A variation of gamblers ruin problem.

[MaxManus]

Homework Statement

A gambler has 2$ and wants to have 10$. To get the money he enters a game where a fair coin is tossed. If he bets on the right side he wins doubles his stake and if he bets wrong he loses his stake. The strategy is to bet everythig if he has 5$ or less and just enough to walk away with 10$ if he has more than 5$

The Attempt at a Solution

X(0) = 2$. He start with 3$
P(X(1) = 4) = 1/2
P(X(1) = 0) = 1/2

P(X(2) = 8 l X(1) = 4) = 1/2

Bets 3 $
P(X(3) = 10 l X(2) = 8) = 1/2
P(X(3) = 6 l X(2) = 8) = 1/2But this is probably ot the way to solve it because you can continiue for ever.
The problem is from a markow chain chapter in my statistics book.

 

[lanedance]

what are you trying to do? find the expectation value of the strategy?

 

[lanedance]

however you could set it up as markov chain, with 5x5 transition matrix as follows, let the 5 states be (0,2,4,8,16)

P(x_j|x_i) =
1 if xi = xj = 16 (walk away), otherwise 0 if xi = 16
0.5 if x_i = x_j-1
0.5 if x_j = 0
0, otherwise

 

[MaxManus]

Sorry, I want to find the probability that the gambler achives 10$ and the expected number of tosses until the gambler either walks away with 10$ or 0 $

 

[MaxManus]

however you could set it up as markov chain, with 5x5 transition matrix as follows, let the 5 states be (0,2,4,8,16)

P(x_j|x_i) =
1 if xi = xj = 16 (walk away), otherwise 0 if xi = 16
0.5 if x_i = x_j-1
0.5 if x_j = 0
0, otherwise

Thanks, but the states are (0,1,2,3,4,5,6,7,8,9,10), the gambler only bets all when he has 5$ or less.

 

[lanedance]

ok i get it now, well you i think i have given you enough to come up with the transition matrix for those states

 

[Dick]

Thanks, but the states are (0,1,2,3,4,5,6,7,8,9,10), the gambler only bets all when he has 5$ or less.

MaxManus, have you drawn out the Markov chain? Not all of those states are possible. You should notice by the fourth or fifth step (depending on how you count steps), you have either won, lost or returned to the initial state of $2? The chain is infinite, but it's a simple form of infinite.

 

[MaxManus]

Thanks I see it now, states are (0,2,4,6,8,10). P = <br /> <br /> \begin{bmatrix} 1 &amp; 0 &amp; 0 &amp; 0 &amp; 0 &amp; 0 \\<br /> .5 &amp; 0 &amp; .5 &amp; 0 &amp; 0 &amp; 0 \\<br /> .5 &amp; 0 &amp; 0 &amp; 0 &amp; .5 &amp; 0 \\ 0 &amp; .5 &amp; 0 &amp; 0 &amp; 0 &amp; .5 \\ 0 &amp; 0 &amp; 0 &amp; .5 &amp; 0 &amp; .5 \\ 0 &amp; 0 &amp; 0 &amp; 0 &amp; 0 &amp; 1 \end{bmatrix}


First part: Find the probability that the gambler with initial capital 2$ will achieve his aim .

But I'm not sure what to do now.

 

[lanedance]

well, if you start out in the state P($2=1)=(0,1,0,0,0,0) and contunually play the game this is equivalent to repeated multiplication by P so what is the limit of P^n?

 

[MaxManus]

Thanks, but it seems like I have the wrong P . My P^n converges to [1 0 0 0 0 0;.8 0 0 0 0 .2;.6 0 0 0 0 .4;.4 0 0 0 0 .6;.2 0 0 0 0.8;0 0 0 0 0 1]
where space is next column and ; is next row. Matlab.

Edtit: Sorry, it seems to converge to a the right values. You either end up with 0 or 15 and the more money you start with the more likely you end up with 15.

Thanks againg

 

[MaxManus]

I got P from

when you have 0 you have 0 in the next period
when you have 2 you either have 0 or 4 in the next period
when you have 4 you either have 0 or 8 in the next period
when you have 8 you either have 6 or 10 in the next period
when you have 6 you either have 2 or 10 in the next period.
when you have 10 you have 10 in the next period.

 

[lanedance]

i think your P is ok, what do you mean by 15?