Rincewind

07-01-2004, 11:45 AM

I've tried to extricate this thread from the gambling on the Results of the Aussie Champs thread. It began (as most thread do) with a slinging match between Bill and Matt. But I'll start reproducing posts from the part where I got involved...

Question:

A bag contains an unknown number of marbles >1.

Each marble is either black or white.

You draw out one marble, it is white and you put it in your pocket.

What colour is most likely to be pulled out in the next draw, and why?

The correct answer to this is that you have insufficent information to answer the question. All you know is that your chance of drawing a white marble has decreased and your chance of drawing a black marble has increased (provided the number of marbles in the bag is finite). But without knowing the initial mix and number of marbles you can't say either colour is more likely than another.

Here are two possible answers based on an unprovided assumption which lead to opposite answers.

Assumption 1: The initial mix of black to white was 1:1

Answer 1: Black is now more likely as there are now 1 more black marble than white. However the change in probability is inversely proportional to the initial number of marbles in the bag.

Assumption 2: The initial mix of black to white was significantly different from 1:1 and there were a reasonably large number of marbles in the bag.

Answer 2: White was pulled out first so it is more likely that there were more white than black. Due to the large number of marble the non-replacement has not affected probablilities too greatly and therefore White is again the more likely result.

However, neither answer is correct unless the assumptions are accepted.

To illustrate the point for you here are some questions which have answers.

1. You have a bag with black and white marbles, 12 of one colour and 8 of the other. A marble is drawn at random (it happens to be white) and put into your pocket. Is it now more likely that the next marble drawn will be black or white? (4 points)

2. Provide the probability to 4 decimal places. (2 points)

Almost true. We can always answer. I practical terms. the answer is "a white marble". However, your arguments and examples are 100% correct! I concur fully with you. We must make some assumptions.

I put it to you that practically, the bag it not of infinite size. Further that the proportion of white to black is unknown, and thus in practical terms normally distributed unless otherwise known. Therfore, of all the possible actual black and actual white marbles in the bag there are proportional very few that have B=W. This means that the number of balck and white are nearly always unequil. Now we can say, as you have already explained, that what ever draw out first is the best prediction for the second marble's colour.

Not so. We can answer or we can guess. When assuming you are making a guess as to what is reasonable. You cannot say what the probability of your guess being correct is. You may as well flip a coin.

Here you are making some additional assumptions as to the likely distribution of marbles. Furthermore, your logic is flawed. The reliability of the "same again" strategy is improved the more biased the sample of marbles and the greater the number of marbles, however, the first rule of Probability is do not trust your intuition.

BTW did you have a crack at my question? (hint)

Marbles are distributed 12/8, does the "same again" strategy perform better or worse than the "opposite" strategy?

There was an attempted solution submitted by Matt at this point but he clearly understood I was giving the distribution as Black:12, White:8 whereas I didn't intend this to be inferred from the way the questoin was worded. Anyway a second solution was submitted...

#1. The most likely colour of the second marble is white.

#2. There are 4 possible outcomes for the two marble draw without replacement:

Majority marble, Majority marble

Majority marble, Minority marble

Minority marble, Majority marble

Minority marble, Minority marble

The probablity of these are :

12/20 x 11/19 = 0.3474

12/20 x 8/19 = 0.2526

8/20 x 12/19 = 0.2526

8/20 x 7/19 = 0.1474

SUM =1=P

The P of a Major coming out second is

(12/20 x 11/20) + (8/20 x 12/19) = 0.6000

The P of a Minor coming out is

1 - P(Maj) = 1 - 0.6000 = 0.4000

The first marble drawn was White,

and the P of drawing a Maj in the first draw is 12/20 = 0.6000,

hense the P of White being the Maj is also 0.6000.

It has been shown that that the porobabilty of drawing a Maj in the second draw is also 0.6. In deed, in the absence of information of what has already been draw, the P of drawing a Maj in any/every draw is always 0.6

Therefore the chance of the second marble being White is also 0.6000

Sorry Barry, your flipped coin and going against your instinct means the laser fitted sharks have just made yum cha of your girlfriend. :evil: :twisted:

Actually your score is

Q1 - 0 points - harsh but fair, the correct answer is Black

Q2 - 1 point - generous but some of your working is right, although your logic is confused.

1/6 is not a good start but don't lose heart, there will be future assignment with which you can earn credits prior to exam day.

The correct solution is as follows.

The colour of the marbles is largely irrelevant as we don't know if white or black is the majority. The easiest thing to do is to calculate the probablity of the second marble being the same colour as the first marble.

That is Maj then Maj, or Min then Min.

The probability (P1) is 12/20 * 11/19 + 8/20 * 7/19 = (132 + 56)/380 = 188/380

For completeness, possibilities for differing colours are Maj then Min or Min then Maj

The probability (P2) is 12/20 * 8/19 + 8/20 * 12/19 = (96 + 96)/380 = 192/380

NB: P1 + P2 = 1

NMB: P1 < P2

So the most likely result (although only ever so slightly) is that second marble will differ in colour from the first. The first was White, therefore the second will most likely be Black.

The correct answers are

Q1. It is more likely that the second marble will be Black

Q2. Black (P2) = 192/380 ~ 0.5053

White (P1) = 188/380 ~ 0.4947

Feel free to asks questions, but next time you're working on a probability problem, don't forget to check your intuition in at the cloak room as it usually leads one astray. ;)

If you don't believe me do a web search on the "Monty Hall Problem". That one is a beauty.

I did this independently and got exactly the same answer as Barry, though I wouldn't have explained it as well as he did. Very neat example indeed. :D

Thanks, as an footnote, the inequality that is satisfied for differing colours to be more likely that same colours is quite neat. Here it is...

(x1 - x2) ^ 2 < (x1 + x2)

where x1 and x2 are the numbers of marbles of each colour.

Or to express in words, the second marble is more likely to differ in colour from the first when the square of the difference in number of the marbles of each colour is less than the total number of marbles.

In this example x1 = 12, x2 = 8

RHS = 12 + 8 = 20 (total number of marbles)

LHS = (12 - 8 ) ^ 2 = 4 ^ 2 = 16 (square of the difference (in this case 4))

Anyway, LHS is less than RHS in this example so different marbles rule. ;)

Kevin, if you haven't already you might like to derive this inequality to see if I have this right.

That is a really tough marking system :shock: . I would have though maybe 3/6 . 2 for using a matrix, 2 for calculating their probabilities, -2 for logic error, 0 for answer, bonus 1 for showing that all draws have exactly the same chance of a Maj being picked.

Yes, I can see how I made it hard for myself, and picked the wrong matrix pair. I like how your more eligant approach did not lead to the logical complexities that I stuffed up. :D That Monty Hall Problem is a cracker. You bloody professional mathamaticians come up with fabulous ways to fool people.

However, you still have yet to say why you picked Black as the second colour in the absence of sufficient information. ;)

You could reasonably assume the ratio of Maj to Min is normaly distributed. Futhermore, you might assume that estimates of the number of marbles on the bag would have an asymetric distribution. Putting those equations and your personbal estimate of the number of marbles, together with the inequality you gave {(x1 - x2) ^ 2 < (x1 + x2)} you might have found the solution, while your girlfriend was dangling - but I doubt it. So why did you pick black.

Question:

A bag contains an unknown number of marbles >1.

Each marble is either black or white.

You draw out one marble, it is white and you put it in your pocket.

What colour is most likely to be pulled out in the next draw, and why?

The correct answer to this is that you have insufficent information to answer the question. All you know is that your chance of drawing a white marble has decreased and your chance of drawing a black marble has increased (provided the number of marbles in the bag is finite). But without knowing the initial mix and number of marbles you can't say either colour is more likely than another.

Here are two possible answers based on an unprovided assumption which lead to opposite answers.

Assumption 1: The initial mix of black to white was 1:1

Answer 1: Black is now more likely as there are now 1 more black marble than white. However the change in probability is inversely proportional to the initial number of marbles in the bag.

Assumption 2: The initial mix of black to white was significantly different from 1:1 and there were a reasonably large number of marbles in the bag.

Answer 2: White was pulled out first so it is more likely that there were more white than black. Due to the large number of marble the non-replacement has not affected probablilities too greatly and therefore White is again the more likely result.

However, neither answer is correct unless the assumptions are accepted.

To illustrate the point for you here are some questions which have answers.

1. You have a bag with black and white marbles, 12 of one colour and 8 of the other. A marble is drawn at random (it happens to be white) and put into your pocket. Is it now more likely that the next marble drawn will be black or white? (4 points)

2. Provide the probability to 4 decimal places. (2 points)

Almost true. We can always answer. I practical terms. the answer is "a white marble". However, your arguments and examples are 100% correct! I concur fully with you. We must make some assumptions.

I put it to you that practically, the bag it not of infinite size. Further that the proportion of white to black is unknown, and thus in practical terms normally distributed unless otherwise known. Therfore, of all the possible actual black and actual white marbles in the bag there are proportional very few that have B=W. This means that the number of balck and white are nearly always unequil. Now we can say, as you have already explained, that what ever draw out first is the best prediction for the second marble's colour.

Not so. We can answer or we can guess. When assuming you are making a guess as to what is reasonable. You cannot say what the probability of your guess being correct is. You may as well flip a coin.

Here you are making some additional assumptions as to the likely distribution of marbles. Furthermore, your logic is flawed. The reliability of the "same again" strategy is improved the more biased the sample of marbles and the greater the number of marbles, however, the first rule of Probability is do not trust your intuition.

BTW did you have a crack at my question? (hint)

Marbles are distributed 12/8, does the "same again" strategy perform better or worse than the "opposite" strategy?

There was an attempted solution submitted by Matt at this point but he clearly understood I was giving the distribution as Black:12, White:8 whereas I didn't intend this to be inferred from the way the questoin was worded. Anyway a second solution was submitted...

#1. The most likely colour of the second marble is white.

#2. There are 4 possible outcomes for the two marble draw without replacement:

Majority marble, Majority marble

Majority marble, Minority marble

Minority marble, Majority marble

Minority marble, Minority marble

The probablity of these are :

12/20 x 11/19 = 0.3474

12/20 x 8/19 = 0.2526

8/20 x 12/19 = 0.2526

8/20 x 7/19 = 0.1474

SUM =1=P

The P of a Major coming out second is

(12/20 x 11/20) + (8/20 x 12/19) = 0.6000

The P of a Minor coming out is

1 - P(Maj) = 1 - 0.6000 = 0.4000

The first marble drawn was White,

and the P of drawing a Maj in the first draw is 12/20 = 0.6000,

hense the P of White being the Maj is also 0.6000.

It has been shown that that the porobabilty of drawing a Maj in the second draw is also 0.6. In deed, in the absence of information of what has already been draw, the P of drawing a Maj in any/every draw is always 0.6

Therefore the chance of the second marble being White is also 0.6000

Sorry Barry, your flipped coin and going against your instinct means the laser fitted sharks have just made yum cha of your girlfriend. :evil: :twisted:

Actually your score is

Q1 - 0 points - harsh but fair, the correct answer is Black

Q2 - 1 point - generous but some of your working is right, although your logic is confused.

1/6 is not a good start but don't lose heart, there will be future assignment with which you can earn credits prior to exam day.

The correct solution is as follows.

The colour of the marbles is largely irrelevant as we don't know if white or black is the majority. The easiest thing to do is to calculate the probablity of the second marble being the same colour as the first marble.

That is Maj then Maj, or Min then Min.

The probability (P1) is 12/20 * 11/19 + 8/20 * 7/19 = (132 + 56)/380 = 188/380

For completeness, possibilities for differing colours are Maj then Min or Min then Maj

The probability (P2) is 12/20 * 8/19 + 8/20 * 12/19 = (96 + 96)/380 = 192/380

NB: P1 + P2 = 1

NMB: P1 < P2

So the most likely result (although only ever so slightly) is that second marble will differ in colour from the first. The first was White, therefore the second will most likely be Black.

The correct answers are

Q1. It is more likely that the second marble will be Black

Q2. Black (P2) = 192/380 ~ 0.5053

White (P1) = 188/380 ~ 0.4947

Feel free to asks questions, but next time you're working on a probability problem, don't forget to check your intuition in at the cloak room as it usually leads one astray. ;)

If you don't believe me do a web search on the "Monty Hall Problem". That one is a beauty.

I did this independently and got exactly the same answer as Barry, though I wouldn't have explained it as well as he did. Very neat example indeed. :D

Thanks, as an footnote, the inequality that is satisfied for differing colours to be more likely that same colours is quite neat. Here it is...

(x1 - x2) ^ 2 < (x1 + x2)

where x1 and x2 are the numbers of marbles of each colour.

Or to express in words, the second marble is more likely to differ in colour from the first when the square of the difference in number of the marbles of each colour is less than the total number of marbles.

In this example x1 = 12, x2 = 8

RHS = 12 + 8 = 20 (total number of marbles)

LHS = (12 - 8 ) ^ 2 = 4 ^ 2 = 16 (square of the difference (in this case 4))

Anyway, LHS is less than RHS in this example so different marbles rule. ;)

Kevin, if you haven't already you might like to derive this inequality to see if I have this right.

That is a really tough marking system :shock: . I would have though maybe 3/6 . 2 for using a matrix, 2 for calculating their probabilities, -2 for logic error, 0 for answer, bonus 1 for showing that all draws have exactly the same chance of a Maj being picked.

Yes, I can see how I made it hard for myself, and picked the wrong matrix pair. I like how your more eligant approach did not lead to the logical complexities that I stuffed up. :D That Monty Hall Problem is a cracker. You bloody professional mathamaticians come up with fabulous ways to fool people.

However, you still have yet to say why you picked Black as the second colour in the absence of sufficient information. ;)

You could reasonably assume the ratio of Maj to Min is normaly distributed. Futhermore, you might assume that estimates of the number of marbles on the bag would have an asymetric distribution. Putting those equations and your personbal estimate of the number of marbles, together with the inequality you gave {(x1 - x2) ^ 2 < (x1 + x2)} you might have found the solution, while your girlfriend was dangling - but I doubt it. So why did you pick black.