# Math Q&A Part 2

#### mobel

You have 3 parameters: number of urns (u), number of balls (b) and maximum balls in any urn (m). (I think it's been established that the size of the urns does not matter).

For any specific u, b and m, you can solve it using my technique, but it can get very long and tedious.

It was okay for u = 14, b = 20 and m = 2. But when m = 3, it would take a lot of calculations.

If you are looking for a general formula invovling u, b and m, I think that would be hard.
No there is misunderstanding about what I tried to say.
Any complicated formula assuming that the formula exists could be simplified to give NOT the exact result but a result nearing the exact solution.

How to afford a problem like this one by simplifying the solution.
We have 14 urns we want to place 20 balls on those urns and we wnat to know what is the probability that one urn contains al least 3 balls (or 4 balls).
How can we restate the problem without altering or changing it and find out the solution quickly using a simple reasoning?

#### CRGreathouse

Forum Staff
My goal in fact was how to avoid simulations or even a complicate and full solution.
Can we use elementary reasoning to obtain an approximated solution not only for this problem in particular but for others too ?
You could use the Poisson distribution with $\lambda=20/14.$ Under that model the chance of getting 2 or fewer balls in a single urn is
$$e^{-\lambda}\left(\frac{\lambda^0}{0!}+\frac{\lambda^1}{1!}+\frac{\lambda^2}{2!}\right) =\frac{1+\lambda+\lambda^2/2}{e^{\lambda}} =\frac{169}{49e^{10/7}}\approx0.8265$$
and so the chance that all 14 have 2 or fewer is that probability to the power of 14, giving a 93.05% chance (under this model) of getting 3 or more.

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#### Pero

You could use the Poisson distribution with $\lambda=20/14.$ Under that model the chance of getting 2 or fewer balls in a single urn is
$$e^{-\lambda}\left(\frac{\lambda^0}{0!}+\frac{\lambda^1}{1!}+\frac{\lambda^2}{2!}\right) =\frac{1+\lambda+\lambda^2/2}{e^{\lambda}} =\frac{169}{49e^{10/7}}\approx0.8265$$
and so the chance that all 14 have 2 or fewer is that probability to the power of 14, giving a 93.05% chance (under this model) of getting 3 or more.
There is the lack of independence to consider. You could vary $\lambda$ at each step somehow to reflect this.

#### CRGreathouse

Forum Staff
There is the lack of independence to consider. You could vary $\lambda$ at each step somehow to reflect this.
Yes, I agree. But I was asked to find an approximation with elementary reasoning, so I gave a simple one.

#### Hoempa

Math Team
At the risk of not fully understanding the question, I guess Pero gets the next question?

#### CRGreathouse

Forum Staff
At the risk of not fully understanding the question, I guess Pero gets the next question?
Sounds good to me.

#### Pero

Gambler's Purgatory (Uncountable Solution)

A gambler is sent to hell and given one coin and he must play a devilish slot machine. Every time he puts in a coin, he gets a countable infinity of coins out. And he remains in Purgatory until he has got rid of all his coins. He can, however, put in his coins arbitrarily quickly.

a) Find a solution for the gambler to get out of Purgatory.

b) Find an uncountable solution where the gambler puts in a coin at every instant. (I.e. if he puts his first coin in at t = 0 and has got rid of his coins by t = T, then find a solution where he puts in a coin at every instant in [0, T).)

#### agentredlum

Math Team
He can't get out simply because every time he puts in a coin he gets at least 1 coin back so how is he going to get rid of all coins? #### Olinguito

To get rid of the first infinite lot of coins, put the first coin in at $t=0$, the next at $t=0.5$, the next after at $t=0.75$, and so on. By $t=1$ you should have got rid of the first infinite lot of coins. For the second infinite lot of coins, repeat the process but twice as quickly. So you get rid of the second infinite lot between $t=1$ and $t=1.5$. For the third infinite lot, do it four times as quickly, getting rid of it between $t=1.5$ and $t=1.75$, and so on. Thus by $t=2$ you have got rid of an infinite number of infinite lots of coins.

Well, thatâ€™s just my idea. :ninja:

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#### CRGreathouse

Forum Staff
So you got rid of $\omega^2$ coins, but didn't you get $\omega^3$ coins back in return? :unsure:

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