Heuristic Bias Example (CFA Reading)

ok, I don’t know why but I really want to know how to actually solve the ‘anchoring-and-adjustment conservatism’ example in R7 (CFAI) pg 9. "Imagine 100 book bags, each of whoch contains 1,000 poker chips. Forty-five bags contain 700 black chips and 300 red chips. The other 55 bags contain 300 black chips and 700 red chips. You cannot see inside either bag. One of the bags is selected at random by means of a coin toss. Consider the following two questions about the book bag. 1. what probability would you assing to the event that the selected bag contains predominantly black chips? 2. Imagine taht 12 chips are drawn, with replacement, from the selected bag. These 12 draws produce 8 blacks and 4 reds. Would you use the new information about the drawing of chips to revise your probability that the selected bag contains predominantly black chips? If so, what new probability would you assing? The answer to 1. is easy - .45 But the answer to the second one is supposed to be 96.04%. I would love to know how to get to this number. thanks!

I may come up with a more elegant way to explain this, but for now this is what I have. Assuming the bag has more black chips: This is a case of binomial sampling (CFA level I :-)) where 12 identical trials, each a Bernoulli trial results in one of two mutually exclusive outcomes (red or black). The probability of getting black = .7 (700/1000). The probability of drawing 8 black out of 12 chips = 12! /( 8!*4!)* .7^8 * (1-.7)^(12-8) = .2311397 If the bag has more red chips, the equivalent probability will be 12! /( 8!*4!)* .7^(12-8) * (1-.7)^(8) = .007797 Thus the chance for bag having more blacks = .2311397 /(.2311397+.007797) = 96.74%, A more direct calculation would be [.7^8 * (1-.7)^(12-8)] / {[.7^8 * (1-.7)^(12-8)] + [(.7^(12-8) * (1-.7)^(8)]} not exactly the same answer in the text 96.04%, but close enough, probably due to rounding errors.

elcfa explained it well. I’m getting 96.74% with the binomial as well. Here’s how CFAI is getting 96.04%…though I don’t believe it is the correct thing to do. In elcfa’s desctiption (and what I’m getting) the key numbers are: .2311397 (black dominant bag) and .007797 (red dominant bag). CFAI is taking .2311397 and multiplying by 45% to get 0.104013 and taking .007797 and multiplying by 55% to get 0.004289 now, 0.104013 / (0.104013 + 0.004289) = 96.04% I just don’t think that is the correct thing to do. Anyone have any clarification on this little discrepancy?

we already discussed this and have resolved it, the thread is here maybe about a month back if you look

Thanks, DoubleDip. we already discussed this and have resolved it, > the thread is here maybe about a month back if you > look

Hey, calm down! I didn’t deserve your attack, if you pay attention I am one who tries to help on these forums and I never say ‘search’ and so on. You are reading something into my response that is not there. Did it not occur to you that maybe I wasn’t able to pull up the link when I posted before - maybe I was posting from a bbery and I was trying to help as quickly as I could - but now that I can find the link, here you go. http://www.analystforum.com/phorums/read.php?13,1069468

Thanks everyone for explaining! I am pulling up my L1 readings… DoubleDip - I promise to search through the forums next time, I was just 100% positive no one would be interested in this :slight_smile:

I’ll take sometime to explain the solution to this problem. It’s a wonderful illustration of the Baye’s theorem. Define two events A and B as: A: Selecting a bag that’s got predominantly black chips. B: Drawing 12 chips and ending up with 8 black and 4 red chips. It’s our job to find out P(A given B). Now P(A given B) = P(A or B)/P(B) = P(B given A)*P(A)/P(B) We’re left to find out the values of each of the individual probabilities that go into the above equation. The easiest first -> P(A) = 0.45 P(B given A) is nothing but landing with 8 “successes” (defined as the event where a black chip is obtained in the selection) with the probability of “success” being 0.7. This equals 12C8 * 0.7^8 * 0.3^4 = 0.23114 P(B) = P(B given A)*P(A) + P(B given A’)*P(A’) = 0.23114*0.45 + [12C8* 0.3^8 * 0.7^4]*0.55 = 0.10832 Hence, P(A given B) = 0.45*0.23114/0.10832 = 96.04% (exact) Elcfa has made a conceptual mistake…

I agreed that I did not apply Baye’s theory. I knew about it myself soon after the posting, but did not have access to Internet to correct the posting until very late. By then, others have made the point, so it became superfluous for me to repeat it. My calculation would have been correct if the prior probability had been 0.5, instead of 0.45.

I’m glad others are interested in this, I also was very curious when I came across it and HAD to figure it out.