1

There is an urn with 20 red (R) and 10 black (B) balls. The first ball is drawn. If it is black,one wins and the game stops. Otherwise, a second draw is made (without reinsertion of the first ball). If it is black, you win and the game stops. Otherwise,you continue with a third and final draw (without reinsertion of the firstand two balls). If it is black, you win; otherwise you lose. Construct a probability space to describe the game.

My attempt: $$ \Omega=\{N,RN,RRN,RRR\} $$

Regarding the $\sigma$-algebra, since the text only says:

Construct a probability space to describe the game.

i have assumed to be interested in finding the probability of E

$$ E=``Winning \ the \ game''=\{N,RN,RRN\}$$ Because the $\sigma$-algebra is the family of events to which we want to assing probabilities then a possible $\sigma$-algebra is $$\mathcal{F}=\{\emptyset,E,\overline{E},\Omega\}$$

As per my understanding now I must come up with a probability distribution which assigns a probability to all the elements in the $\sigma$-algebra in order to construct a probability space$(\Omega,\mathcal{F},\mathbb{P})$.

I'm stuck thinking how to come up with a probability distribution.

Giuseppe
  • 55
  • The rules are not clear. Is "one" different from "you"? If there is only one player, then what is it we are computing the probability of? Did you just mean to ask for the probability distribution of the first black draw? If so, just say that. – lulu Sep 13 '23 at 12:55
  • This is a quintessential example of when you might use a probability tree diagram. Winning on the first turn happens with probability $\frac{10}{30}$. Winning on the second turn happens by pulling a red ball followed by a black with probability $\frac{20}{30}\times\frac{10}{29}$ and so on... – JMoravitz Sep 13 '23 at 13:01
  • It seems to me that you may be spending too much time overthinking the sample space and $\sigma$-algebra portions of the problem. – JMoravitz Sep 13 '23 at 13:03
  • @JMoravitz "Winning on the first turn happens with probability $10/30$" but you are using the naive definition of probability which assumes that elementary outcomes are equally like and in this case given the above sample space I don't think so. Am I right? – Giuseppe Sep 13 '23 at 13:19
  • It should be assumed that each of the thirty balls in the bag are equally likely to be drawn in the first round, and if continuing on to later rounds, each of the remaining balls are equally likely to be drawn. Without that assumption then there is of course not enough information to solve... but this assumption is so incredibly common for problems of this nature that authors often neglect to include it. Should it have been explicitly stated? Sure, but it doesn't need to be. – JMoravitz Sep 13 '23 at 13:22
  • @JMoravitz At first I was tempted to use a probability tree to solve this exercise (as you suggested) but if the exercise clearly states that I must come up with a $(\Omega,\mathcal{F},\mathbb{P})$ then given that I've chose "$\Omega={N,RN,RRN,RRR}$" then the elemntary outcomes aren't equally like – Giuseppe Sep 13 '23 at 13:29
  • The outcomes of N, RN, RRN... no, of course those aren't equally likely but that doesn't stop the outcomes of the balls for each step from being equally likely. There are many choices of sample spaces. Some are more convenient than others. – JMoravitz Sep 13 '23 at 13:30
  • 1
    Again... if you were to talk about the possibility of an extreme scenario where the red balls are sticky or are all at the top of the urn or are much larger and influence you to be more likely to select them, then all bets are off and there is no way to answer this problem due to lack of information... but the traditional interpretation of this problem is that the balls in the urn are "mixed" and otherwise indistinguishable until pulled out, and that each of the 30 balls individually are equally likely to be drawn. This is a standard problem type and this is a standard assumption to make. – JMoravitz Sep 13 '23 at 13:34
  • @JMoravitz You make me realize that my problem was to think that, given an experiment, there is only one possible $\Omega$ and I wrote immediately the one which seemed more intuitive (given the statement of the problem.). So, there are many sample spaces that we can choose for the same experiment but we tend to choose ,if it's posiible, the ones which allow us to consider the elementary outcomes equally like? – Giuseppe Sep 13 '23 at 13:44
  • 1
    @Giuseppe https://math.stackexchange.com/questions/3321361/why-is-flipping-a-head-then-a-tail-a-different-outcome-than-flipping-a-tail-then/3321368#3321368 – JMoravitz Sep 13 '23 at 13:52
  • Let us continue this discussion in chat. – Giuseppe Sep 13 '23 at 14:34

1 Answers1

0

For the purposes of probability computation, it is better to have a sample space where every outcome is equally likely. This means you can compute probabilities by counting the number of favorable outcomes, and dividing by the number of total outcomes. Therefore, I suggest using $$ \Omega=\{(b_1,b_2,b_3)\mid b_1,b_2,b_3\in \{1,\dots,30\}, b_1\neq b_2\neq b_3\neq b_1\} $$ The interpretation is that the balls are numbered $1$ to $30$, and each outcome describes the full sequence of $3$ balls drawn. This probability space has extra information, because it gives a full sequence of three balls even in the case where the experiment only demands drawing one ball. This is fine to do, and is a common strategy. For example, if you want a probability space to describe a five-card unordered poker hand from a shuffled deck, it is perfectly valid to let $\Omega$ be the set of $52!$ ways to shuffle the deck, knowing that you will only pay attention to the first $5$ cards of the shuffled deck, and ignoring their order.

Let's say you want to calculate the probability of winning. We need to specify which of the 30 balls are black; I will arbitrarily say that balls numbered $1$ to $10$ are black, and the remaining balls are red. Let $E$ be the event that you win the game. Then $$ E=\{(b_1,b_2,b_3)\in \Omega\mid \exists i \;b_i\in \{1,\dots,10\}\} $$ That is, as long as one of the first three balls you draw is black you will surely win.

It is easier to find the complementary probability that you lose the game, and subtract from $1$. The set theoretic description of the complementary event is $$ E^c=\{(b_1,b_2,b_3)\in \Omega\mid b_1,b_2,b_3\in \{11,\dots,30\}\} $$

It is then a routine combinatorial problem to show that $|\Omega|=30\cdot 29\cdot 28$ and $|E^c|=20\cdot 19\cdot 18$, so we conclude $$P(|E|)=1-P(E^c)=1-|E^c|/|\Omega|=1-20\cdot 19\cdot 18/(30\cdot 29\cdot 28).$$

Mike Earnest
  • 75,930