1
$\begingroup$

There's a gambler with initial money $k=100\$$. He plays till bankruptcy or till having $N=500\$$. In every game he wins $100\$$ with probability $p=\frac{1}{2}$, loses $100\$$ with probability $q=\frac{1}{4}$ or the game ends in tie with probability $r=\frac{1}{4}$.

(a) Calculate the probability of racking up 500\$,

(b) Calculate the mean time of the game.

I've seen exercises with gamblers ruin, with $p$ and $q=1-p$, but in my case there's also $r$. Does it change much, or do I just "forget" about the r and calculate the ruin of the gambler? I'm obviously referring to "Unfair coin flipping" -> https://en.wikipedia.org/wiki/Gambler%27s_ruin. There is a formula, and then I just have to solve a differential equation.

According to (b) I have no idea how should I calculate the above.

Please help with both sub-points. Any help will be much appreciated.

$\endgroup$
  • 3
    $\begingroup$ Notice that probability of winning = $\frac{\textrm{probability of winning}}{\textrm{probability of not tieing}}=\frac23$, and likewise, probability of losing is $\frac13$. This reduces your problem to a typical gambler's ruin problem. $\endgroup$ – Don Thousand Oct 27 '18 at 13:52
  • 1
    $\begingroup$ A good, general, way to approach questions like this is to work with states. Let $\psi_i$ be the probability of getting to $5$ given that you start with $i$, so the answer you want is $\psi_1$. Of course $\psi_0=0,\psi_5=1$. For the other $i$ we have $\psi_i=\frac 12\times \psi_{i+1}+\frac 14\times \psi_{i-1}+\frac 14\times \psi_i$ and you can just solve that system. A similar calculation works for the expected time. $\endgroup$ – lulu Oct 27 '18 at 14:30
  • $\begingroup$ For (b), the probability of a win or lose game is $\frac34$ and of a tied game $\frac14$, i.e. in the ratio $3:1$. So if you can calculate the expected number of win or lose games, then the expected total games is $\frac43$ of this $\endgroup$ – Henry Oct 27 '18 at 14:49
1
$\begingroup$

Lulu has given you a method for the probabilities of reaching $500$ before reaching $0$: dividing the values by $100$ to give you labels, you want to find the solution to $$\psi_i=\tfrac 12 \psi_{i+1} +\tfrac 14 \psi_i +\tfrac 14 \psi_{i-1} \\ \text{ i.e. } 2\psi_{i+1} -3 \psi_i + \psi_{i-1}=0$$ with the boundary conditions $\psi_0=0$ and $\psi_5=1$. In particular you are trying to find $\psi_{1}$. As Rushabh Mehta says, you would get the same answer ignoring ties by starting with $\psi_i=\frac 23 \psi_{i+1} +\frac 13 \psi_{i-1}$

For the expected time, you do something similar but this times have $$\tau_i=1+\frac 12 \tau_{i+1} +\frac 14 \tau_i +\tfrac 14 \tau_{i-1} \\ \text{ i.e. } 4+2\tau_{i+1} -3 \tau_i + \tau_{i-1}=0$$ with the boundary conditions $\tau_0=0$ and $\tau_5=0$, and in particular you are trying to find $\tau_{1}$. My comment was that if you ignore ties by changing this to $t_i=1+\frac 23 t_{i+1} +\frac 13 t_{i-1}$ then you would get $\tau_i = \frac43 t_i$

$\endgroup$
  • $\begingroup$ Shouldn't the initial value for (a) be $\phi_0=1$ and $\phi_N=0$, where $N=5$? Because the probability of gambler's ruin given that the player starts with $0$ money is $1$? Also I don't understand your boundary conditions for (b). $\endgroup$ – MacAbra Nov 1 '18 at 8:12
  • 1
    $\begingroup$ @MacAbra $\psi_0$ is the probability of winning the match when you have $\$0$, which is $0$ since you are bankrupt. $\psi_5$ is the probability of winning the match when you have $\$500$, which is $1$ since you have hit your target. $\tau_0$ is the expected number of future rounds when you have $\$0$, which is $0$ since you are bankrupt. $\tau_5$ is the expected number of future rounds when you have $\$500$, which is $0$ since you have hit your target. $\endgroup$ – Henry Nov 1 '18 at 8:19
  • $\begingroup$ Henry, thank you for the complete explanation. Now I'm sure that I understand the initial values. But I have one more question. What if gambler wins $200\$$ instead of $100\$$ with the same probability and loses $100\$$ with given probability? Obviously there's a chance for a tie too (just like in my exercise). $\endgroup$ – MacAbra Nov 1 '18 at 8:26
  • 1
    $\begingroup$ @MacAbra - If the gambler wins $200$ instead of $100$, then you get terms like $\psi_{i+2}$ and also the possibility of reaching $\psi_6$ $\endgroup$ – Henry Nov 1 '18 at 14:12

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.