# Probability: Picking M&M's

• Oct 12th 2010, 04:27 PM
RageSpectate
Probability: Picking M&M's
I am a student in an Intro. Stat. class at a local community college.

I recently got a quiz back in which the teacher marked me wrong and upon questioning with more explicit detail he still told me I was wrong.

The question:
1) The Masterfoods company manufactures bags of Peanuts Butter M&M's. They report that they make 10% each brown and red candies, and 20% each yellow, blue, and orange candies. The rest of the candies are green.

--the last question on the quiz--

c. After picking 10 M&M's in a row, you still have not picked a red one. A friend says that you should have a better chance of getting a red candy on your next pick since you have yet to see one. Comment on your friends statement.

I answered that the friend is correct that in ideal circumstances probability has increased. (ideal, such as you didn't receive a bag with 0 red M&M's) This is covered by the friends diction of "should" any how.

The teacher said the correct answer is: Your friend is speaking nonsense.

My rationale:

P(picking a red) = N(number of red M&M's) / D(total number of M&M's)

or simplified to P = N/D.

After you have drawn out 10 M&M's previously the equation can be understood as P = N / (D-10)

It will not matter that you don't know what N or D is because you know that N has not changed but that the denominator has, no matter how many M&M's are in the bag the probability will increase. Substitute any non-zero number for the variables and the probability will increase.

If that is true then the friend is correct in assuming that you have a better chance of picking a red M&M.

This seems blatantly obvious to me. When I pressed my teacher he said "But you don't know the denominator" and I told him "that doesn't matter" and his response was "You are thinking to hard."

So, who is right? Me or the teacher?
• Oct 12th 2010, 04:49 PM
josh_amsterdam
The answer depends on how you look at it. Your teacher simplifies the question by assuming the company manufactures infinitely many M&M's, while you assume they make a countable number. If they make infinite M&M's then you're chance does not increase, but if they make a non-infinite (but high) number then your chances do increase (if ever so slightly). Seeing how they, probably, make countless M&M's your odds increase so very slightly that you might call it none. It all depends on whether you abstract that part of the question or not. But for all intends and purposes the answer is that there is no increase, while being mathematically precise you can only state the number approaches zero but will never reach it.
• Oct 12th 2010, 05:01 PM
RageSpectate
Quote:

Originally Posted by josh_amsterdam
The answer depends on how you look at it. Your teacher simplifies the question by assuming the company manufactures infinitely many M&M's, while you assume they make a countable number. If they make infinite M&M's then you're chance does not increase, but if they make a non-infinite (but high) number then your chances do increase (if ever so slightly). Seeing how they, probably, make countless M&M's your odds increase so very slightly that you might call it none. It all depends on whether you abstract that part of the question or not. But for all intends and purposes the answer is that there is no increase, while being mathematically precise you can only state the number approaches zero but will never reach it.

The only way that D is infinite is if you are pulling the M&M's off the conveyor belt but as per the first portion of the question we are speaking of a bag of M&M's which contains a finite number of M&M's.
• Oct 12th 2010, 05:51 PM
awkward
It is not necessary to suppose an infinite-size bag for the teacher's answer to be correct.

Consider a simplified version of the problem: Suppose candies come in two colors, Blue and Red, with equal probabilities, and suppose the bag contains two candies. If the first candy drawn is Blue, what is the probability that the second is Blue?

If we list the colors in the order (first, second), the initial contents of the bag are one of (B,B), (B,R), (R,B), and (R,R), all with probability 1/4. If the first drawn is Blue, then the possibilities are (R,B) and (R,R), each with probability 1/2. So the probability that the second candy drawn is Blue, given that the first is Blue, is

$\displaystyle \frac{P(\text{both are Blue})} {P(\text{first is Blue})} = \frac{1/4}{1/2} = \frac{1}{2}$

which is the same as the unconditional probability that the second drawn is Blue.
• Oct 12th 2010, 05:58 PM
RageSpectate
Quote:

Originally Posted by awkward
It is not necessary to suppose an infinite-size bag for the teacher's answer to be correct.

Consider a simplified version of the problem: Suppose candies come in two colors, Blue and Red, with equal probabilities, and suppose the bag contains two candies. If the first candy drawn is Blue, what is the probability that the second is Blue?

If we list the colors in the order (first, second), the initial contents of the bag are one of (B,B), (B,R), (R,B), and (R,R), all with probability 1/4. If the first drawn is Blue, then the possibilities are (R,B) and (R,R), each with probability 1/2. So the probability that the second candy drawn is Blue, given that the first is Blue, is

$\displaystyle \frac{P(\text{both are Blue})} {P(\text{first is Blue})} = \frac{1/4}{1/2} = \frac{1}{2}$

which is the same as the unconditional probability that the second drawn is Blue.

Ok, I think here in lies what would seem to me to be your misunderstanding.

"After picking 10 M&M's in a row, you still have not picked a red one. A friend says that you should have a better chance of getting a red candy on your next pick since you have yet to see one."

So there you are yanking the M&M's out of the bag and you have YET to pull out a red one. Your friend says, ok, well NOW (10 later) you should have a better chance of getting a red then you did previously. So this being the case, we need not worry as to how accurate the initial actual probability of picking red from the initial sample was. The friend isn't saying "You have a better then 10% chance of picking" but rather "you have a better chance then you did initially".
• Oct 13th 2010, 03:02 AM
josh_amsterdam
Quote:

Originally Posted by RageSpectate
The only way that D is infinite is if you are pulling the M&M's off the conveyor belt but as per the first portion of the question we are speaking of a bag of M&M's which contains a finite number of M&M's.

The bag is filled with random M&M's coming from a pile of M&M's in the factory. You might just as well say you're taking a number of M&M's from the factory as from the bag.

But ignore my blah about it being an infinite number, I now realise it doesn't even have to be. The chance of every M&M's colour is given. Now look at this, easier, problem. The odds of a person being male is 50%, the odds of a person being female is 50%.

You gather the first 11 random people who walk past you. The first ten all happen to be female, are your odds of the tenth being male now higher then they initially were? No. The 11th person is still a random person walking by who happens to be the eleventh person walking past you. As for any random person his odds of being male/female are 50/50. This may at first be a weird concept to grasp, but it works like this because the chances are independent from each other. Consider it like this, there are two options now, either you get 1(f)-2(f)...10(f)-11(f) or you get 1(f)-2(f)...10(f)-11(m), the odds of getting either option are equal.

The mistake that you are making is that the odds of getting this many woman in this scenario are very small. This is absolutely true! On average you'd get 5.5 man and 5.5 woman, but in this specific scenario you already HAVE 10 woman in a row. The odds of this situation arising may be low, this doesn't matter, you are here now. The odds of the next person being M/F are still 50/50
The same goes for the bag of M&M's. Every M&M has a given chance of what colour it is, this is independent on what colour other M&M's are.

I hope you understand now.
• Oct 13th 2010, 05:54 AM
RageSpectate
If you proffer real numbers then P = N/D becomes P = 1/11 prior to any person walking by you.

After 10 walk by you it becomes P = 1/1.