5
votes
Dividing $100$ coins into $7$ groups such that I can choose any number of coins by selecting whole groups
You are forced to powers of $2$ when the total number of coins is $2^k-1$ and you split them into $k$ groups. You have groups from $2^0=1$ to $2^{k-1}$ and the sum of these is $2^k-1$. To find the ...
- 378k
4
votes
Accepted
Creative Algebra Net Problem Solving Question
Since you have $4$ linear equations in $3$ unknowns, it's an overdetermined system and, thus, potentially inconsistent, but this answer shows this isn't the case. First, to avoid dealing with ...
- 49.7k
3
votes
Probability of 3 darts landing in the same half of the board
The “darts on a board” version is equivalent to the “points on a circle” version. When you throw darts on a board, you can radially project the darts onto the circumference of the dartboard; these ...
- 78.3k
2
votes
Expected Number of Letters Typed Until MOO is Typed When Letters Are Typed Randomly
I disagree with the official solution. Further, I think that the original poster's first approach is both workable and elegant. However, the math needs to be nailed down.
Assume that the character ...
- 37.3k
2
votes
Number of Chess Games Possible: Parity Discussion
My ideas aren't completely thought-through, but here's what I've come up with so far:
$\Large \text{Switching colors}$
The idea here is to "pair up" games to prove that the total number of ...
- 21
1
vote
Russian roulette with re-spin intuition for asymmetric solution
Each run is independent in a certain abstract sense.
We can imagine an indefinite sequence of spins of the cylinder,
always with one live bullet and pulling the trigger after each spin.
In this ...
- 100k
1
vote
Question 4.31 Heard on the Street - Probability/Game Theory Question
As you say, if B plays red/blue with equal probability, it is better for A to play red with probability 1; but then it is better for B to play blue with probability 1; etc; but then it is better for A ...
- 3,576
1
vote
Accepted
expected value of high-low guessing game
You can get one number (either $50$ or $51$) on your first guess, then either of two numbers on your second guess, then any of four numbers on your third guess, and so forth, except that your seventh ...
- 34.5k
1
vote
Relationship of the Egg Riddle solution to bisection and $\log(N)$ function?
Defining the function of minimum number of drops required with $k$ eggs for $n$ floors:$$f(n, k) = \begin{cases}
0 & \text{ if } n=0\\
\infty & \text{ otherwise if } k=0 \\
\min\limits_{x=1}^{...
- 1,181
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