Math Challenge - November 2019

[Not anonymous]

Close, but no. You used Bayes' theorem but didn't formalize it, so I have difficulties to pin down the error. My guess is, that e.g. if Mr. Smith boards first, then there aren't 50 choices left, only 49. It can be solved without Bayes, or better with far fewer cases.

Thanks fresh_42, for reviewing my answer! I am not as strong on intuitive thinking as other people on this forum and will need to think a lot more to find a simpler approach that doesn't use Bayes' theorem. Since you mentioned that my previous answer was close, I make one more attempt at solving the problem and along the same lines as before. Smith will definitely have the choice of 50 seats if he is the 1st passenger to board, so I guess it should be assumed that Smith is NOT the last passenger to board, either because it is a trivial case and there is no choice to be made, or because the wording of the question says "printed on his boarding pass", so the last passenger has a pass and therefore is not Mr. Smith. If that assumption is made, only the final step in my earlier solution, and the probability of Smith being the $i^{th}$ passenger change. The probability of Smith being the $i^{th}$ passenger would be $1/49$ where $i \in {1, 2, .., 49}$. The final step therefore becomes:

Spoiler: Q14

Final unconditional probability of last passenger getting his/her originally allotted seat is $\sum_{i=1}^{49} 1/49 \times P_{i} = 49 \times 1/49 \times 1/2 = 1/2 = 0.5$

Note that $P_{50} = 1$ is still used in deriving $P_{49}, .., P_{1}$ and is a valid value because it represents the situation where the NONE of the passengers before the 50th passenger have occupied a seat that is different from what was originally allotted to them.

When I get some free time, I will continue pondering over this problem in the hope of finding a simpler, more elegant way to solve the problem. It requires more creative thinking and that doesn't come naturally to all

 

[fresh_42]

Thanks fresh_42, for reviewing my answer! I am not as strong on intuitive thinking as other people on this forum and will need to think a lot more to find a simpler approach that doesn't use Bayes' theorem.

Well, probabilities isn't my favorite subject either. I tend to count wrongly. The solution I have also uses more words than formulas, so yours is not bad, and 0.5 is the correct answer.

 

[member 587159]

Well, probabilities isn't my favorite subject either. I tend to count wrongly. The solution I have also uses more words than formulas, so yours is not bad, and 0.5 is the correct answer.

Yes, this is also what I hate about probability. The analytical side of the subject is nice though.

 

[Antarres]

Spoiler: Q7

Based on the equivalence relation given in the exercise, we find that this $n$-dimensional projective space is quotient space of $\mathbb{R}^{n+1}$ where points on every ray that is given by a $(n+1)$-dimensional vector from the origin are equivalent. Therefore, every equivalence class is one ray in $\mathbb{R}^{n+1}$, and quotient map with respect to which we define the topology, maps every point in $\mathbb{R}^{n+1}$ onto it's equivalence class. It is trivial to see that this map is surjective. We will use this map to induce quotient topology on the projective space. 'Geometrically', this space will then be given by a certain set of points on $S^n$, the $n$-dimensional hypersphere of unit radius with the center at the origin. Each point on the hypersphere will be mapped with inverse quotient map into a ray that is defined by the vector in $\mathbb{R}^{n+1}$. We notice that the points which are connected by the diameter of the hypersphere are also identified, that is, rays of opposite directions are also identified. So, bearing this in mind, we will define geometrical representation of $\mathbb{R}P^n$ in $\mathbb{R}^{n+1}$ with the following convention(it will make the treatment more palpable, at least to me):
1. We take $(0, 0, \dots , 0, 1)$ to be the north pole of $S^n$(All points/vectors we are mentioning are $\mathbb{R}^{n+1}$ coordinates notation).
2. Then $\mathbb{R}P^n$ is represented by a set of points on $S^n$ with positive coordinate $0 < x_{n+1} \leq 1$(northern hemisphere), and points $x_{n+1} = 0$ on $S^n$(equator), for which $x_1>0$, and points $x_{n+1}=0 , x_1 = 0$ for which $x_2>0$, etc. inductively, up until the point $(0, 0, \dots , 0, 1, 0)$.
Three-dimensional analog of this would be north hemisphere of a 3-dimensional sphere with equator circle given by points defined by $\phi \in [0,\pi)$, where $\phi$ is the azimuthal angle on the equator with $\phi=0$ defined to be measured counterclockwise from the point $(x =1 , y = 0, z=0)$.
We will call this geometrical representation of $\mathbb{R}P^n$, $S$, and it is homeomorphic to $\mathbb{R}P^n$.

The quotient topology is then induced in a standard way. Open sets in $R\mathbb{P}^n$ are sets who's inverse images under quotient map are open in $\mathbb{R}^{n+1}$. Let's take $P$ to be the set of all $n$-dimensional hyperplanes in $\mathbb{R}^{n+1}$ that intersect $S^n$. Their intersections will be $(n-1)$-dimensional spheres on the surface of this hypersphere. These $(n-1)$-dimensional spheres are contained in $S$, or they're intersecting $S$ or they're disjunct from $S$. We will consider the first two cases, since the third one isn't interesting to us.

In the first case, if we name interiors of those $(n-1)$-dimensional spheres, $U_i$, we find that inverse image of $U_i$ is the space bounded by a circular conical surface in $\mathbb{R}^{n+1}$, whose axis direction is given by the direction of the vector perpendicular to the hyperplane that generated the boundary of $U_i$ by intersecting with $S$. This set is obviously open in $\mathbb{R}^{n+1}$, so it should be open in the quotient topology.

In the second case, the $(n-1)$-dimensional hypersphere is intersecting $S$ and the intersection is an $(n-1)$ dimensional 'arc', with ends at the equator(If it's end is on the open-ended part of the equator, the arc is open-ended on that side, while if it's on the part of the equator that is included in $S$, then that end of the arc is closed(not in the sense of topology but in the sense that it includes the point from the equator)). The equator of $S^n$(or of $S$) is mapped by inverse quotient map into a hyperplane passing through the origin, which is open in $\mathbb{R}^{n+1}$, hence the equator of $S$ is open in quotient topology. However, the interior bounded by the arc and the equator, which we may call $W_i$ is mapped with inverse quotient map into space that is bounded on one side by part of a conical surface that is the image of the arc the same way as in case 1, and the hyperplane that is the image of the equator.
Also, it can be that the arc is actually half of the 'great circle' of $S$, in which case it is mapped into a hyperplane in $\mathbb{R}^{n+1}$ so in that case the inverse image of $W_i$ is the space between two intersecting hyperplanes passing through the origin(the direction of 'between' is given by any vector whose end is mapped into $W_i$). So this set in the second case is also open in $\mathbb{R}^{n+1}$, and therefore should be open in the quotient topology. So the union of $W_i$ and the part of the equator that is the boundary of $W_i$, which we can call $V_i$ is open in $\mathbb{R}^{n+1}$. We assert that the sets $U_i$ and $V_i$ consist the basis of quotient topology.
We need to check the two conditions for the basis:
1. For every $x \in S$ we have a basis element that contains it.
If we take an arbitrary $x \in S$, then the vector given by coordinates of $x$ in $\mathbb{R}^{n+1}$ defines the hyperplane that is perpendicular to it. We choose this hyperplane to intersect $S^n$ at the half the length of this vector, and obtain one of the two intersections mentioned above. Then such an intersection with $S$ defines a basis element $U_i$ or $V_i$ that contains $x$. The notion of length of this vector is not needed for this claim, since the point of intersection of the chosen hyperplane with the ray is defined with all the coordinates of $x$ halved.
2. If $x$ belongs to the intersection of two basis elements $B_1$ and $B_2$, then there exists a basis element $B_3 \subset B_1 \cap B_2$ containing $x$.
Take two basis elements containing a chosen point $x$ in $S$, such that they are not subsets of one another(in which case the condition follows trivially). Every basis element can be defined by the hyperplane that generated it, to which there corresponds a unique point in $S$ whose inverse image is a ray perpendicular to this hyperplane. Choose a hyperplane corresponding to $x$, such that it is tangent to the boundary of $B_1$. Name the basis element generated by it $B'_1$. Choose another hyperplane corresponding to $x$ such that it is tangent to the boundary of $B_2$. Name the basis element generated by it $B'_2$. Then both $B'_1$ and $B'_2$ contain $x$, and either $B'_1 \subseteq B'2$ or $B'_2 \subseteq B'_1$, since they are generated by two parallel hyperplanes. Without loss of generality, choose the first option(proving the condition in the second option is the same). Since the boundary of $B'_2$ is tangent to the boundary of $B_2$, it may intersect the boundary $B_1$. But then since $B'_1 \subseteq B'_2$, it cannot have a non-empty intersection with $B_2$. Since it's boundary tangents the boundary of $B_1$, it's interior is then contained in $B_1 \cap B_2$. Therefore $B'_1$ is the sought for basis element, by construction. In the second case it would be $B'_2$.

This has proven that our collection of sets $U_i$ and $V_i$ is a basis, and therefore, the quotient topology is generated by it.

To prove that $\mathbb{R}P^n$ is Hausdorff, we need to show that two points $x_1$ and $x_2$ of $S$ may have disjunct neighbourhoods.
Choose two points $x_1$ and $x_2$ and hyperplanes corresponding to them in such a way, that those hyperplanes intersect outside $S$, and generate basis elements belonging to the collection $U_i$. Then those two basis elements $U_1$ and $U_2$ are disjunct trivially.
To prove that $\mathbb{R}P^n$ is second-countable, we may parametrize every basis element with the radius of the $(n-1)$-dimensional sphere that is it's boundary. Then we can choose for the basis of the quotient topology those basis elements that have rational radii. Such a basis is countable, so $\mathbb{R}P^n$ is second-countable.

I might have overcomplicated the solution, but I hope it is correct. I've only recently started improving in topology, so this was a nice exercise.

 

[member 587159]

Spoiler: Q7

Based on the equivalence relation given in the exercise, we find that this $n$-dimensional projective space is quotient space of $\mathbb{R}^{n+1}$ where points on every ray that is given by a $(n+1)$-dimensional vector from the origin are equivalent. Therefore, every equivalence class is one ray in $\mathbb{R}^{n+1}$, and quotient map with respect to which we define the topology, maps every point in $\mathbb{R}^{n+1}$ onto it's equivalence class. It is trivial to see that this map is surjective. We will use this map to induce quotient topology on the projective space. 'Geometrically', this space will then be given by a certain set of points on $S^n$, the $n$-dimensional hypersphere of unit radius with the center at the origin. Each point on the hypersphere will be mapped with inverse quotient map into a ray that is defined by the vector in $\mathbb{R}^{n+1}$. We notice that the points which are connected by the diameter of the hypersphere are also identified, that is, rays of opposite directions are also identified. So, bearing this in mind, we will define geometrical representation of $\mathbb{R}P^n$ in $\mathbb{R}^{n+1}$ with the following convention(it will make the treatment more palpable, at least to me):
1. We take $(0, 0, \dots , 0, 1)$ to be the north pole of $S^n$(All points/vectors we are mentioning are $\mathbb{R}^{n+1}$ coordinates notation).
2. Then $\mathbb{R}P^n$ is represented by a set of points on $S^n$ with positive coordinate $0 < x_{n+1} \leq 1$(northern hemisphere), and points $x_{n+1} = 0$ on $S^n$(equator), for which $x_1>0$, and points $x_{n+1}=0 , x_1 = 0$ for which $x_2>0$, etc. inductively, up until the point $(0, 0, \dots , 0, 1, 0)$.
Three-dimensional analog of this would be north hemisphere of a 3-dimensional sphere with equator circle given by points defined by $\phi \in [0,\pi)$, where $\phi$ is the azimuthal angle on the equator with $\phi=0$ defined to be measured counterclockwise from the point $(x =1 , y = 0, z=0)$.
We will call this geometrical representation of $\mathbb{R}P^n$, $S$, and it is homeomorphic to $\mathbb{R}P^n$.

The quotient topology is then induced in a standard way. Open sets in $R\mathbb{P}^n$ are sets who's inverse images under quotient map are open in $\mathbb{R}^{n+1}$. Let's take $P$ to be the set of all $n$-dimensional hyperplanes in $\mathbb{R}^{n+1}$ that intersect $S^n$. Their intersections will be $(n-1)$-dimensional spheres on the surface of this hypersphere. These $(n-1)$-dimensional spheres are contained in $S$, or they're intersecting $S$ or they're disjunct from $S$. We will consider the first two cases, since the third one isn't interesting to us.

In the first case, if we name interiors of those $(n-1)$-dimensional spheres, $U_i$, we find that inverse image of $U_i$ is the space bounded by a circular conical surface in $\mathbb{R}^{n+1}$, whose axis direction is given by the direction of the vector perpendicular to the hyperplane that generated the boundary of $U_i$ by intersecting with $S$. This set is obviously open in $\mathbb{R}^{n+1}$, so it should be open in the quotient topology.

In the second case, the $(n-1)$-dimensional hypersphere is intersecting $S$ and the intersection is an $(n-1)$ dimensional 'arc', with ends at the equator(If it's end is on the open-ended part of the equator, the arc is open-ended on that side, while if it's on the part of the equator that is included in $S$, then that end of the arc is closed(not in the sense of topology but in the sense that it includes the point from the equator)). The equator of $S^n$(or of $S$) is mapped by inverse quotient map into a hyperplane passing through the origin, which is open in $\mathbb{R}^{n+1}$, hence the equator of $S$ is open in quotient topology. However, the interior bounded by the arc and the equator, which we may call $W_i$ is mapped with inverse quotient map into space that is bounded on one side by part of a conical surface that is the image of the arc the same way as in case 1, and the hyperplane that is the image of the equator.
Also, it can be that the arc is actually half of the 'great circle' of $S$, in which case it is mapped into a hyperplane in $\mathbb{R}^{n+1}$ so in that case the inverse image of $W_i$ is the space between two intersecting hyperplanes passing through the origin(the direction of 'between' is given by any vector whose end is mapped into $W_i$). So this set in the second case is also open in $\mathbb{R}^{n+1}$, and therefore should be open in the quotient topology. So the union of $W_i$ and the part of the equator that is the boundary of $W_i$, which we can call $V_i$ is open in $\mathbb{R}^{n+1}$. We assert that the sets $U_i$ and $V_i$ consist the basis of quotient topology.
We need to check the two conditions for the basis:
1. For every $x \in S$ we have a basis element that contains it.
If we take an arbitrary $x \in S$, then the vector given by coordinates of $x$ in $\mathbb{R}^{n+1}$ defines the hyperplane that is perpendicular to it. We choose this hyperplane to intersect $S^n$ at the half the length of this vector, and obtain one of the two intersections mentioned above. Then such an intersection with $S$ defines a basis element $U_i$ or $V_i$ that contains $x$. The notion of length of this vector is not needed for this claim, since the point of intersection of the chosen hyperplane with the ray is defined with all the coordinates of $x$ halved.
2. If $x$ belongs to the intersection of two basis elements $B_1$ and $B_2$, then there exists a basis element $B_3 \subset B_1 \cap B_2$ containing $x$.
Take two basis elements containing a chosen point $x$ in $S$, such that they are not subsets of one another(in which case the condition follows trivially). Every basis element can be defined by the hyperplane that generated it, to which there corresponds a unique point in $S$ whose inverse image is a ray perpendicular to this hyperplane. Choose a hyperplane corresponding to $x$, such that it is tangent to the boundary of $B_1$. Name the basis element generated by it $B'_1$. Choose another hyperplane corresponding to $x$ such that it is tangent to the boundary of $B_2$. Name the basis element generated by it $B'_2$. Then both $B'_1$ and $B'_2$ contain $x$, and either $B'_1 \subseteq B'2$ or $B'_2 \subseteq B'_1$, since they are generated by two parallel hyperplanes. Without loss of generality, choose the first option(proving the condition in the second option is the same). Since the boundary of $B'_2$ is tangent to the boundary of $B_2$, it may intersect the boundary $B_1$. But then since $B'_1 \subseteq B'_2$, it cannot have a non-empty intersection with $B_2$. Since it's boundary tangents the boundary of $B_1$, it's interior is then contained in $B_1 \cap B_2$. Therefore $B'_1$ is the sought for basis element, by construction. In the second case it would be $B'_2$.

This has proven that our collection of sets $U_i$ and $V_i$ is a basis, and therefore, the quotient topology is generated by it.

To prove that $\mathbb{R}P^n$ is Hausdorff, we need to show that two points $x_1$ and $x_2$ of $S$ may have disjunct neighbourhoods.
Choose two points $x_1$ and $x_2$ and hyperplanes corresponding to them in such a way, that those hyperplanes intersect outside $S$, and generate basis elements belonging to the collection $U_i$. Then those two basis elements $U_1$ and $U_2$ are disjunct trivially.
To prove that $\mathbb{R}P^n$ is second-countable, we may parametrize every basis element with the radius of the $(n-1)$-dimensional sphere that is it's boundary. Then we can choose for the basis of the quotient topology those basis elements that have rational radii. Such a basis is countable, so $\mathbb{R}P^n$ is second-countable.

I might have overcomplicated the solution, but I hope it is correct. I've only recently started improving in topology, so this was a nice exercise.

Thanks for the answer! Looking at the projective space as a sphere with antipodal points identified is good intuition. My answer does not use this though but I looked up other answers and this is definitely a way to approach the problem.

For starters, it needs a proof that this construct $S$ is homeomorphic to my definition of the projective space.

To be quite explicit, describe the sphere with antipodal points identified somewhat more carefully (I guess you will need an equivalence relation for this), the topology you place on it and exhibit an explicit homeomorphism between your construct $S$ and the projective space.

 

[PeroK]

Thanks fresh_42, for reviewing my answer! I am not as strong on intuitive thinking as other people on this forum and will need to think a lot more to find a simpler approach that doesn't use Bayes' theorem. Since you mentioned that my previous answer was close, I make one more attempt at solving the problem and along the same lines as before. Smith will definitely have the choice of 50 seats if he is the 1st passenger to board, so I guess it should be assumed that Smith is NOT the last passenger to board, either because it is a trivial case and there is no choice to be made, or because the wording of the question says "printed on his boarding pass", so the last passenger has a pass and therefore is not Mr. Smith. If that assumption is made, only the final step in my earlier solution, and the probability of Smith being the $i^{th}$ passenger change. The probability of Smith being the $i^{th}$ passenger would be $1/49$ where $i \in {1, 2, .., 49}$. The final step therefore becomes:

Spoiler: Q14

Final unconditional probability of last passenger getting his/her originally allotted seat is $\sum_{i=1}^{49} 1/49 \times P_{i} = 49 \times 1/49 \times 1/2 = 1/2 = 0.5$

Note that $P_{50} = 1$ is still used in deriving $P_{49}, .., P_{1}$ and is a valid value because it represents the situation where the NONE of the passengers before the 50th passenger have occupied a seat that is different from what was originally allotted to them.

When I get some free time, I will continue pondering over this problem in the hope of finding a simpler, more elegant way to solve the problem. It requires more creative thinking and that doesn't come naturally to all

If it's not too late:

Spoiler

For a plane with any number of seats. Let's call Mr Smith's seat $S$ and the last to board's seat $X$.

When Mr Smith enters the plane, some seats are taken. But not $S$ or $X$. If Mr Smith sits elsewhere, then one and only one of seats $S$ or $X$ is left for the last passenger. To see this, imagine Mr Smith sits in seat $Y$. There are no problems until passenger $Y$ boards. If he/she sits in another seat $Z$, then there are no problems until passenger $Z$ boards. Sooner or later one of these displaced passengers sits in seat $S$ or $X$ and then there are no problems until $X$ boards.

As $S$ and $X$ are equally likely to be the one taken, the probability is equal that $X$ has his seat or Mr Smith's left.

If Mr Smith sits in his seat, $S$, then $X$ gets his correct seat. But, if Mr Smith sits in seat $X$, then $X$ gets Mr Smith's seat. As these happen with equal likelihood (whenever Mr Smith boards), again $X$ gets his seat or Mr Smith's with equal probability in these cases.

This assumes, as above, that Mr Smith is not the last passenger.

 

[PeroK]

14. (solved by @Not anonymous and by @PeroK ) Mr. Smith on a full up flight with 50 passengers on a CRJ 100 had lost his boarding pass. The flight attendant tells him to sit anywhere. All other passengers sit on their booked seats, unless it is already occupied, in which case they randomly choose another seat just like Mr. Smith did. What are the chances that the last passenger gets the seat printed on his boarding pass?

I have a generalisation of this for any passenger. Assume Mr Smith boards first. Suppose there are $N$ seats on the flight and consider the $K$th last to board and the set of $K + 1$ seats that are Mr Smith's seat and the $K$ seats of the last $K$ passengers to board. Note that the $K$th last to board gets their seat as long as one of the other $K$ seats in that list gets occupied before their seat. The probability they get their seat is, therefore, $K/(K+1)$.

To see this, Mr Smith is equally likely to sit in anyone of these $K +1$ seats. If he does, then the remaining passengers gets their correct seats until possibly the $K$th last. If Mr Smith picks another seat, then the passenger whose seat he has taken is again equally likely to sit in anyone of those $K + 1$ seats. And, if he/she sits in another seat, then the same applies to the next displaced person etc.

In any case, when the $K$th last passenger boards, there is a probability of only $1/(K+1)$ that their seat is occupied.

And, even if Mr Smith does not board first then the same probabilities apply to all passengers who board after Mr Smith. I.e. everyone before Mr Smith gets their seat with probability $1$; and, everyone after Mr Smith gets their seat with probability $K/(K+1)$, if they are the $K$th last to board.

 

[fresh_42]

If it's not too late:

Spoiler

For a plane with any number of seats. Let's call Mr Smith's seat $S$ and the last to board's seat $X$.

When Mr Smith enters the plane, some seats are taken. But not $S$ or $X$. If Mr Smith sits elsewhere, then one and only one of seats $S$ or $X$ is left for the last passenger. To see this, imagine Mr Smith sits in seat $Y$. There are no problems until passenger $Y$ boards. If he/she sits in another seat $Z$, then there are no problems until passenger $Z$ boards. Sooner or later one of these displaced passengers sits in seat $S$ or $X$ and then there are no problems until $X$ boards.

As $S$ and $X$ are equally likely to be the one taken, the probability is equal that $X$ has his seat or Mr Smith's left.

If Mr Smith sits in his seat, $S$, then $X$ gets his correct seat. But, if Mr Smith sits in seat $X$, then $X$ gets Mr Smith's seat. As these happen with equal likelihood (whenever Mr Smith boards), again $X$ gets his seat or Mr Smith's with equal probability in these cases.

This assumes, as above, that Mr Smith is not the last passenger.

It's never too late for a second solution, or in this case, a third since the solutions have been published.