Get our expert's

answer on brainly

SEE EXPERT ANSWER

Get your **free** account and access **expert** answers to this and **thousands** of other questions.

See more answers at brainly.com

Get this expert

answer on brainly

SEE EXPERT ANSWER

Get your **free** account and access **expert** answers to this and **thousands** of other questions

I am terrible at probability, but everyone always asks me about it.

I'll be curious to hear what others say though

So am I. Took me like 3 days to solve this. But I do have the solution.

I think this is geometric distribution?

Well, no I don't think so.

@Mariantheduke You might be interested. This problem came from the physics dept.

I do not think this solution works.

1/n * (n)/n + 1/n *(n-1)/n + 1/n * (n-2)/n+...

1/n^2 (n+n-1+n-2+n-3+...n-n)

on the right track, but not it.

Did I missed something?

I'm sorry, that's the probability for the first player to lose. Wait a second.

The probability that the first player wins is\[1-\left(1-\frac1n\right)^n\]where \(r=0\).

I got |dw:1349578761373:dw|

What is r ?

The roll just made.

Well, goodnight, Sir. I will see you another day perhaps.

WELL GOOD NIGHT

@badreferences : No man! I can't do probability. :S