 Imagine that you have a loaded die (that’s the singular of ‘dice’). This means the die is biased. It’s not fair. If you roll it, the probability that you’ll get a $4$ is higher than the probability of getting any other number. You roll the die a few times and analyze the data.

Decide which of the following is more likely to happen:

$A. \quad 2, 5, 3, 4, 6$ $B. \quad 4, 2, 5, 3, 4, 6$

Drop a comment below with your answer, and please do not explain your answer because I don’t want anyone to get influenced by other peoples’ comments. Just a simple $A$ or a $B$ will suffice.

I’m going to bed now and when I wake up the next day I hope to see a lot of comments! :)

Until then! Note by Mursalin Habib
5 years, 4 months ago

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Option 1: A

Upvote this comment if you think it is right.

- 5 years, 4 months ago

Option 2: B

Upvote this comment if you think it is right.

- 5 years, 4 months ago

Option B

- 5 years, 4 months ago

I say A.

- 5 years, 4 months ago

B

- 5 years, 4 months ago

I see it's possible that both are equally likely to happen. But one option cannot be more likely than the other; it's either equally likely or less likely depending on the probability distribution.

- 5 years, 4 months ago

But one option cannot be more likely than the other;

Why is that? If something is less likely to happen then something else, then that something else is more likely to happen than the first something.

- 5 years, 4 months ago

I mean one particular option cannot be more likely than the other one, not any one option cannot be more likely. Ambiguity; my bad.

- 5 years, 4 months ago

B

- 5 years, 4 months ago

Option A

- 5 years, 4 months ago

@Mursalin Habib It has been a day!

- 5 years, 4 months ago

A

- 5 years, 4 months ago

Because apparently this one person's "day" meaning 15 days in real world, I decide to screw it and give my reasoning. Here, $P(A), P(B), P(4)$ are probabilities of getting the sequence $A$, the sequence $B$, and the throw $4$ in that order.

$A$ is more likely. Note that $P(B) = P(4) \cdot P(A) \le 1 \cdot P(A) = P(A)$, so the probability of getting $B$ is less than or equal to $A$.

- 5 years, 4 months ago