probability question

[Azazello1313]

Here I have a couple of urns. The one on the left contains 70 red balls and 30 black. The one on the right contains 30 red and 70 black.

 

While you weren’t looking, I reached into one of these urns and randomly drew out a dozen balls. As you can see, 4 of them were red and 8 were black.

 

Here are three questions that I think you ought to be able to answer if you want to be in the business of assessing evidence:

 

1.If you had to guess, which urn would you guess I drew from?

2.What’s your estimate of the odds that you’re right?

3.Do you think you’re right beyond a reasonable doubt?

[driveby]

puddy

[matt770]

Here I have a couple of urns. The one on the left contains 70 red balls and 30 black. The one on the right contains 30 red and 70 black.

 

While you weren’t looking, I reached into one of these urns and randomly drew out a dozen balls. As you can see, 4 of them were red and 8 were black.

 

Here are three questions that I think you ought to be able to answer if you want to be in the business of assessing evidence:

 

1.If you had to guess, which urn would you guess I drew from?

2.What’s your estimate of the odds that you’re right?

3.Do you think you’re right beyond a reasonable doubt?

 

1. The right urn.

2. 100%

3. There is no way in hell those balls came out of the left urn.

[cliaz]

Here I have a couple of urns. The one on the left contains 70 red balls and 30 black. The one on the right contains 30 red and 70 black.

 

While you weren’t looking, I reached into one of these urns and randomly drew out a dozen balls. As you can see, 4 of them were red and 8 were black.

 

Here are three questions that I think you ought to be able to answer if you want to be in the business of assessing evidence:

 

1.If you had to guess, which urn would you guess I drew from?

2.What’s your estimate of the odds that you’re right?

3.Do you think you’re right beyond a reasonable doubt?

 

Me likes.

 

1. You pulled the balls out of the urn on the right hand side. I say this because the probability of pulling out a black ball each time in 7 out of 10 and your probability of pulling out a red ball is 3 out of 10 (70 black balls to 30 red balls).

 

2. I would estimate I am probably 70% correct. Even though there is a 70/30 split (Black to Red) on the urn on the right, each time you stick your hand in to the earn you have an equal change of grabbing the lesser count (for either the left or right) since they both have an equal amount of the minority color of balls. So, you could have pulled them from the urn on the left but it would be a high change you drew them from the urn on the right.

 

3. Not for my definition of beyond reasonable doubt. I need the following:

 

-See you pull the balls from the urn on the right

-Hear you pulling balls from the right direction (right as in urn on the right)

-See a lower ball count in the earn

-See the weight of each urn filled with 100 balls and then see the weight of each urn after you pull your balls out

-Note any differences in pigment colors between urns (red to red and black to back) and visually inspect the balls you've pulled out and laid out in front of me

-How many different testicle jokes will come from my response?

 

Reasonable doubt needs to be established (for me) through proxy definitive evidence like I listed above or incredibly strong circumstantial evidence (e.g. like if the urns were space differently, how long did it take you each time you put your hand in there to pull out a ball).

[Perchoutofwater]

1. I'd guess you pulled it out of the right urn

2. My odds of being correct are 50/50

3. No, I do not think I'm right beyond a reasonable doubt.

[Big Country]

1. I'd guess you pulled it out of the right urn

2. My odds of being correct are 50/50

3. No, I do not think I'm right beyond a reasonable doubt.

 

 

This.

 

1. Based on distribution, this seems most likely.

 

2. It is 50/50 as there are only 2 choices.

 

3. 50/50 is not beyond reasonable doubt in my book.

[Azazello1313]

2. My odds of being correct are 50/50

 

2. It is 50/50 as there are only 2 choices.

 

 

I expect this answer from club ("it's 50/50, like a parachute")....but come on, fellas.

 

anyway, I'll post some more on this tonight or tomorrow...

[Perchoutofwater]

85%

[Furd]

Its an interesting question.

 

I understand the basic principles behind it, but I'm not sure how it relates to the law.

[billay]

I'd guess you drew from the right urn.

I'd say the chances that I'm right are 55-65%

It would not be a beyond reasonable doubt based on a single run of the experiment.

[Big Country]

 

I expect this answer from club ("it's 50/50, like a parachute")....but come on, fellas.

 

anyway, I'll post some more on this tonight or tomorrow...

 

 

The odds that I am right are indeed 50/50, if I am reading the questions correctly.

 

The question is not what are the odds that this particular arrangment of balls came from either side.

 

The options for question 2 are are you right or wrong. The person either drew from the left or he drew from the right.

 

The answer to number one is where we would figure the odds that this particular arrangement of drawn balls came from either urn and thus gave us our guess.

 

It is clearly evident that based on the makeup of each urn, a draw of 8 black and 4 red is more likely to come from the urn on the right. Thus we answer the right for number one.

 

Question 2 is still 50/50, either it was from the right or it was from the left.

[billay]

Question 2 is still 50/50, either it was from the right or it was from the left.

Given the breakdown inside the urns, the probability may be greater than 50%.

[Big Country]

Given the breakdown inside the urns, the probability may be greater than 50%.

 

Maybe I'm reading too much into a "trick" question aspect.

 

I think the probability that the draw was created by pulling from the right is higher than the probability it came from the left, thus for question 1 I guess the answer is the right.

 

 

For question 2, either it is putting a figure to how much more likely it is that it came from the right than it is that it came from the left, or it truly is a question of are you right or wrong. Two choices, 50/50 shot. I was thinking about the Let's Make a Deal paradox when looking at the question.

 

I may be offbase, in which case I would say that yes, the probability that I am right in my guess is greater than 50/50, but, the reailty is that just because it is more likely that this particular draw came from the makeup of the right, it doesn;t change the fact that we have one of two choices and are either right or wrong.

[Azazello1313]

Question 2 is still 50/50, either it was from the right or it was from the left.

 

the fact that there are two possible outcomes certainly does not equate to a 50/50 breakdown in probabilities between them.

 

think of it like a heads up poker hand. after the turn, you have 3 aces and your opponent has 3 jacks, and the last jack is still in the deck. if the question is who wins the hand, there are two possible answers, but it sure as chit ain't a 50/50 proposition at that point.

[Big Country]

the fact that there are two possible outcomes certainly does not equate to a 50/50 breakdown in probabilities between them.

 

think of it like a heads up poker hand. after the turn, you have 3 aces and your opponent has 3 jacks, and the last jack is still in the deck. if the question is who wins the hand, there are two possible answers, but it sure as chit ain't a 50/50 proposition at that point.

 

 

Like I said above, came down to my interpretation of the question and thinking it was a trick one somehow.

 

 

Yes, the odds that it is the irght one are clearly greater than the odds of it being the left one.

 

I took the 2nd one to be a bit tricky and the answer is 50/50, either I am wrong or I am right. Other option is to look at chances of drawing that combo of balls from the left urn and chances of drawing the combination from the right one, and then compariing those to give the probability between the two choices and that is the answer to #2.

 

For simplicity sake, if that pull is made 1 in 10 times from the right urn and 1 in 20 times from the left, then my probability of being right is 67%.

[muck]

the fact that there are two possible outcomes certainly does not equate to a 50/50 breakdown in probabilities between them.

 

think of it like a heads up poker hand. after the turn, you have 3 aces and your opponent has 3 jacks, and the last jack is still in the deck. if the question is who wins the hand, there are two possible answers, but it sure as chit ain't a 50/50 proposition at that point.

 

Correct, as I'm already sitting on four nines.

 

[Azazello1313]

I believe this is the correct math:

 

To get the probability of picking BBBBBBBBRRRR (in exactly that order) from the right urn, the numerator is 70*69*68*67*66*65*64*63*30*29*28*27 and the denominator is 100*99*98*97*96*95*94*93*92*91*90*89.

 

The probability of picking BBBBBBBBRRRR (in exactly that order) from the left urn can be calculated similarly, but it’s 70 through 67 and 30 through 23 in the numerator instead of 70 through 63 and 30 through 27.

 

The ratio of the first to the second is about 48.2 to 1, which means the probabilities are about 98% and 2%, respectively.

 

(The “exactly that order” stuff doesn’t matter in this case because the number of possible orders is the same for the left and right urns, and multiplying each of the two probabilities by the same number does not affect the ratio between them.)

[Jackass]

I believe this is the correct math:

 

It's not. nowhere in the initial question did you say anything about exact order.

 

ETA: Ok, just read your last sentence. No way is that correct.

Edited November 10, 2010 by Jackass

[keggerz]

1. I'd guess you pulled it out of the right urn

2. My odds of being correct are 50/50

3. No, I do not think I'm right beyond a reasonable doubt.

this

 

This.

 

1. Based on distribution, this seems most likely.

 

2. It is 50/50 as there are only 2 choices.

 

3. 50/50 is not beyond reasonable doubt in my book.

and this and I agree with the stance BC has taken as to why too

[Azazello1313]

It's not. nowhere in the initial question did you say anything about exact order.

 

ETA: Ok, just read your last sentence. No way is that correct.

 

I am pretty certain it is. if you're multiplying a bunch of numbers, it doesn't matter what order they are in. the only reason he frames it that way (I agree it's slightly confusing) is to make note of the fact that the probability of selecting a red or black ball change slightly after each ball is drawn.

 

98% is the correct answer, that much I am sure of beyond a reasonable doubt.

[Jackass]

I am pretty certain it is. if you're multiplying a bunch of numbers, it doesn't matter what order they are in. the only reason he frames it that way (I agree it's slightly confusing) is to make note of the fact that the probability of selecting a red or black ball change slightly after each ball is drawn.

 

98% is the correct answer, that much I am sure of beyond a reasonable doubt.

 

here's my problem with this math. and maybe the answer is 98% - seems high but possible. but the first 4 balls are either red or black - that's a 100% certainty. so that's where the math should start - with the next 8 balls. there's probably a formula to figure this out that involves factorials.

[Big Country]

here's my problem with this math. and maybe the answer is 98% - seems high but possible. but the first 4 balls are either red or black - that's a 100% certainty. so that's where the math should start - with the next 8 balls. there's probably a formula to figure this out that involves factorials.

 

Problem is that the makeup of those first 4 balls has to be known in order to accurately do the math by reducing the number of available balls of a particular color for each step of the equation. I'm sure it can be done, but it's beyond my limited capabilities at the end of the day to do it.

 

But, like you noted, order does not matter, so the actual odds of pulling this exact combination will be higher in each case, but, I think Az's last sentence accurately notes that because the number of possible orders is the same for both sides, the ratio will end up being the same whether the odds for each urn are calculated with any order possible or for a specific order.

[Azazello1313]

but the first 4 balls are either red or black - that's a 100% certainty.

 

[wiegie]

Az's math looks fine to me.

 

But from my point of view, the math is the least interesting thing about Az's post.

 

The most interesting thing is the question: Does a 98% chance that something happened equate to "it happened beyond a reasonable doubt"?

[satelliteoflovegm]

Have you visited Australia? Then you clearly could not choose the urn on the right!