Mathematical and Physical Journal
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Problem B. 4419. (January 2012)

B. 4419. Jack Potter is an addicted gambler. Yesterday, he threw 20 000 forints (HUF, Hungarian currency) in a one-armed bandit machine without his family knowing about it. To make things worse, he took that money from the amount that the family reserved for food. To avoid the affair being revealed, today he is taking the remaining 40 000 forints of the family, too, and goes back to the casino to play roulette. Since he does not want to risk too much, he is betting 1000 forints on red or on black in each game. If he wins, which has a probability of 18/37, then he gets another 1000 forints. Otherwise he loses his bet. He stops playing either when he has gathered a total of 60 000 forints -- in that case he can put all the money back in its place -- or when he has lost all the money. What is the probability that Jack Potter will manage to gather the 60 000 forints?

(Based on the idea of D. Pálvölgyi, Budapest)

(5 pont)

Deadline expired on February 10, 2012.

Sorry, the solution is available only in Hungarian. Google translation

Megoldás. Legyen \(\displaystyle p=18/37\), és \(\displaystyle k=0,1,\ldots,60\) esetén jelölje \(\displaystyle a_k\) annak a valószínüségét, hogy ha Kassza Blankának éppen \(\displaystyle k\)-szor 1000 forintja van, akkor sikerül összegyüjtenie a 60000 forintot. Nyilván \(\displaystyle a_0=0\), hiszen ha semmi pénze nincs, akkor már nem tud tovább játszani, és \(\displaystyle a_{60}=1\). Továbbá \(\displaystyle 1\le k\le 59\) esetén érvényes az \(\displaystyle a_k=pa_{k+1}+(1-p)a_{k-1}\) összefüggés, aminek alapján felírhatjuk az \(\displaystyle (a_k)\) sorozatra érvényes

\(\displaystyle a_{k}=\frac{1}{p}a_{k-1}-\frac{1-p}{p}a_{k-2},\quad k=2,3,\ldots,60\)

rekurziót. Mivel az \(\displaystyle x^2-(1/p)x+(1-p)/p=0\) másodfokú egyenlet megoldásai \(\displaystyle x_1=(1-p)/p\) és \(\displaystyle x_2=1\), a lineáris rekurziók elmélete alapján alkalmas \(\displaystyle \alpha,\beta\) valós számokkal

\(\displaystyle a_k=\alpha x_1^k+\beta x_2^k=\alpha\left(\frac{1-p}{p}\right)^k+\beta\)

teljesül \(\displaystyle k=0,1,\ldots,60\) esetén. Az \(\displaystyle a_0=0\) feltétel alapján \(\displaystyle \beta=-\alpha\), amit az \(\displaystyle a_{60}=1\) feltételbe behelyettesítve

\(\displaystyle \alpha=\frac{1}{\left(\frac{1-p}{p}\right)^{60}-1}\)

adódik, ahonnan

\(\displaystyle a_{k}= \frac{\left(\frac{1-p}{p}\right)^{k}-1}{\left(\frac{1-p}{p}\right)^{60}-1}= \frac{\left(\frac{19}{18}\right)^{k}-1}{\left(\frac{19}{18}\right)^{60}-1}= \frac{e^{k\ln(19/18)}-1}{e^{60\ln(19/18)}-1}.\)

Mivel \(\displaystyle \ln(19/18)\approx 0.05406722\), \(\displaystyle 40\ln(19/18)\approx2.1626888\), \(\displaystyle 60\ln(19/18)\approx 3.2440332\), azt kapjuk, hogy a keresett valószínüség,

\(\displaystyle a_{40}\approx\frac{e^{2.1626888}-1}{e^{3.2440332}-1}\approx \frac{8.6945-1}{25.637-1}\approx0.3123.\)


36 students sent a solution.
5 points:Ágoston Péter, Ágoston Tamás, Czipó Bence, Czövek Márton, Demeter Dániel, Fehér Zsombor, Fonyó Viktória, Forrás Bence, Havasi 0 Márton, Herczeg József, Janzer Olivér, Kabos Eszter, Kovács-Deák Máté, Maga Balázs, Medek Ákos, Mester Márton, Mihálykó András, Nagy Róbert, Ódor Gergely, Papp Roland, Szabó 928 Attila, Tardos Jakab, Tossenberger Tamás, Varnyú József, Viharos Andor.
4 points:Strenner Péter, Zilahi Tamás.
2 points:1 student.
1 point:7 students.
0 point:1 student.

Problems in Mathematics of KöMaL, January 2012