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visits member for 1 year, 8 months
seen Jul 9 at 8:16

Financial mathematician-beginner.


Feb
2
answered Probability of rolling a “1” on a die conditioned on when all rolls are different.
Feb
2
comment Odds of picking the same number
Yes, you're right. However, it seemed to me that since he is confused about a basic question like this, it is in fact understanding rather than a formula what he's really after.
Feb
1
awarded  Popular Question
Jan
31
comment Simplify $\frac yx-\frac xy \over \frac 1y- \frac 1x$
Diagonal cancelling lines, neat! Didn't know about those.
Jan
31
answered Very silly permutation question
Jan
31
comment Very silly permutation question
Sadly I can't now, but others will surely do that and help you. By the way, no need to call this a silly question, it is a perfectly understandable confusion when one deals with combinatorics for the first time :)
Jan
31
comment Very silly permutation question
Short answer: because you've got twice as many. That's why you need to divide by $2!=2$. If you had, say AAAB, you'd have $3!=6$times as much and would need to divide by that.
Jan
31
awarded  Custodian
Jan
31
reviewed Reviewed Coin flipping probability game ; 7 flips vs 8 flips
Jan
31
answered Finding Eigenvectors of 3 x 3 matrix
Jan
31
revised Odds of picking the same number
added 395 characters in body
Jan
31
comment Odds of picking the same number
Yes you are correct, although your result is written (and possibly thought of) in a rather complicated manner (I will elaborate in the answer). Now, do you know what a complementary event is and how to calculate it?
Jan
31
revised Odds of picking the same number
added 3 characters in body
Jan
31
answered Odds of picking the same number
Jan
27
answered I'm trying to understand an equality
Jan
27
comment Solving a limit with logarithm and GIF
About using other methods - how about using the equivalence of Heine-Cauchy definition of a limit of a function/sequence and seeing this as a limit of a sequence?
Jan
27
comment Solving a limit with logarithm and GIF
Please correct the notation, this way it is unclear what the actual limit is.
Jan
26
comment Limits at infinity
By the way, what you did isn't "wrong", it just doesn't help you anyhow in this case.
Jan
23
accepted How do I see that $x^5+x-1=(x^2-x+1)(x^3+x^2-1)$
Jan
23
comment How do I see that $x^5+x-1=(x^2-x+1)(x^3+x^2-1)$
Thanks for an alternative route, I was thinking something similar, will try.