Probability: Flawed Assumptions in Picking M&M's

[bpet]

...It is often handy to give those somewhat awkward P(R|H_i and E)P(H_i|E) a name. In many texts and articles this name is P(R|H_i,E).

Ok thanks for taking the time to explain that subtle point about the notation. It's unfortunate that the literature has chosen this convention when we have $$P(R|H_i \mbox{ and } E)P(H_i|E) = P(R \mbox{ and } H_i|E)$$

 

[diggy]

The bag is superfluous in this problem.

So yes, original-poster, you are specifically correct, every time you don't find a red, your chance of drawing a red from a finite set that has a finite set of reds increases.

This is not the same as a gambler's fallacy, where the picks are reset every time (like a coin or a roulette wheel).

But your teacher is big-picture correct that the difference is so small, that it is effectively zero.

{{Another way of thinking about the bag being meaningless (if that bothers you) is that every time it was filled 1 M&M at a time, the odds of what the next M&M would be changed. So as its filled it becomes more and more improbable that the bag will be filled with all non-red M&M's. But this is the same as you picking directly from the factory set, not the bag.}}

 

[bpet]

...There is no need any reference to Bayes' Theorem or to the particularity of the production of M&M's etc.

The bag is superfluous in this problem...

That was my first thought as well, but it's incorrect. See e.g. posts #23 and #28 to see why the mixing model does matter.

Edit: see also posts #35, 36 regarding the difference in notation between D H's table and mine (our posts are too old to edit now).

 

[D H]

The bag is superfluous in this problem.

So yes, original-poster, you are specifically correct, every time you don't find a red, your chance of drawing a red from a finite set that has a finite set of reds increases.

Try reading the replies. The bag is anything but superfluous in this problem. It is central to it.

 

[diggy]

Try reading the replies. The bag is anything but superfluous in this problem. It is central to it.

Which post are you considering the correct interpretation?

 

[diggy]

Just to simplify life -- assume the factory that made (10) M&M's, (1) of which was red. They are put in (2) bags with (5) M&M's in each. You eat your first M&M and its blue.

I assume we all agree on the next M&M being more likely to be red, yes?

 

[CRGreathouse]

Just to simplify life -- assume the factory that made (10) M&M's, (1) of which was red. They are put in (2) bags with (5) M&M's in each. You eat your first M&M and its blue.

Alternate assumption: there are 3486784401 bags with 0 red M&Ms in them, 3874204890 bags with 1 red M&M in them, 1937102445 bags with 2 red M&Ms in them, 573956280 bags with red 3 M&Ms in them, 111602610 bags with 4 red M&Ms in them, 14880348 bags with 5 red M&Ms in them, 1377810 bags with 6 red M&Ms in them, 87480 bags with 7 red M&Ms in them, 3645 bags with 8 red M&Ms in them, 90 bags with 9 red M&Ms in them, and 1 bag with red 10 M&Ms in it. (All bags have 10 M&Ms in total.) You pick a bag and eat the first M&M and it's blue. Is the second one more likely to be red?

 

[diggy]

Alternate assumption: there are 3486784401 bags with 0 red M&Ms in them, 3874204890 bags with 1 red M&M in them, 1937102445 bags with 2 red M&Ms in them, 573956280 bags with red 3 M&Ms in them, 111602610 bags with 4 red M&Ms in them, 14880348 bags with 5 red M&Ms in them, 1377810 bags with 6 red M&Ms in them, 87480 bags with 7 red M&Ms in them, 3645 bags with 8 red M&Ms in them, 90 bags with 9 red M&Ms in them, and 1 bag with red 10 M&Ms in it. (All bags have 10 M&Ms in total.) You eat the first M&M and it's blue. Is the second one more likely to be red?

Yes.

 

[bpet]

Just to simplify life -- assume the factory that made (10) M&M's, (1) of which was red. They are put in (2) bags with (5) M&M's in each. You eat your first M&M and its blue.

I assume we all agree on the next M&M being more likely to be red, yes?

Now try the case where the factory makes 20 M&M's of which 2 are red. There are two possibilities: (a) the 2 reds go into one bag or (b) the 2 reds go into separate bags.

Without knowing how the bagging machine works (i.e the probabilities of (a) vs (b)), we've got a kind of Schrodinger's cat scenario where the probability of next being red can either increase or decrease.

 

[D H]

Yes.

Try again.

Before you open the bag, what can you say about the probability of getting a red M&M as the first candy from that bag? (No peeking now!) Does opening the bag but not peeking inside change that number one iota?

Hint: Per CRGreathouse's setup, there are 10 billion bags each containing 10 candies (100 billion candies total). Of these 100 billion candies, 10 billion are red.



Now you draw a blue candy as the first candy from that bag. What is the probability the next candy is red?

 

[diggy]

Ah -- so you are saying *if the bags aren't randomly filled (in the i.i.d. sense)* then it's possible that the likelihood of drawing a red goes down once you draw a blue. Yes totally agree.

Edit:
Sorry I thought you were just getting caught up in the bayesian inversion and not accounting for the density of red states increasing as you filtered out the blue-rich bags.