

That is it word for word (including the hypen!). I might not have noticed it, but it is interesting the way the probability is over half!

 

Thanks again to Daniel0, and anyone who has info about this!

 

Josh.

[corbin]

What an awesome text book lol.

[Daniel0]

Uh... I never liked statistics and probability much. The calculations are often annoying to make, and it's easy making mistakes.

 

Anyways, if we let [imath]E_n[/imath] be the event that letter [imath]n[/imath] is in the correct envelope and we just consider having 3 envelopes (because I don't want to write this for 50 or one million), then [imath]P(E_1)=\frac{1}{3}[/imath] and [imath]P(E_1 \cap E_2) = \frac{1}{3}\cdot\frac{1}{2}[/imath], right?

 

It is common to imagine the probability space as a set [imath]U[/imath] having cardinality 1, so [imath]\forall E_i \in U:0\leq|E_i|\leq1[/imath]. We say that the probability of an event occurring is the cardinality of the corresponding set. Note that you normally do not talk about sets that have a cardinality that's not a natural number because it normally not makes sense to say that a set has e.g. 0.42 elements.

 

Iff two sets [imath]A[/imath] and [imath]B[/imath] are disjoint, the cardinality of their union will be the sum of each set's cardinality, i.e. [imath]A\cap B = \emptyset \Leftrightarrow |A\cup B|=|A|+|B|[/imath]. In probability terms, two disjoint sets correspond to two events that cannot possibly occur at the same time, and the union of two sets corresponds to the probability that both events will occur.

 

In our case, it is indeed possible that two or more of our events can occur at the same time, so they're not disjoint. To find the cardinality of two non-disjoint sets [imath]A[/imath] and [imath]B[/imath] you do like before, but you have to remove the common elements (the intersection) to prevent double counting (you can draw a Venn diagram to see that this is true). So [imath]|A\cup B| = |A|+|B| - |A \cap B|[/imath]. This holds generally because if two sets are disjoint, their intersection will be empty (thus having cardinality 0).

 

So this means, [imath]P(A \cup B) = P(A) + P(B) - P(A\cap B)[/imath] for some events [imath]A,B\in U[/imath]. Note that the union corresponds to "or" and the intersection corresponds to "and".

 

For [imath]n=3[/imath] envelopes, we can then calculate [imath]P(E_1 \cup E_2 \cup E_3)[/imath], the probability that at least one of the letters are in the right envelope.

 

[math]\begin{split} P(E_1 \cup E_2 \cup E_3) &= P(E_1) + P(E_2) + P(E_3) - P(E_1 \cap E_2) - P(E_1 \cap E_3) - P(E_2 \cap E_3) + P(E_1 \cap E_2 \cap E_3) \\ &= 1 - 3\left(\frac{1}{3}\cdot\frac{1}{2}\right) + \frac{1}{3}\cdot\frac{1}{2}\cdot\frac{1}{1} \\ &= \frac{2}{3} \\ &\approx 0.667 \end{split}[/math]

 

This is pretty close to your 0.632 for n=50. My guess is that the probability converges towards that number as n goes to infinity (or I made an error during my calculations).

 

 

Note to self: Need to install jsMath for better LaTeX support on the forums.

Edit: Installed jsMath and updated post to use it.

[Zane]

note to self... don't try to understand anything marked as...question for daniel0

[seventheyejosh]

Interesting. I'll have to dissect this once I get off work

 

Thank you VERY much, Daniel.

[corbin]

>.<