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The problem is as follows:

Let X be the winnings of a gambler. Let $p(i)=P(X=i)$ and suppose that
$p(0)=1/3;\\ p(1)=p(-1)=13/55;\\p(2)=p(-2)=1/11;\\p(3)=p(-3)=1/165$.
Compute the conditional probability that the gambler wins $i$, $i=1,2,3$ given that he wins a positive amount.

I am not looking for a solution, but just be pointed in the right direction to begin solving this question.

I don't think this has been posted before, at least I was unable to find this question. If it has, I apologise.

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    $\begingroup$ $Pr(A\mid B) = \frac{Pr(A\cap B)}{Pr(B)}$. Let event $B$ be that he wins a positive amount. What is the probability that he wins a positive amount? What is the probability that he wins an amount of $i$ where $i$ is positive? $\endgroup$ – JMoravitz Jul 23 '16 at 21:22
  • $\begingroup$ Would $B$ just be the sum of the probabilities associated with a positive $i$ and then the conditional probability for each $i$ would be the probability of that $i$ divided by what the sum of $B$? $\endgroup$ – Gearboxx Jul 23 '16 at 21:33
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    $\begingroup$ $B$ is an event, and is the union of the events associated with positive $i$. The probability $Pr(B)$ is the sum of the probabilities associated with events with positive $i$, otherwise yes, so for example $Pr(X=1\mid X>0) = \frac{13/55}{13/55 + 1/11 + 1/165}$ $\endgroup$ – JMoravitz Jul 23 '16 at 21:39
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Since you seem to have figured it out from the prod in the comments earlier, here is a fuller explanation.

Here we note two important things. First, the definition of conditional probability: the probability of event $A$ occurring given that we know that event $B$ occurs is

$$Pr(A\mid B) = \frac{Pr(A\cap B)}{Pr(B)}$$

Second, we remember one of the axioms of probability:

$$\text{If}~E\cap F = \emptyset~~\text{then}~~Pr(E\cup F) = Pr(E)+Pr(F)$$

In this problem, our sample space can be partitioned into the possible outcomes of the amount won by the gambler. We can describe it as $\Omega = \{-3,-2,-1,0,1,2,3\}$ with $Pr(X=i)$ given in the problem statement.

Letting $B$ be the event that the outcome is positive, we recognize that $B=\{1,2,3\}=\{1\}\cup \{2\}\cup \{3\}$ where the final representation is as a union of disjoint sets, allowing us to use the aforementioned axiom.

Thus, $Pr(B)=Pr(\{1\}\cup\{2\}\cup \{3\}) = Pr(\{1\})+Pr(\{2\})+Pr(\{3\})=\frac{13}{55}+\frac{1}{11}+\frac{1}{165} = \frac{1}{3}$

Now, for each of the outcomes in question, $Pr(X=1\mid X>0) = Pr(\{1\}\mid \{1,2,3\}) = \frac{Pr(\{1\}\cap \{1,2,3\})}{Pr(\{1,2,3\})} = \frac{Pr(\{1\})}{Pr(\{1,2,3\})}$ since $\{1\}\subseteq \{1,2,3\}$. Plugging in the values given in the question gives $Pr(X=1\mid X>0) = \frac{13/55}{1/3} = \frac{39}{55}$

Similarly one can plug in the remaining values for the other outcomes to find those probabilities.

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Intuitively, you're looking for the probability that you win a certain positive amount, say 1, given that you win a positive amount.

We are given that

$$p(1)=13/55$$

if you pick 1 from $\{-3,-2,-1,0,1,2,3\}$.

What if we do away with nonpositive values?

It's like asking for the probability you pick 1 from $\{1,2,3\}$.

Then we have:

$$p(1 | 1,2,3)=\frac{13/55}{13/55+1/11+1/165}$$

Similarly, we have

$$\\p(2 | 1,2,3)=\frac{1/11}{13/55+1/11+1/165};\\p(3 | 1,2,3)=\frac{1/165}{13/55+1/11+1/165}$$

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