Post History
#3: Post edited
by
Hudjefa
·
2024-01-16T18:14:29Z (12 months ago)
- My approach is, I suppose, simpler.
- There are 4 voters: A, B, C, D. They vote randomly
- There are 4 candidates vying for A, B, C, D's attention. Let's call them W, X, Y, Z
- How many different possible electorate results are there?
- 1 candidate gets all 4 votes: $^4C_1 = 4$
- 1 candidate gets 3 votes: $^4C_1 \times ^3C_1 = 12$
1 candidate gets 2 votes (half the votes): $^4C_1 \times ^3C_2 = 12$All candidates get 1 vote each: $^4P_4 = 24$Total = $4 + 12 + 12 + 24 = 52$- Probability that one candidate gets at least 50% of the votes = $P(50\%)$
$P(50\%) = \frac{28}{52} = \frac{7}{13}$- I don't know how to generalize this "finding" though. What I can perhaps say that I know is the electorate is going to be myriad and the candidates, comparatively few.
- How does that affect the probability? Feels very tower of Hanoish and I don't have what it takes to solve that problem.
- Side quest: Despite the fact that this is a discrete distribution (1 or 2 votes, not 1.5 votes), the OP has chosen a continuous model (area). Does the fact that the answer is in combinatorics form, $\frac{n!}{(n - 1)!}$ mean anything?
- My approach is, I suppose, simpler.
- There are 4 voters: A, B, C, D. They vote randomly
- There are 4 candidates vying for A, B, C, D's attention. Let's call them W, X, Y, Z
- How many different possible electorate results are there?
- 1 candidate gets all 4 votes: $^4C_1 = 4$
- 1 candidate gets 3 votes: $^4C_1 \times ^3C_1 = 12$
- 1 candidate gets 2 votes (half the votes): $^4C_1(^3C_1 + ^3C_2) = 24$
- All candidates get 1 vote each: $^4C_4 = 1$
- Total = $4 + 12 + 24 + 1 = 41$
- Probability that one candidate gets at least 50% of the votes = $P(50\%)$
- $P(50\%) = \frac{40}{41}$
- I don't know how to generalize this "finding" though. What I can perhaps say that I know is the electorate is going to be myriad and the candidates, comparatively few.
- How does that affect the probability? Feels very tower of Hanoish and I don't have what it takes to solve that problem.
- Side quest: Despite the fact that this is a discrete distribution (1 or 2 votes, not 1.5 votes), the OP has chosen a continuous model (area). Does the fact that the answer is in combinatorics form, $\frac{n!}{(n - 1)!}$ mean anything?
#2: Post edited
by
Hudjefa
·
2024-01-16T13:08:31Z (12 months ago)
- My approach is, I suppose, simpler.
- There are 4 voters: A, B, C, D. They vote randomly
There are 3 candidates vying for A, B, C, D's attention. Let's call them W, X, Y, Z- How many different possible electorate results are there?
- 1 candidate gets all 4 votes: $^4C_1 = 4$
- 1 candidate gets 3 votes: $^4C_1 \times ^3C_1 = 12$
- 1 candidate gets 2 votes (half the votes): $^4C_1 \times ^3C_2 = 12$
- All candidates get 1 vote each: $^4P_4 = 24$
- Total = $4 + 12 + 12 + 24 = 52$
- Probability that one candidate gets at least 50% of the votes = $P(50\%)$
- $P(50\%) = \frac{28}{52} = \frac{7}{13}$
- I don't know how to generalize this "finding" though. What I can perhaps say that I know is the electorate is going to be myriad and the candidates, comparatively few.
- How does that affect the probability? Feels very tower of Hanoish and I don't have what it takes to solve that problem.
- Side quest: Despite the fact that this is a discrete distribution (1 or 2 votes, not 1.5 votes), the OP has chosen a continuous model (area). Does the fact that the answer is in combinatorics form, $\frac{n!}{(n - 1)!}$ mean anything?
- My approach is, I suppose, simpler.
- There are 4 voters: A, B, C, D. They vote randomly
- There are 4 candidates vying for A, B, C, D's attention. Let's call them W, X, Y, Z
- How many different possible electorate results are there?
- 1 candidate gets all 4 votes: $^4C_1 = 4$
- 1 candidate gets 3 votes: $^4C_1 \times ^3C_1 = 12$
- 1 candidate gets 2 votes (half the votes): $^4C_1 \times ^3C_2 = 12$
- All candidates get 1 vote each: $^4P_4 = 24$
- Total = $4 + 12 + 12 + 24 = 52$
- Probability that one candidate gets at least 50% of the votes = $P(50\%)$
- $P(50\%) = \frac{28}{52} = \frac{7}{13}$
- I don't know how to generalize this "finding" though. What I can perhaps say that I know is the electorate is going to be myriad and the candidates, comparatively few.
- How does that affect the probability? Feels very tower of Hanoish and I don't have what it takes to solve that problem.
- Side quest: Despite the fact that this is a discrete distribution (1 or 2 votes, not 1.5 votes), the OP has chosen a continuous model (area). Does the fact that the answer is in combinatorics form, $\frac{n!}{(n - 1)!}$ mean anything?
#1: Initial revision
by
Hudjefa
·
2024-01-16T13:07:51Z (12 months ago)
My approach is, I suppose, simpler.
There are 4 voters: A, B, C, D. They vote randomly
There are 3 candidates vying for A, B, C, D's attention. Let's call them W, X, Y, Z
How many different possible electorate results are there?
1 candidate gets all 4 votes: $^4C_1 = 4$
1 candidate gets 3 votes: $^4C_1 \times ^3C_1 = 12$
1 candidate gets 2 votes (half the votes): $^4C_1 \times ^3C_2 = 12$
All candidates get 1 vote each: $^4P_4 = 24$
Total = $4 + 12 + 12 + 24 = 52$
Probability that one candidate gets at least 50% of the votes = $P(50\%)$
$P(50\%) = \frac{28}{52} = \frac{7}{13}$
I don't know how to generalize this "finding" though. What I can perhaps say that I know is the electorate is going to be myriad and the candidates, comparatively few.
How does that affect the probability? Feels very tower of Hanoish and I don't have what it takes to solve that problem.
Side quest: Despite the fact that this is a discrete distribution (1 or 2 votes, not 1.5 votes), the OP has chosen a continuous model (area). Does the fact that the answer is in combinatorics form, $\frac{n!}{(n - 1)!}$ mean anything?