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Dec
18
awarded  Yearling
Dec
18
awarded  Yearling
Jan
7
awarded  Nice Answer
Jan
6
comment Asymptotic equivalent of the law of lotto minimal value
Oh god. I was pretty sure that using your $\ell$ trick it was possible to prove $$1 - \left( 1 - {m\over n-\ell} \right)^\ell <\mathbb P(X\le \ell) < 1-(1-p)^\ell.$$ I’am missing something, I’ll have to go back on this later.
Jan
6
awarded  Scholar
Jan
6
accepted Asymptotic equivalent of the law of lotto minimal value
Jan
6
comment Asymptotic equivalent of the law of lotto minimal value
In fact, $\left(1-\frac m{n-l}\right)^{k-1}\frac m{n-l} < \mathbb P(X=k) < (1-p)^{k-1} p$, so you get easily a framing of $\mathh b \P(X \le \ell)$ that allows to conclude with the squeeze theorem. This is nice.
Jan
6
comment Asymptotic equivalent of the law of lotto minimal value
My question was more general, but I think in this way it will be possible to concludre concerning $\mathbb P(X \le k)$... the $\ell$ trick is beautiful, many thanks.
Jan
6
comment Probability of an odd number in 10/20 lotto
You might be interested by this question I asked today, related to this problem...
Jan
6
revised Asymptotic equivalent of the law of lotto minimal value
typo
Jan
6
asked Asymptotic equivalent of the law of lotto minimal value
Jan
4
revised Probability that n numbered objects placed in random order have p objects in the correct position
typo
Jan
4
revised Probability that n numbered objects placed in random order have p objects in the correct position
notation clarified + asymptotics
Jan
4
answered Probability that n numbered objects placed in random order have p objects in the correct position
Jan
4
comment About finding the function such that $f(xy)=f(x)f(y)-f(x+y)+1$
This is homework, right? Starting point: show that $f(0) = 1$. Try to continue with integers, then with rationals of the form $1\over q$, finish with $\mathbb Q$.
Jan
3
comment Probability of an odd number in 10/20 lotto
@PatrickDaSilva This is the "recursive formula" I was alluding to. Besides that, no offense, don’t worry!
Jan
3
comment Probability of an odd number in 10/20 lotto
This is for even $m$ :) I am confident you won’t have problem to modify it for odd $m$
Jan
3
comment Probability of an odd number in 10/20 lotto
@PatrickDaSilva Please note that I detailed the computation for readibility for the layman (I teach a lot of elementary probabilities), of course the general formula is $$\sum_{k \text{ odd }, k < n} { {k\choose m-1} \over {n\choose m} }$$ with the convention ${a \choose b} = 0$ when $a<b$. The identity with joriki’s formula is a direct consequence of the recursive formula well known by "we mathematicians" :)
Jan
3
comment Probability of an odd number in 10/20 lotto
Now an other question: you know for sure that the answer is $p = {k \over {20\choose 10}} = {k \over 184756}$, but you feel unable to compute $k$. Instead, you want to find it by Monte Carlo simulation. How many values do you need to simulate before having a 95% CI for $184756 p$ containing only one integer? Answer: around 117 billions :)