tsmid, here's a simple analogue so you can try it yourself using a deck of cards and a piece of paper (unless, of course, you actually have some pool balls available and can do the original experiment): Take out the Ace of Spades (rock on!) and the Ace of Hearts (and any jokers), and place the Ace of Hearts somewhere safe. Keep the Ace of Spades. The experiment then goes as follows. 1. Shuffle the remaining deck well, and deal off one card without looking at it. You now have two cards. One of them is known to be black (the Ace of Spades), the other is pretty much randomly chosen to be either black or red with equal probability. 2. You now shuffle those two cards, and deal the first card. 3a. If this first card is a red suit, you don't count this particular experiment towards determining the probabilities empirically, because this case falls outside the stated problem, which was that the first ball pulled out of the bag was solid = the first card dealt is black. 3b. If the card is a black suit (not necessarily the Ace of Spades, but just any black card), you deal the second card, and tick off "Second Ball Solid" if this second card is black, "Second Ball Striped" if it is red. 4. Repeat a good number of number of times to decrease variance. 5. Report your findings back here. I urge you to try this. The (rock solid) math is, for whatever reason, unconvincing to you, but in this case the math relates to a real world problem that we can actually try. Maybe experiment will convince you.
This caveat isn't necessary (though it doesn't hurt). The solid balls can easily be distinguishable (say, pool balls nr. 1 and 8). The requirement merely says that the first ball is solid. Let's be explicit: The eight-ball is always in the bag and ball nr. 1 is added half the time, ball nr. 15 is added the other half. Let's now draw the two balls from the bag: Case nr. 1: The eight-ball is drawn first. Ball nr. 1 is drawn second. This happens 1/4 of the time. Case nr. 2: The eight-ball is drawn first. Ball nr. 15 is drawn second. This happens 1/4 of the time. Case nr. 3: Ball nr. 1 is drawn first. The eight-ball is drawn second. This happens 1/4 of the time. Case nr. 4: Ball nr. 15 is drawn first. The eight-ball is drawn second. This happens 1/4 of the time. The question is now as follows: Given that case 1, 2 or 3, happened, what is the probability that it was case 2?
There is no case 3, there is only a case 1 (the random ball is a solid ball) and a case 2 (the random ball is a striped ball). These are the only cases that need to be distinguished. Everything else is fixed by the conditions. Dinosaur's sampling argument 30 times a solid ball was transferred, and 30 times you drew a solid ball from the bag. 30 times a striped ball was transferred. In 15 of the 30 cases, you expect to draw a solid ball the second time. In 15 cases, you expect to draw a striped ball. is in fact incorrect. In the latter case (30 striped balls are transferred), it is not possible to draw a solid ball at the second draw as it has already been removed in the first draw by definition. So in the latter case you will draw a striped ball as the second ball also 30 out of 30 times, so overall the probability for the second ball being striped/solid is 50/50. Thomas
Case number 4 is disallowed by the condition that a solid has to be drawn first (the task specifically requires "A ball is now selected randomly from the bag and it turns out to be a solid"). This must hold for all cases considered. So you have to put the nr.15 back and repeat the first draw until you draw the nr.8 first and the nr.15 second. You then have twice the nr.15 drawn second and once the nrs. 1 and 8, so a 50/50 split between solid and striped (see also my reply to Przyk above). Thomas
If event A is "the random ball is a solid" and event B is "the random ball is a stripe". The information that event A is as likely as event B before the first ball is drawn does not imply that event Y = "the random ball is a solid and a solid was drawn first" and event Z = "the random ball is a stripe and a solid ball was drawn first" are equally likely. The core of your argument is a non-sequitur. It is possible for the first ball drawn to be a stripe. We are just told that on the particular occasion we are considering, this did not occur. Conditional probability is all about using information obtained during a trial to re-assess the probabilities of possible events. Examples like this problem are often presented as introductory examples to this branch of probability theory.
You are changing the rules here - this was not done in the original problem. No attempt was made to force the first ball drawn to be solid.
Exactly. That's what "Given that case 1,2 or 3 happened" means. Absolutely not. That's in no way what the setup says. That's simply wrong.
The rule is "A ball is now selected randomly from the bag and it turns out to be a solid". This clearly does not allow you to draw a striped ball first and then simply ignore the draw (as you are doing). So the only way to obtain the solid ball in this case by means of a random selection is to put the striped ball back and try again (which unlike your method is not against the rules). Thomas
Yes. "A ball is now selected randomly from the bag and turns out to be a solid" and not "steps are taken to ensure that the first ball removed from the bag is solid". No. Contrary to your claims, this does alter the original rules, as well as distorting the results. You can repeat the trial as stated in the original problem, but not individual operations within each trial. The first ball drawn "turns out" to be a solid. It wasn't known that this would happen until after the draw, and a repeat of the experiment would not guarantee the same result. This is why the results that do not meet the "first drawn turns out to be solid" criterion are discarded - it's the only way of getting an experimental answer to the problem without interfering with the trial itself. Also your criticism against discarding the results that don't apply is baseless and contradicts the way statistical surveys are conducted in real life. If you want to find the proportion of apples that are red, you explicitly disregard any available information about oranges. It's exactly the same with this problem. We want to find the proportion of stripes picked on the second draw where the first draw (again to quote the original statement of the problem) "turns out" to be a solid, so we perform multiple trials, then focus on the results where the first ball turned out to be solid. Not every investigation can be planned in such a way that will ensure only relevant results are obtained.
The question is, a bag has a solid pool ball in it, and another ball is added, equally likely to be a stripe or a solid. What is the probability of the second ball drawn a stripe given that the first ball drawn is a solid? That's it. The problem doesn't consider the number of solid and striped pool balls that come with a pool table (8 solid, 7 striped, 1 cue ball). It also doesn't factor in bias from the people doing the test. It is just asking about this scenario. The first ball is a solid.
this is wrong the scenario is EXACTLY as follows start here> you put a solid in the bag you gather 20 ball 1/2 solid, 1/2 striped mix them up draw one ball and put it in the bag with the known solid now draw one ball out of the bag. if the ball is striped empty the bag and go back to start here now draw the second ball. the above is the procedure for detirmining the outcome of the problem stated in the first post. quit a few posters says there is a 1/3 chance of drawing a stripe on the second draw. although pete has tried to explain it i still fail to see how 1/3 is arrived at unless there are 3 possibilities (2 solid and 1 stripe)
Hi leopold, You have to be careful with statements like this. This implies a rule that forces the first ball drawn to be solid (ie the results of the first draw are known before the first draw is made). If this is the case, eg. if the problem were stated a little differently like this: "An assistant looks in the bag and deliberately (ie. not randomly) picks a solid ball. What is the probability that the second ball in the bag is a stripe?" the answer would be 1/2. This is an easier problem than the one Pi-Sudoku originally posted.
the first ball MUST be solid, the op states that. if the first ball is stripe then the results are void
I am serious, and that's why I am applying the given constraint strictly, which is that the first draw is 100% biased towards the solid ball. This is evidently different from having two unbiased draws and then biasing the result towards the solid ball. Apart from the repeat method, there is actually a further way to satisfy the constraint, namely if you simply assume that, due to some freak coincidence, you draw only solid balls with the first draw in the first place. This may be an unlikely scenario, but it is the only scenario that satisfies the constraint, and if you evaluate the second draws in this case, you will find that they are distributed 50/50 between solid and striped. Thomas
You don't understand the concept of conditional probabilities. You are the one biasing the results. The odds are 1/3 that the second ball drawn from the bag is striped. The problem can be recast to eliminate the conditional probability: A con man puts a striped ball and a solid ball in a hat. He asks you to pick a ball at random without looking at it and transfer it to an empty bag. He then adds a solid ball to the bag. Now he offers you a bet: he will take a solid ball out of the bag. You get the remaining ball. He asks you what the odds are that your ball will be striped. You say fifty-fifty. He says, I bet that it is solid. I'll even give you 60-40 odds: If the ball is striped, I'll pay you $60, but if it is solid, you only pay me $40. Will you take the bet?
