Alice has to buy food for a party while Bob is sending invitations. Alice initially expects 10 guests to arrive and plans to buy corresponding amount of food. For simplicity let’s suppose that having 1-person worth of food more than needed bears the same cost as having the same amount less than needed, and Alice’s prior probability distribution is symmetric around 10, so she intends to buy 10 units of food.
But before going to the shop, she asks Bob how many people are coming. Bob is reluctant to answer, so Alice offers an arbitrary threshold 40. Actually only 15 guests are going to attend, but Bob doesn’t know this. His prior expectations were more or less the same as Alice’s, but by power of anchoring, he gives the answer “more than 40” with p=0.3. This answer causes Alice to buy 45 (the lowest round number greater than 40) units of food. If Bob answers “not more than 40″, Alice will stick with her previous guess. Which means the cost is 0.3 30 + 0.7 5 = 12.5.
Without asking, Alice would buy 10 units of food and the cost of error would be only 5.
I dont really think it would work like that. My understanding of anchoring is that they think 40 and then search for the nearest reasonable answer and whether that’s above or below the number correlates to their initial internal beliefs about the value in question. That is to say, if he originally thought it was ten, he would think 40, and then consider (subconsciously) all numbers between 40 and zero in decreasing order. I expect he would stop no earlier than 20 (or maybe as low as 15 :D).
I assumed he was unwilling to tell an exact number before, so he would probably not do it afterwards. He will only tell whether it is higher or lower than 40, and I suppose anchoring will move the median of the probability distribution towards the anchor, thus increasing the probability of answering “more than 40” in this case.
I think about making a poll here to test the hypothesis, only not sure how to do that.
I disagree. I suppose it may slightly increase the probability of the “more than 40” answer, but his prior expectations have to play some role. Since there would be no other real factors, the effect of his prior expectation would in my opinion shift his answer downwards, yielding the “less than 40″ answer. Anchoring pulls you towards the suggested value, but I’m not sure in this case it would pull it ABOVE the suggested value, especially given the difference between his original beliefs and the anchor.
Depends. A convoluted counterexample:
Alice has to buy food for a party while Bob is sending invitations. Alice initially expects 10 guests to arrive and plans to buy corresponding amount of food. For simplicity let’s suppose that having 1-person worth of food more than needed bears the same cost as having the same amount less than needed, and Alice’s prior probability distribution is symmetric around 10, so she intends to buy 10 units of food.
But before going to the shop, she asks Bob how many people are coming. Bob is reluctant to answer, so Alice offers an arbitrary threshold 40. Actually only 15 guests are going to attend, but Bob doesn’t know this. His prior expectations were more or less the same as Alice’s, but by power of anchoring, he gives the answer “more than 40” with p=0.3. This answer causes Alice to buy 45 (the lowest round number greater than 40) units of food. If Bob answers “not more than 40″, Alice will stick with her previous guess. Which means the cost is 0.3 30 + 0.7 5 = 12.5.
Without asking, Alice would buy 10 units of food and the cost of error would be only 5.
I dont really think it would work like that. My understanding of anchoring is that they think 40 and then search for the nearest reasonable answer and whether that’s above or below the number correlates to their initial internal beliefs about the value in question. That is to say, if he originally thought it was ten, he would think 40, and then consider (subconsciously) all numbers between 40 and zero in decreasing order. I expect he would stop no earlier than 20 (or maybe as low as 15 :D).
I assumed he was unwilling to tell an exact number before, so he would probably not do it afterwards. He will only tell whether it is higher or lower than 40, and I suppose anchoring will move the median of the probability distribution towards the anchor, thus increasing the probability of answering “more than 40” in this case.
I think about making a poll here to test the hypothesis, only not sure how to do that.
I disagree. I suppose it may slightly increase the probability of the “more than 40” answer, but his prior expectations have to play some role. Since there would be no other real factors, the effect of his prior expectation would in my opinion shift his answer downwards, yielding the “less than 40″ answer. Anchoring pulls you towards the suggested value, but I’m not sure in this case it would pull it ABOVE the suggested value, especially given the difference between his original beliefs and the anchor.
I have tested the hypothesis experimentally, and it seems that the probability is increased quite substantially. See here.
Doesn’t “more than 40 with p=0.3” actually mean “less than or equal to 40 with p=0.7″ ?
Absolutely.
Does your remark mean that you have spotted a mistake in my parent comment?
Edit: Ah, I have omitted the “or equal” part. Fixed.