The seven dwarves have seven beds in a dormitory room. One night, Sleepy is so tired that he falls asleep on the first bed that he sees. This has happened before, so the other dwarves follow a standard procedure as they enter the dormitory room later that night: if a dwarf can sleep in his own bed, he does so, but if not, he selects a random bed and uses that one. On the night in question, Grumpy is the fourth dwarf to enter the room to go to bed. What's the probability that he sleeps in his own bed?
[Math] The Seven Dwarves (of Snow White fame)
combinatoricsprobability
Related Solutions
These questions are conditional probabilities.
There is useful information in this regard on Wikipedia, and on Wolfram, but basically, conditional probabilities go like this:
$P(A|B)$ is the probability that event $A$ occurs given that event $B$ has already occurred. If you take the entire sample space, $S$, (the set of all possible events, or outcomes), then $A$ and $B$ represent sub-sets of that space.
Visually, you can think of these as Venn diagrams where $S$ is the entire area, and $A$ and $B$ are two smaller areas within the space. The intersecting, or shared space, between $A$ and $B$ is notated $AB$ (or $A\cap B$), while the combination of $A$ and $B$ is notated $A\cup B$.
Well, if we have that $B$ has occurred, and we want the probability that $A$ will occur, this must relate to the $AB$ sub-set, the set of events in which both $A$ and $B$ occur. We want to measure this relative to the initial probability that $B$ occurred, so:
$P(A|B) = \frac{P(AB)}{P(B)}$
Based on this, we can look at part (a) of the question.
a. What is the probability that the dwarf named Bashful gets kissed first on Monday?
Let's assign $A$ as the event that Bashful gets kissed first, and $B$ as the event that it is Monday. We are then looking for $P(A|B)$, the probability of Bashful being kissed first, given that it is Monday.
What is $P(AB)$? It is the probability that it is both Monday ($\frac{1}{5}$) and that Bashful is first to be kissed ($\frac{1}{7}$).
What is $P(B)$? It is the probability that it is Monday ($\frac{1}{5}$).
Therefore, $P(A|B)$ = $\frac{\frac{1}{5}\times \frac{1}{7}}{\frac{1}{5}}$ = $\frac{1}{7}$.
The remaining parts of the question follow the same process.
Parity: Dwarves agree that black = 1 and white = 0. First dwarf adds up all the hats he sees mod 2 and calls that out. The rest of the dwarves need to rememeber this. Each dwarf adds up the hats he sees mod 2 and if it's the same, considers his hat white, if it's not the same parity, then he considers his hat black.
After that each dwarf in turn calls out his presumption of hat color, except that the listening dwarves must change their guess each time someone before them calls "black" (after the first dwarf). That way they only have to add up the hats once, and do not need to remember each call, only to change their color when "black" is called.
All dwarves but the first are guarenteed to survive.
Best Answer
Suppose that there are $n$ dwarfs, $D_1,\dots,D_n$, and they enter in reverse order (i.e., $D_n$ first and $D_1$ last). I claim that the probability that $D_k$ gets his own bed is $\frac{k}{k+1}$ for every $k<n$; in the present problem $k=4$, so the answer is $\frac45$.
When $D_k$ enters the room, there are $k$ beds still free. Since each dwarf except $D_n$ takes his own bed if it’s available, the beds belonging to $D_{k+1},\dots,D_{n-1}$ must all be occupied, and the free beds must be $k$ of the $k+1$ beds belonging to $D_1,\dots,D_k$, and $D_n$. When a dwarf picks a bed at random, the beds are in effect indistinguishable, so each of the $k+1$ beds belonging to these $k+1$ dwarfs is equally likely to be the one that’s already occupied. In particular, the probability that $D_k$’s bed is already occupied is $\frac1{k+1}$, so the probability that he gets his own bed is $\frac{k}{k+1}$.