All Questions
28
questions
3
votes
0
answers
35
views
What is (the length of) the longest string that has no $k$ contiguous repeat substrings? [duplicate]
I'm not sure if "concatenated" or "contiguous" is the better word here, or what the best way of phrasing it is, so let me give an example.
Consider strings of $\ 0'$s and $\ 1'$s, ...
- 20.7k
2
votes
1
answer
103
views
A game of identifying real coins and weighing them.
Here is my problem as follows.
There are 2 counterfeit coins among 5 coins that look
identical. Both counterfeit coins have the same weight and the other
three real coins have the same weight. The ...
- 375
4
votes
1
answer
86
views
Probability that N i.i.d. draws from a multinomial distribution have made all events appear
Consider a multinomial distribution $\mathbb{P}$ on $S$ states $\{s_1,\dots,s_S\}$ where $S\in \mathbb{N}$ and $S\geq 2$, with probabilities $\mathbb{P}(s_i)=:p_i$. Now consider $N$ i.i.d. draws $X_1,...
- 6,779
8
votes
2
answers
483
views
Probability that the sum of two integers in $\{1,\dots,n\}$ equals a perfect square
I found the following question in MIT OCW's Mathematical Problem Solving and I'd like to know if my solution is ok:
"Let $p_n$ be the probability that $c+d$ is a perfect square when the integers $...
- 247
3
votes
1
answer
573
views
Combinatorics question - why am I over counting?
A group of 12 people want to go to a concert. They can travel in a small car that takes one driver and one passenger and two cars each taking one driver and 4 passengers. If there are five drivers in ...
- 310
1
vote
1
answer
39
views
Why does $\sum_{i=n}^{2n-1}\binom{i-1}{n-1}2^{1-i}$ computes the probability of $n$ head or tails
$\sum_{i=n}^{2n-1}\binom{i-1}{n-1}2^{1-i}$
For $i = n,n+1,\ldots, 2n - 1$, the sum above computes $P(E_i)$, the probability that i
tosses of a fair coin are required before obtaining $n$ heads or $n$ ...
user675768
0
votes
3
answers
110
views
I have this identity that I'd like to prove. $\sum_{k=0}^{n}\left(\frac{n-2k}{n}\binom{n}{k}\right)^2=\frac{2}{n}\binom{2n-2}{n-1}$
I have this identity that I'd like to prove.
$$\displaystyle{\sum_{k=0}^{n}\bigg(\dfrac{n-2k}{n}\binom{n}{k}}\bigg)^2=\dfrac{2}{n}\binom{2n-2}{n-1}$$
Here's what I have done so far: (using a binomial ...
user675768
0
votes
2
answers
88
views
Show that $\binom{n}{1}-3\binom{n}{3}+3^2\binom{n}{5}\cdots=0$
Show that if $n\equiv 0\pmod 6$ (although the statement holds true for $n\equiv 0\pmod 3$)
$\binom{n}{1}-3\binom{n}{3}+3^2\binom{n}{5}\cdots=0$
I am having trouble finding the appropriate polynomial ...
user675768
4
votes
1
answer
241
views
Coin Flip Problem: a competition based on the probability of flipping heads
I encountered a difficult coin problem and I wasn't sure how to solve the problem.
A has 30 coins and B has 20 coins. Each coin is only flipped once, and the winner is the individual which received ...
- 43
1
vote
1
answer
80
views
Restoring permutation from differences of adjacent elements
Suppose a permutation $\pi \in S_n$ is encoded by a list of integers $P=(p_1, p_2, ... p_{n-1})$, where $p_i = \pi(i+1) - \pi(i)$, i.e. $P$ is the list of differences of adjacent elements. Now, given $...
- 765
1
vote
0
answers
102
views
Odds of winning a prize in a weighted, random raffle
There is a raffle a state holds annually to assign salmon fishing licenses to fisherman; in a specific stetch of river, to control harvest and limit access, to ensure resource management and a high ...
- 11
2
votes
1
answer
198
views
Coloured partitions
I have $2n$ elements, $n$ of which are blue and the other $n$ are orange. Other than sharing a colour, they are distinct, i.e. each of those $2n$ objects is different and recognizable from any other.
...
- 143
3
votes
1
answer
194
views
Identifying distinct colored cubes from faces of $2 \times 2 \times 2$ arrangement
Note: The answer to Question 1 is "never." But Question 2 remains open!
Consider cubes that can have their faces colored white or red, and let us say that two colorings of a cube are equivalent if ...
- 15k
2
votes
2
answers
691
views
How many paths are there from point P to point Q if each step has to go closer to point Q.
Each of the six faces of a solid cube is divided into four squares as indicated in the diagram. Starting from vertex P paths can be travelled to vertex Q along connected line segments. Each movement ...
- 195
0
votes
1
answer
57
views
How do you sum a series when individual elements themselves may be series?
Okay Math SE, I've got a problem to fry your brains over. Let's see who can get this.
Consider this:
A cube exists in the euclidean space, it seemingly has the power to divide itself into a copy. ...
- 3