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Question Number 4166 by Filup last updated on 30/Dec/15

Commented by prakash jain last updated on 31/Dec/15

$To go broke : k wins, 2 + k losses$ $\underset{k = 0}{\sum^{\infty}} {(\frac{3}{4})}^{k} {(\frac{1}{4})}^{2 + k} = \frac{{(\frac{1}{2})}^{2}}{1 - \frac{3}{16}} = \frac{1}{13} ?$
Commented by Filup last updated on 31/Dec/15

$I submitted thw answer and it was wrong .$ $Hmm . . . this is a tricky question$
Commented by Filup last updated on 31/Dec/15

Commented by prakash jain last updated on 31/Dec/15

$What is the correct answer ?$
Commented by Yozzii last updated on 31/Dec/15

$L e t n = e v e n t b r o k e a f t e r n t h r o u n d$ $g_{n - 1} = e v e n t o f h a v i n g $ 1 a f t e r (n - 1) t h r o u n d$ $h = e v e n t y o u l o s e $ 1 i n a r o u n d$ $∴ P (n) = P (h) P (g_{n - 1}) = \frac{1}{4} P (g_{n - 1})$ $n = 2 : P (2) = 0.25 P (g_{1}) = 0.25 \times 0.25 = 0 . 25^{2}$ $n = 3 : P (3) = 0.25 P (g_{2}) = 0.25 \times 0$ $n = 4 : P (4) = 0.25 P (g_{3}) = 0.25 \times . . .$ $s t a r t e v e n (n = 0), n = 1 \to o d d \Rightarrow n = 2 \to e v e n (c a n n o t h a v e $ 1 \Rightarrow P (g_{e v e n}) = 0)$ $I^{'} m t h i n k i n g t h a t$ $P (g o b r o k e) = \underset{n = 2}{\sum^{\infty}} P (n) = 0.25 \underset{n = 1}{\sum^{\infty}} P (g_{n - 1}) .$
Commented by Yozzii last updated on 31/Dec/15

Commented by Yozzii last updated on 31/Dec/15

$I^{'} v e c r e a t e d a c r a z y^{'} e v e n t {s e q u e n c e}^{'}$ $d i a g r a m . T h e r e d p o i n t s r e p r e s e n t$ $h a v i n g $ 1 a f t e r t h e n t h r o u n d,$ $w h e r e t h e n t h r o u n d i s g i v e n b y$ $t h e n t h c i r c l e G r e e n d o t s r e p r e s e n t$ $$ 2 . D e e p p u r p l e d o t s r e p r e s e n t$ $t e r m i n a t i o n o f a g a m e .$ $A l l r e m a i n i n g d o t s c o u l d b e $ 3 o r$ $$ 4, e t c . . . F o l l o w i n g t h e c o l o u r c o d i n g h e l p s$ $t o t h i n k o f w h a t c o u l d p o s s i b l y$ $l e a d t o $ 2 f o r e x a m p l e . I^{'} m o n l y$ $c o n c e r n e d w i t h c o u n t i n g r e s u l t i n g r e d d o t s .$ $S o, f o r n = 1 : y o u c a n g e t $ 1 i n 1 w a y$ $f o r n = 3, y o u g e t $ 1 i n 2 w a y s$ $f o r n = 5 y o u g e t $ 1 i n 5 w a y s$ $f o r n = 7, y o u g e t $ 1 i n 14 w a y s$ $L e t a (n) = n u m b e r o f w a y s o f g e t t i n g$ $$ 1 a f t e r (2 n - 1) t h r o u n d .$ $\Rightarrow a (1) = 1, a (2) = 2, a (3) = 5, a (4) = 14$ $a (2) - a (1) = 2 - 1 = 1 = 3^{0}$ $a (3) - a (2) = 5 - 2 = 3^{1}$ $a (4) - a (3) = 14 - 5 = 9 = 3^{2}$ $a (n + 1) - a (n) = 3^{n - 1} ?$ $T h e s e a r e i m p o r t a n t i n f i n d i n g$ $t h e m u l t i p l e s f o r e a c h e v e n t$ $p r o b a b i l i t y (g e t t i n g $ 1 o n t h e n t h r o u n d) .$
Commented by prakash jain last updated on 31/Dec/15

$Mistake in my previous argument :$ $k = 0 : {(\frac{1}{4})}^{2}$ $k = 1 : (\frac{3}{4}) {(\frac{1}{4})}^{3} + {(\frac{1}{4})}^{2} (\frac{3}{4}) (\frac{1}{4}) = 2 (\frac{3}{4}) {(\frac{1}{4})}^{3}$ $k = 2 :^{2 k} C_{k} {(\frac{3}{4})}^{k} {(\frac{1}{4})}^{k + 2}$ $k wins can occur^{2 k} C_{k} different ways .$ $P = \underset{k = 0}{\sum^{\infty}}^{2 k} C_{k} {(\frac{3}{4})}^{k} {(\frac{1}{4})}^{k + 2}$ $Treating one dollar loss is easier :$
Commented by prakash jain last updated on 31/Dec/15

$For losing one dollar$ $\underset{k = 0}{\sum^{\infty}} {(\frac{3}{4})}^{k} {(\frac{1}{4})}^{k + 1} = \frac{4}{13}$ $In this case the only possibility is k$ $consecutive wins, followed by k + 1$ $consecutive losses to lose 1 dollar in 2 k + 1$ $moves .$ $So P = \frac{16}{169}$
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