A Problem my teacher set me

[tsmid]

We have the two bags in front of us, 1,000,000 balls in each, one all solid one 999,999 stripe and one solid, bags are closed and we can't tell which is which.

We flip a coin and pick one bag to pull balls out of. We pull out a solid ball. Here's the wager. If the next ball pulled out of this bag is striped, I will give you $1000, if it is solid, you give me $1. Willing to play?

You have apparently still not understood it: you obviously have to flip the coin again to determine the bag for the second draw. Still willing to play then?

Thomas

 

[shmoe]

Count me in, I would love to join too.

The bouncers at my casino have instructions to not let you in, sorry

 

[shmoe]

You have apparently still not understood it: you obviously have to flip the coin again to determine the bag for the second draw. Still willing to play then?

Thomas

No flipping to determine the second bag. We're drawing from the same bag as the first ball as per pryzk's game.

 

[tsmid]

They are NOT equally likely. Before you draw anything, each column has is equally likely, i.e. 1/4 chance of occuring. After the first draw, we are given more information, we know we are in the first 3 columns. Each *column* is still equally likely, i.e. we have a 1/3 of being in each. It's equivalent to just drawing 2 balls a bunch of times and throwing away any of the draws that give a stripe first, what proportion of the remaining draws would have a stripe in them? Seruiously, get some balls and try it.

If you are averaging over the independent variables (Second ball added is solid/striped) then these are equally likely by definition.
Pete in his table has averaged over the final possibilities and here you have to give the red field twice the weight exactly because of the conditional probability. So the table effectively should look more like this

<table border="1" cellspacing="0" cellpadding="4"><tr><td colspan="2"><p align="center">Second ball added is solid</td><td width="63" style="border-top: none; border-bottom: none;" rowspan="5">&nbsp;</td><td colspan="2"><p align="center">Second ball added is stripe</td></tr><tr><td colspan="2" valign="top"><p align="center">Ball drawn 1^st</p></td><td colspan="2" valign="top"><p align="center">Ball drawn 1^st</p></td></tr><tr><td><p align="center">Ball 1</p></td><td><p align="center">Ball 2</p></td><td><p align="center">Ball 1</p></td><td><p align="center">Ball 1</p></td></tr><tr><td style="background: #FFFFA0">Solid,Solid</td><td style="background: #FFFFA0">Solid,Solid</td><td style="background: #FFA0A0">Solid,Stripe</td><td style="background: #FFA0A0">Solid, Stripe</td></tr></table>

If you now calculate the probability from the final columns, you also get 50% probability for a striped ball.

But as should be evident from what I said above, you don't actually have to bother about the conditional probabilities. It is sufficient here merely to consider the initial probabilities for putting the second ball in.

Thomas

 

[tsmid]

No flipping to determine the second bag. We're drawing from the same bag as the first ball as per pryzk's game.

That wasn't przyk's game nor the one related to the original problem. If you use two bags then you have to flip the coin again, otherwise you are playing a different game.

Thomas

 

[shmoe]

That wasn't przyk's game nor the one related to the original problem. If you use two bags then you have to flip the coin again, otherwise you are playing a different game.

Thomas

Incorrect. The two bags was just a way to include the "50% chance of a bag with 1 million solids and 50% chance of 999,999 stripes and one solid". In both pryzk's and the original problem, the bag was loaded and fixed then you started drawing. You pulled both balls out of the same bag, it wasn't altered after your first draw.

If you prefer, forget two bags we'll just have pryzk come over (I'll pay him 50 cents for his services) with his bag that he randomly filled. Note a key feature implicit in the setup of this game- if the first ball drawn is a stripe, we don't play. We send him out to reload. We keep doing this until we get a solid on the first go, then we wager. This is exactly the setup of the original problem (well more balls of course), we are only playing given that the first ball was a solid.

 

[shmoe]

So the table effectively should look more like this

You have the same event in the last two columns, you don't give it more weight because of 'conditional probability'. If you think otherwise, lets see some math.

 

[leopold]

so you can completely take it out of the equation.

putting a solid in and drawing a solid out is no guarentee it's the same ball, it could have been the random ball.

 

[Pete]

Here we go...

All we need now is to start a thread on the "other child" probability problem, and invite Eldon Moritz along.

 

[LaidBack]

My maths teacher set this problem the other day:

(Just clarifying that everyone knows how pool balls work, they are either stripey or solid colour)

A bag contains one pool ball that is known to be a solid, a second ball is chosen in such a way that it has equal chance of being a solid or a stripe, this ball is then placed in the bag.

A ball is now selected randomly from the bag and it turns out to be a solid.

If somebody now selects a ball from the bag what is the probability of it being a stripe?


I worked out an answer and i was told that it was wrong. I will post my answer shortly so as not to influence your decisions.

Please could people have a go at this and post their reasoning.

If the striped ball is the only ball left, it has a 100% chance of being plucked out of the bag.

 

[Pete]

That's right, LaidBack. But there might be another solid ball in the bag, not a striped one.

 

[tsmid]

Incorrect. The two bags was just a way to include the "50% chance of a bag with 1 million solids and 50% chance of 999,999 stripes and one solid". In both pryzk's and the original problem, the bag was loaded and fixed then you started drawing. You pulled both balls out of the same bag, it wasn't altered after your first draw.

Drawing the second ball from the same bag as the first ball is different from the problem discussed here because the second draw is supposed to operate also on all balls available.

Just put the 1,000,001 solid balls and 999,999 striped balls in one bag, pull out two solid balls (two because the two sets have been added together), and then ask again what the probability is to draw a striped ball at the next attempt.

Thomas

 

[tsmid]

putting a solid in and drawing a solid out is no guarentee it's the same ball, it could have been the random ball.

That's irrelevant. The balls are indistiguishable and you can replace one with the other.

Let's go through it step by step again, and to make it even clearer, assume that you add the random ball first to the bag and then the solid:

case 1: The random ball is a solid ball which you put it in the bag. You add the second solid ball, then take a solid ball out, which leaves a solid ball (whichever of the two balls you take). The result is thus the same as if you had not put the second solid ball in at all.

case 2: The random ball is a striped ball which you put it in the bag. You add the solid ball, then take a solid ball out, which leaves the striped ball. Again, the result is the same as if you had not put the solid ball in at all.

It is thus obvious that the probability for the second draw is simply given by the probability for the random ball, which is supposed to be 50%/50%.

If this is still unconvincing to anybody, then I really can't help it.

Thomas

 

[leopold]

Just put the 1000001 solid balls and 999,999 striped balls in one bag, pull out one solid ball, and then ask again what the probability is to draw a striped ball at the next attempt.

Thomas

incorrect.
if you put 1,000,001 solid balls into the bag then you must choose 999,999 balls from an equal mixture of solid/stripe.
i don't think this is the same as the OP.

 

[tsmid]

incorrect.
if you put 1,000,001 solid balls into the bag then you must choose 999,999 balls from an equal mixture of solid/stripe.
i don't think this is the same as the OP.

The 999,999 balls are not a mixture, they are striped.

Still, I made a mistake in the post you quoted when I said that 1 solid ball should be pulled out of the bag first. This would obviously leave 1,000,000 solids and 999,999 stripes, which does not quite leave a 1/2 chance. Because you added the two sets together, you have actually to pull out 2 solids first, leaving 999,999 each.

Thomas

 

[przyk]

Drawing the second ball from the same bag as the first ball is different from the problem discussed here because the second draw is supposed to operate also on all balls available.

It does. The balls available are the ones in the bag selected prior to the first ball being drawn. Whether there are two bags available of known contents and one is picked randomly for use, or only one randomly filled one is provided doesn't really change the problem, hence shmoe's suggestion:

If you prefer, forget two bags we'll just have pryzk come over (I'll pay him 75 cents for his services) with his bag that he randomly filled. Note a key feature implicit in the setup of this game- if the first ball drawn is a stripe, we don't play. We send him out to reload. We keep doing this until we get a solid on the first go, then we wager. This is exactly the setup of the original problem (well more balls of course), we are only playing given that the first ball was a solid.

Just put the 1,000,001 solid balls and 999,999 striped balls in one bag, pull out two solid balls (two because the two sets have been added together), and then ask again what the probability is to draw a striped ball at the next attempt.

1/2, not that that really tells you anything. You can't substitute a problem for an unrelated one like this and expect things to magically work out. I you "add the two sets together" you completely change the rules of the problem. I wouldn't even call it "scaling up".

If you still have doubts about the textbook solution for this problem, think about how you might go about finding the answer experimentally.

About your earlier response:

This assumes statistically independent draws. If the condition is that 2 red balls are being picked together, the probability is still 1/2.

Showing that you can pick an interpretation of "scaling up" that doesn't change the answer only further undermines your position: "scaling up" is not well defined.

Alright, take the problem:
"What is the probability of drawing 2 red balls from a bag containing 2 red and 2 blue balls, assuming each ball drawn is returned to the bag before the next one is drawn?" (Ans: 1/4)
and "scale it up" by a factor of 2:
"What is the probability of drawing 4 red balls from a bag containing 4 red and 4 blue balls, assuming each ball drawn is returned to the bag before the next one is drawn?" (Ans: 1/16)

The rules are the same in both cases; only the quantities involved have been doubled. The results are not the same.

 

[przyk]

That's irrelevant. The balls are indistiguishable and you can replace one with the other.

They're distinct - which is significant. The problem could be rephrased in a way that makes this more clear:

The bag initially contains a ball that is known to be black. A second ball is added to this bag. There's a 50% chance that this second ball is white, and a 50% chance that it is blue.

You then pick a ball randomly out of the bag. It looks black, but the problem is, you have an eye problem and can't distinguish between black and blue. So, given that the first ball removed is either black or blue, what is the probability that the second picked is white?

Let's go through it step by step again, and to make it even clearer, assume that you add the random ball first to the bag and then the solid:

case 1: The random ball is a solid ball which you put it in the bag. You add the second solid ball, then take a solid ball out, which leaves a solid ball (whichever of the two balls you take). The result is thus the same as if you had not put the second solid ball in at all.

case 2: The random ball is a striped ball which you put it in the bag. You add the solid ball, then take a solid ball out, which leaves the striped ball. Again, the result is the same as if you had not put the solid ball in at all.

It is thus obvious that the probability for the second draw is simply given by the probability for the random ball, which is supposed to be 50%/50%.

No, it's the probability that we're in case 2, given that we're either in case 1 or in case 2, that we want. Before the draw case 1 is twice as likely as case 2, and this still holds if the first draw rules out case 3 (the random ball is striped, and you take the striped ball out on your first draw). This would become obvious to you if you imagined performing this trial repeatedly, or simply followed Dinosaur's reasoning on page 1.

 

[shmoe]

Drawing the second ball from the same bag as the first ball is different from the problem discussed here because the second draw is supposed to operate also on all balls available.

The original statement:

A bag contains one pool ball that is known to be a solid, a second ball is chosen in such a way that it has equal chance of being a solid or a stripe, this ball is then placed in the bag.

Bag is loaded with two balls. This is equivalent to selected one of 2 bags loaded in the two possible ways this bag could be loaded.

A ball is now selected randomly from the bag and it turns out to be a solid.

If somebody now selects a ball from the bag what is the probability of it being a stripe?

Both selections are from this same bag. It is not tampered with after the first ball is drawn. You don't put more balls in, you just remove the second ball in it.

We want to find the probability that the second ball drawn is a stripe *given that* the first ball remoevd was a solid. This *is* a conditional probabilty. I'm begining to think you either have never seen conditional probability before or misunderstand them in some way given how you claimed it was irrelevant to this problem and how you tried to apply it with absolutely no justification in the last post where you had modified Pete's chart.

 

[tsmid]

The bag initially contains a ball that is known to be black. A second ball is added to this bag. There's a 50% chance that this second ball is white, and a 50% chance that it is blue.

You then pick a ball randomly out of the bag. It looks black, but the problem is, you have an eye problem and can't distinguish between black and blue. So, given that the first ball removed is either black or blue, what is the probability that the second picked is white?

I don't see the point of your suggestion to introduce a third option and then say that it is indistinguishable from the first.

No, it's the probability that we're in case 2, given that we're either in case 1 or in case 2, that we want. Before the draw case 1 is twice as likely as case 2, and this still holds if the first draw rules out case 3 (the random ball is striped, and you take the striped ball out on your first draw).

I don't know why you keep on maintaining that case 1 is twice as likely as case 2. Case 1 is the case where the random ball is the full ball, case 2 where the random ball is the striped ball. This is the only independent parameter here which comes into play (and this is assumed to be 50/50 by definition here). I doesn't matter at all if one of these options splits into several sub-options. As an example, assume for instance two towns A and B connected each by a road to town C. Assume that the road B->C somewhere splits into two roads both leading to C. Now why should this split make any difference for the number of people that arrive at C from B? I doesn't. You could split the road into 100 sub-roads and it wouldn't matter. The ratio of people from A and B that arrive at C is solely given by the ratio of people who set out from A and B in the first place.

Thomas

 

[przyk]

I don't know why you keep on maintaining that case 1 is twice as likely as case 2.

Case 1 has a probability of 50%
Case 2 has a probability of 25%
Case 3 has a probability of 25%

Case 1 is twice as likely as case 2 before ball 1 is drawn. If you know that case 3 did not occur during a particular trial, case 1 is still twice as likely as case 2. If you perform repeated trials, case 1 will occur twice as often as case 2 whether or not you discard all the trials that give case 3. Again, I invite you to follow Dinosaur's sampling argument, which you seem to have ignored.

Case 1 is the case where the random ball is the full ball, case 2 where the random ball is the striped ball. This is the only independent parameter here which comes into play (and this is assumed to be 50/50 by definition here).

Then the case "the random ball is the striped ball and the solid ball was drawn first" is a sub-case of case 2. We know that either case 1 or a sub-case of case 2 occured, so the 50/50 no longer applies, regardless of what your intuition tells you.

I doesn't matter at all if one of these options splits into several sub-options. As an example, assume for instance two towns A and B connected each by a road to town C. Assume that the road B->C somewhere splits into two roads both leading to C. Now why should this split make any difference for the number of people that arrive at C from B? I doesn't. You could split the road into 100 sub-roads and it wouldn't matter. The ratio of people from A and B that arrive at C is solely given by the ratio of people who set out from A and B in the first place.

As a better example, consider this setup:

Town A has 2 roads leading to town C.
Town B has 1 road leading to town C and one road leading to town D.

Towns A and B have equal populations. A random person from this extended population (so there's a 50% chance they're a citizen of town A, and 50% that they're from B) randomly picks one of the two roads away from their town. Now, given that this person arrived at town C, what is the probability that they started at town B?

For comparison with the original problem, towns A and B are this problem's equivalent of the random ball added to the bag being either a solid or a stripe, and the condition of arriving at town C is the equivalent of the "first drawn was solid".