[SFW] Riddles / Puzzles / Brain Teasers - Page 21

think of how you got the first orange ball. You might pick the white ball from white-white or a white ball from orange-white. You keep going until you pick an orange ball. You can see how you will be more likely to pick the orange ball from orange-orange, right? Since the first orange ball is more likely to be from orange-orange, you have a higher chance of picking a second orange ball

It is all in how you interpret the question.

Imagine there are 2 jars, jar 1 has 1000 orange marbles and jar 2 has 10 orange marbles and 990 white marbles.

You close your eyes and pick a random marble from a random jar. When you open your eyes, you have an orange marble in your hand. Which jar do you think you reached into? (~99% chance you have a marble from jar 1) If you pick another random marble from the same jar, what is the chance that it is also orange? (~99%)

Now imagine that you pick a random jar, look into it and pull out an orange marble. Which jar did you most likely choose? (50% chance for either jar) If you pick a random marble from the same jar what is the probability that it is orange? (since you are equally likely to be in the orange jar as you are in the white jar, the chance of pulling an orange marble is about 50%).

there are 12 face up cards in the deck. take 12 cards out and flip it all of em over so the face up ones are now face down and vice versa. for example, if i pick out 4 face up cards in the 12 cards i take out, 8 cards will be face up after i flip em all over. because there are 8 face up cards left in the remaining 40 cards, it should be equal in both piles.

read through the tons of spoiler tags and somebody got it before me sadface

if you want to know if ur answer is correct, you can type it in here: http://diaknaposfeladat.hostei.com/osszekot/ if tha page changes to a new one, u ve done it (captcha required)

Split all sequences of reals into equivalence classes, where the relation is "have the same tail." That is, (a_1 a_2 a_3 ...) ~ (b_1 b_2 b_3 ... ) if there exists N > 0 such that for all n > N, a_n = b_n.

I didn't get this explanation. If the sports bag had 100 pockets, 98 of which had two white balls, 1 of which had an orange ball and a white ball, and 1 of which had two orange balls, and you reach into a pocket and pull out an orange ball- does the problem change in the least bit? Clearly not- the pockets with no orange balls are irrelevant once we know that the pocket that was picked has at least 1 orange ball.

In order to answer "What's the probability of the other ball in the pocket being orange?", the real question that must be answered is, "What is the probability that the pocket with two orange balls was chosen, given that an orange ball was picked at random from a random pocket?"

Let the event of choosing the pocket with 2 orange balls be known as X, the event of choosing the pocket with an orange ball and a white ball be known as Y, the event of picking out an orange ball be known as A, and the event of picking out a white ball be known as B.

P(X|A) = P(X∩A)/P(A)

If we only consider pockets X and Y,

P(X∩A) = 1/2, P(A) = 3/4

Therefore, P(X|A) = P(X∩A)/P(A) = 2/3

So the probability of the other ball in the pocket being orange is, in fact, 2/3.

once you KNOW you drew an orange ball, the math changes.

it does not matter which orange ball you have drawn. the question says only "you drew an orange ball."

so, if you drew an orange ball in the O-O pocket, it's a 100% chance to draw another orange. if you drew an orange ball in the O-W pocket, it's a 0% chance to draw another orange. you cannot ever draw an orange ball in the W-W pocket, so this is discounted for math purposes.

because you have already given the outcome of event #1, you don't multiply two events for probability. it's only one event, and in the given situation, only two outcomes: O-O or O-W pocket. 50/50.

this is wrong because it is NOT 1/2 to get the first orange ball. it is 1/1. 100%. the question has given it already. you do not multiply here for two events because there are not two events. there is only one event.

You're wrong. As I said, the question that you really want to answer is "What is the probability that the pocket with two orange balls was chosen, given that an orange ball was picked at random from a random pocket?"

Yes, there are only two possible outcomes. But those two outcomes aren't equally likely. Put another way: If you were to play a game of BW with Flash and the only possible outcomes would be for you to win or for you to lose... what is the probability that you will win? There are only two possible outcomes, but they are not equally likely.

In order to determine how likely each possibility is, I used conditional probabilities, and showed that the chance that the pocket with 2 orange balls was chosen is 2/3.

do you understand that the fact something was randomly selected is irrelevant to probabilistic calculations IF the question presumes that an outcome has already been determined? the question in the riddle is only concerned with the probability involved with drawing ball 2, given some predetermined guaranteed information.

Sorry, but your interlocutors are correct. Think of it this way:

Suppose you do 60 trials and the results accord perfectly with the probabilities. Then, you should get 30 orange balls and 30 white balls. 1/3 of the time, you will reach into the double orange pocket, and every time you do so you will get an orange, resulting in 20 orange balls from this pocket. 1/3 of the time you will reach into the one-orange one-white pocket and half of that time you will get an orange, resulting in 10 orange balls from that pocket.

When you pull out an orange ball, you learn that you are in one of the 30 trials where an orange ball was pulled out. 2/3 of these are trials in which the double orange pocket was drawn from, so that's the probability that the other is orange.

edit: here's an analogy. There's a competition between a masters player and DRG. There is 1/3 chance that DRG will automatically get the win, 1/3 that the masters player will automatically get the win, and 1/3 chance that they will play a match of SC2 to determine the winner. If you find out that the masters player won, what's the probability that the middle option was chosen? According to your method, it is 50%, because that it is one of the two ways that the masters player can win.

However, this is obviously false. If you find out that the masters player won, you have a very good reason to believe that the middle option was chosen because the odds of winning against DRG in a match are so low. This is perfectly analogous to the present examples. There are two ways to win (i.e., get the orange ball) with an equal probability of occurring, but one of the ways delivers the winning result a higher percentage of the time than the other. In such scenarios, learning that a win has occurred makes it likely that the higher percentage of winning way happened.

Ok, new question, just for you: There are 2 pockets. Pocket A has 100 orange marbles, whereas pocket B has 99 white marbles and 1 orange marble. You draw a random marble from a random pocket. Question 1: What is the probability that you drew the marble from pocket A? Question 2: If the marble you drew is orange, then what is the probability that you drew it from pocket A?

The answer to Question 1 is 1/2; there are two, equally likely answers. The answer to Question 2 is greater than 1/2, because the fact that you drew an orange marble means that you either a) You drew a marble from pocket A, or b) You drew a marble from pocket B AND out of the 100 marbles, you happened to pick the only orange one. In this case, the contents of the container give us additional information about how likely it is that we picked one container or the other.

Further proof: Let's go back to the original problem of 3 pockets and the ping pong balls. Let's say that Pocket X has 2 orange balls, Pocket Y has 1 orange ball and 1 white ball, and pocket Z has 2 white balls:

If we pick a random ball from a random pocket, then there's a 50% chance that we'll pick a white ball, and a 50% chance that we'll pick an orange ball. If what you say is true (Probability of second ball being orange given that first ball is orange = 1/2), then the following is also true:

If first ball is orange (50% chance), then the probability of having picked pocket X = 1/2, probability of having picked pocket Y = 1/2.
If first ball is white (50% chance), then the probability of having picked pocket Z = 1/2, probability of having picked pocket Y = 1/2.

Therefore, if we pick a random ball from a random pocket,
Total probability of having picked a ball from pocket X = (1/2)*(1/2) = 25%
Total probability of having picked a ball from pocket Y = (1/2)*(1/2)+(1/2)*(1/2) = 50%
Total probability of having picked a ball from pocket Z = (1/2)*(1/2) = 25%

The above makes no sense, there's 3 pockets and there should be a probability of 1/3 or having picked any one of them (if you repeat the exercise assuming that the probability of second ball being orange given that the first ball is orange = 2/3, then the numbers check out correctly)