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Unsolved statistics questions from other sites, part II

chisigma

Well-known member
Feb 13, 2012
1,704
Posted some years ago on an Italian math forum and not solved...

A maker for the promotion of his product includes in each pop corn pakage a prize [a colored pencil, a dummy animal, a pitcure card, etc...] randomly choosen among n different types. What is the expected number of pakages to be purchased if one wants to have the entire set of prizes?...

Kind regards


$\chi$ $\sigma$
 

chisigma

Well-known member
Feb 13, 2012
1,704
Posted some years ago on an Italian math forum and not solved...

A maker for the promotion of his product includes in each pop corn pakage a prize [a colored pencil, a dummy animal, a pitcure card, etc...] randomly choosen among n different types. What is the expected number of pakages to be purchased if one wants to have the entire set of prizes?...
In the original post it has been proposed n=6, but it is better to try to analyse the general case. The failure of the attempts to solve the problem is probably due to the fact that nobody realized that in fact this is a Markov Chain problem with n states. In fact in the first purchase in any case one of the price is acquired and the missing prizes are n-1. At the second purchase or one finds the same price ad in the first and no progress is made, or one adds at his collection a new price and the missing prizes are n-2. Proceeding in this way the 'game' finish when the final state n is met. The state diagram is represented in figure...




The transition matrix is...

$\displaystyle P = \left | \begin{matrix} \frac{1}{n}& \frac{n-1}{n} & 0 & 0 & \cdot & \cdot & \cdot & 0 & 0 & 0 \\ 0& \frac{2}{n} & \frac{n-2}{n} & 0 & \cdot & \cdot & \cdot & 0 & 0 & 0 \\ 0 & 0 & \frac{3}{n} & \frac{n-3}{n} & \cdot & \cdot & \cdot & 0 & 0 & 0 \\ \cdot & \cdot & \cdot & \cdot & \cdot & \cdot & \cdot & \cdot & \cdot & \cdot \\ \cdot & \cdot & \cdot & \cdot & \cdot & \cdot & \cdot & \cdot & \cdot & \cdot \\ \cdot & \cdot & \cdot & \cdot & \cdot & \cdot & \cdot & \cdot & \cdot & \cdot \\ 0 & 0 & 0 & 0 & \cdot & \cdot & \cdot & \frac{3}{n} & 0 & 0 \\ 0 & 0 & 0 & 0 & \cdot & \cdot & \cdot & \frac {n-2}{n} & \frac{2}{n} & 0 \\ 0 & 0 & 0 & 0 & \cdot & \cdot & \cdot & 0 & \frac{n-1}{n} & \frac{1}{n} \\ 0& 0 & 0 & 0 & \cdot & \cdot & \cdot & 0 & 0 & 1\end{matrix} \right| $ (1)

Now we proceed as in...

http://mathhelpboards.com/basic-pro...number-questions-win-game-4154.html#post18909

... starting by n=2 and finding A(n), i.e. the mean number of purchase necessary to acquire the entire set of prizes...

n=2

In this case is...

$\displaystyle Q = \frac{1}{2} \implies I - Q = \frac{1}{2} \implies (I_{1} - Q)^{- 1} = 2 \implies A(2)= 1 + 2 = 3$

n=3

In this case is...

$\displaystyle Q = \left | \begin{matrix} \frac{1}{3} & \frac{2}{3} \\ 0 & \frac{2}{3} \end{matrix} \right | \implies I - Q = \left | \begin{matrix} \frac{2}{3} & - \frac{2}{3} \\ 0 & \frac{1}{3} \end{matrix} \right | \implies (I - Q)^{-1} = \left | \begin{matrix} \frac{3}{2} & 3 \\ 0 & 3 \end{matrix} \right | \implies A(3) = 1 + 3 + \frac{3}{2} = \frac{11}{2}$

n=4

In this case is...

$\displaystyle Q = \left | \begin{matrix} \frac{1}{4} & \frac{3}{4} & 0 \\ 0 & \frac{1}{2} & \frac{1}{2} \\ 0 & 0 & \frac{3}{4} \end{matrix} \right | \implies I - Q = \left | \begin{matrix} \frac{3}{4} & - \frac{3}{4} & 0 \\ 0 & \frac{1}{2} & - \frac{1}{2} \\ 0 & 0 & \frac {1}{4} \end{matrix} \right | \implies (I - Q)^{-1} = \left | \begin{matrix} \frac{4}{3} & 2 & 4 \\ 0 & 2 & 4 \\ 0 & 0 & 4 \end{matrix} \right | \implies A(4) = 1 + 2 + 4 + \frac{4}{3} = \frac{25}{3}$


n=5

In this case is...

$\displaystyle Q = \left | \begin{matrix} \frac{1}{5} & \frac{4}{5} & 0 & 0 \\ 0 & \frac{2}{5} & \frac{3}{5} & 0 \\ 0 & 0 & \frac{3}{5} & \frac{2}{5} \\ 0 & 0 & 0 & \frac{4}{5} \end{matrix} \right | \implies I - Q = \left | \begin{matrix} \frac{4}{5} & - \frac{4}{5} & 0 & 0 \\ 0 & \frac{3}{5} & - \frac{3}{5} & 0 \\ 0 & 0 & \frac {2}{5} & - \frac{2}{5} \\ 0 & 0 & 0 & \frac{1}{5} \end{matrix} \right | \implies (I - Q)^{-1} = \left | \begin{matrix} \frac{5}{4} & \frac{5}{3} & \frac{5}{2} & 5 \\ 0 & \frac{5}{3} & \frac{5}{2} & 5 \\ 0 & 0 & \frac{5}{2} & 5 \\ 0 & 0 & 0 & 5 \end{matrix} \right | \implies $

$\displaystyle \implies A(5) = 1 + \frac{5}{4} + \frac{5}{3} + \frac{5}{2} + 5 = \frac{137}{12}$

n=6

In this case is...

$\displaystyle Q = \left | \begin{matrix} \frac{1}{6} & \frac{5}{6} & 0 & 0 & 0 \\ 0 & \frac{1}{3} & \frac{2}{3} & 0 & 0 \\ 0 & 0 & \frac{1}{2} & \frac{1}{2} & 0 \\ 0 & 0 & 0 & \frac{2}{3} & \frac{1}{3} \\ 0 & 0 & 0 & 0 & \frac{5}{6} \end{matrix} \right | \implies I - Q = \left | \begin{matrix} \frac{5}{6} & - \frac{5}{6} & 0 & 0 & 0 \\ 0 & \frac{2}{3} & - \frac{2}{3} & 0 & 0 \\ 0 & 0 & \frac {1}{2} & - \frac{1}{2} & 0 \\ 0 & 0 & 0 & \frac{1}{3} & - \frac{1}{3} \\ 0 & 0 & 0 & 0 & \frac{1}{6} \end{matrix} \right | \implies$

$\displaystyle \implies (I - Q)^{-1} = \left | \begin{matrix} \frac{6}{5} & \frac{3}{2} & 2 & 3 & 6 \\ 0 & \frac{3}{2} & 2 & 3 & 6\\ 0 & 0 & 2 & 3 & 6 \\ 0 & 0 & 0 & 3 & 6 \\ 0 & 0 & 0 & 0 & 6 \end{matrix} \right | \implies A(6) = 1 + \frac{6}{5} + \frac{3}{2} + 2 + 3 + 6 = \frac{147}{10}$

Observing the result we have obtained it seems not to be necessary to proceed with n>6 and with great probability we can conclude that the general result is...

$\displaystyle A(n) = n\ \sum_{k=1}^{n} \frac{1}{k} = n\ H_{n}\ (2)$

... and (2) is an interesting result wich can be useful in many application fields...

Kind regards

$\chi$ $\sigma$
 

chisigma

Well-known member
Feb 13, 2012
1,704
Posted the 03 21 2014 on www.matematicamente.it by the user biglio23 [original in Italian...] and not yet solved...

In a restaurant n people live their umbrellas at the entrance. The first person leaving the restaurant chooses randomly an umbrella. The successive people take their umbrella if they find it, otherwise choose randomly an umbrella. What is the probability that the last person finds his own umbrella?...

Kind regards

Kind regards

$\chi$ $\sigma$
 

chisigma

Well-known member
Feb 13, 2012
1,704
Posted the 03 21 2014 on www.matematicamente.it by the user biglio23 [original in Italian...] and not yet solved...

In a restaurant n people live their umbrellas at the entrance. The first person leaving the restaurant chooses randomly an umbrella. The successive people take their umbrella if they find it, otherwise choose randomly an umbrella. What is the probability that the last person finds his own umbrella?...
Also this problem can be considered as a Markov Chain type. If n is the persons leaving the restaurant, then the number of states in n+1 and the process ends after at least n-1 steps in the adsorbing state OK [the last person finds his own umbrella...] or KO {the last person doesn't...]. The state diagram is shown in the figure...



In the first step we have three possibilities...

a) the first person chooses his own umbrella and the process ends in OK...

b) the first person chooses the umbrella of the last person and the process ends in KO...

c) the first person chooses an umbrella different from a) and b) and the process goes to further step...


What is important to see is that in case c) the remaining process is the same of the original process with n-1 person, so that the final result of the two process are the same, i.e. the result is independent from n. Taking into account that we can choose n=3, in which case the transition matrix is...

$\displaystyle A = \left | \begin{matrix} 0 & \frac{1}{3} & \frac{1}{3} & \frac{1}{3} \\ 0 & 0 & \frac{1}{2} & \frac{1}{2} \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{matrix} \right |\ (1)$

The probability that the process ends in OK is the term $a_{1,4}$ of the matrix $A^{2}$, i.e. ...


$\displaystyle P = 0 + \frac{1}{6} + 0 + \frac{1}{3} = \frac{1}{2}\ (2)$

Kind regards

$\chi$ $\sigma$
 

chisigma

Well-known member
Feb 13, 2012
1,704
Posted the 03 19 2014 on www.artofproblemsolving.com by the user Tetrapak1234 and not yet solved...

Two independent r.v. X and Y are given, both with p.d.f. $\displaystyle f(x) = \frac{1}{\pi\ (1 + x^{2})}$. Let be Z a r.v. defined as...

$\displaystyle Z =\begin{cases}Y &\text{if}\ |Y|>1 \\ - Y &\text{if}\ |Y| \le 1\end{cases}$

... and V = X + Z. Find the p.d.f. $\displaystyle f_{V} (x)$...



Kind regards


$\chi$ $\sigma$
 

chisigma

Well-known member
Feb 13, 2012
1,704
Posted the 03 19 2014 on www.artofproblemsolving.com by the user Tetrapak1234 and not yet solved...

Two independent r.v. X and Y are given, both with p.d.f. $\displaystyle f(x) = \frac{1}{\pi\ (1 + x^{2})}$. Let be Z a r.v. defined as...

$\displaystyle Z =\begin{cases}Y &\text{if}\ |Y|>1 \\ - Y &\text{if}\ |Y| \le 1\end{cases}$

... and V = X + Z. Find the p.d.f. $\displaystyle f_{V} (x)$...
The first part is trivial because, given the simmetry of the p.d.f. of Y around x=0, Y and Z have the same p.d.f. and the problem consists in finding the p.d.f. of the sum of X and Y. The Fourier Transform of $f_{X}$ is given by...


$\displaystyle \mathcal{F} \{f_{X}(x)\} = \frac{2}{\pi}\ \int_{0}^{\infty} \frac{\cos (\omega\ x)}{1+x^{2}}\ dx = e^{- |\omega|}\ (1) $

...so that is...

$\displaystyle f_{V} (x) = \mathcal{F}^{-1} \{e^{-2\ |\omega|}\} = \frac{1}{\pi} \int_{0}^{\infty} e^{- 2\ \omega}\ \cos (\omega\ x)\ d \omega = \frac{1}{\pi\ [1 + (\frac{x}{2})^{2}]}\ (2)$

Kind regards

$\chi$ $\sigma$
 

chisigma

Well-known member
Feb 13, 2012
1,704
Posted [in a bit different form...] the 03 17 2014 on www.artofproblemsolving.com by the user herrmann and not yet solved...


You are playing a following game: In every move you roll a regular dice (1-6) and your current account is sum of all dice rolls. For example, if in first roll you get 3, in second roll 5, then your current account is 8. In every move you can either re-roll the dice or you can take all from your current account and finish the game. The only problem is that, if after some roll you have amount on your account that is a multiple of 6, you lose everything. What is the first number after that it is better to take everything from your account than to re-roll, and what is the expected value of this game if played optimal?...

Kind regards

$\chi$ $\sigma$
 

chisigma

Well-known member
Feb 13, 2012
1,704
Posted [in a bit different form...] the 03 17 2014 on www.artofproblemsolving.com by the user herrmann and not yet solved...


You are playing a following game: In every move you roll a regular dice (1-6) and your current account is sum of all dice rolls. For example, if in first roll you get 3, in second roll 5, then your current account is 8. In every move you can either re-roll the dice or you can take all from your current account and finish the game. The only problem is that, if after some roll you have amount on your account that is a multiple of 6, you lose everything. What is the first number after that it is better to take everything from your account than to re-roll, and what is the expected value of this game if played optimal?...
After n dice rolls the probability to be 'play on' is $\displaystyle P_{n} = (\frac{5}{6})^{n}$, so that the expected value of the gain in this situation is...


$\displaystyle G_{1} = \frac{15}{5}\ P_{1} = \frac{5}{2} = 2.5$

$\displaystyle G_{2} = \frac {15 - 1 + 45}{9}\ P_{2} = \frac{1475}{324} = 4.552...$

$\displaystyle G_{3} = \frac{15 - 3 + 45 + 75}{13}\ P_{3} = \frac{16500}{2808} = 5.876...$

$\displaystyle G_{4} = \frac{15-6 + 45 + 75 + 105}{17}\ P_{4} = \frac{146250}{22032} = 6.638...$

$\displaystyle G_{5} = \frac{15 - 10 + 45 + 75 + 105 + 135}{21}\ P_{5} = \frac{1141625}{163296} = 6.985...$

$\displaystyle G_{6} = \frac{45 + 75 + 105 + 135+ 195}{25}\ P_{6} = \frac{8671875}{1166400} = 7.4347...$

$\displaystyle G_{7} = \frac{45 - 7 + 75 + 105 + 135 + 195 + 15 + 216}{29}\ P_{7} = \frac{60859375}{8118144} = 7.496...$

$\displaystyle G_{8} = \frac{45 - 15 + 75 + 105 + 135 + 195 + 231 + 15 + 240}{33}\ P_{8} = \frac{400781250}{55427328} = 7.23...$

At this point our job is finshed and we can coclude that the optimal strategy' is to stop the play after the 7-th dice roll because if n=8 the expected gain decreases...

Kind regards

$\chi$ $\sigma$
 

maxkor

Member
Jul 22, 2014
62
Posted the Jan 28, 2013 on Art of Problem Solving (AoPS) by the user thugzmath10 and not yet solved

$A$ and $B$ are points on a circle centered at $O$ and radius $2$. An arbitrary point $X$ lies on the major arc $AB$. Determine the probability that $[AXB]\geq \sqrt{6}$.

Note: $[AXB]$ denotes the area of triangle $AXB$.