Do lower weights affect choices. (This question is based off one when thinking about college applications with binding decisions)

Question

Three events $A, B, C$. They are yes/no events. They are independent. $A$ is of greater weight/worth than $B$, and $B$ than $C$. In cases where multiple occur, only the greatest weight matters. You have the chance to increase the probability of $B$ at the cost that if $A$ and $B$ occur ($A\ B$ and $C$ or $A\ B$), you must go with the lower weight $B$. Does the weight of $C$ matter in your choice of whether or not you accept the proposition to increase the probability of $B$ with its costs? (Sorry for the poor wording—I'm not a probability guy).

Answer 1

Yes, it matters. In fact, the less worth $C$ has, the higher your tendency to prefer the new proposition. The more worth $C$ has, the higher your tendency to prefer the old proposition.

Basically the reason for this is, with the new proposition, the only way $C$'s influence changes is that you decrease the chance of going with $C$. So the higher $C$ is worth, then all else equal, the worse the new proposition looks to you.

For example, if $A$ had worth $100$, $B$ had worth $70$, and $C$ had worth $20$, the new proposition will look much better to you compared to if $A$ had worth $100$, $B$ had worth $70$, and $C$ had worth $68$.

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Here are some mathematical details, if you want to read them.

Suppose the utility (aka "worth") you get if you end up taking $A$ is $u(A)$, and similarly $u(B), u(C)$ if you go with $B$ and $C$ respectively. Let $p_A,p_B,p_C$ be the probability of $A,B,C$ being successful (respectively).

Initially, you can take the highest-utility event that occurs. Your expected utility with the initial choice is then $$\begin{align*} u_{initial} &= u(A) P(A \text{ occurs}) + u(B) P(A\text{ does not occur but }B\text{ does}) + u(C) P(\text{only }C\text{ occurs}) \\ &= u(A)p_A + u(B)(1-p_A)p_B + u(C) (1-p_A)(1-p_B)p_C.\end{align*}$$

From my interpretation: With the new proposition, you can change $p_B$ to some new value $b$ such that $p_B < b < p_A$. With the new proposition's costs,

• you go with $A$ iff $A$ occurs but $B$ does not (note: you didn't say that if $A$ and $C$ occur, we take $C$), which happens with probability $p_A (1-b)$;
• you go with $B$ iff $B$ occurs, which happens with probability $b$; and
• you go with $C$ iff only $C$ occurs, which occurs with probability $(1-p_A)(1-b)p_C$ (for you to go with $C$, we need $C$ to occur, and no others to occur, since otherwise you would go with either $A$ or $B$).

Therefore, in the new proposition, your expected utility is $$u_{new} = u(A)p_A (1-b) + u(B)b + u(C)p_C(1-p_A)(1-b).$$

Therefore, the change in expected utility, $u_{new} - u_{initial}$, is

$$ \begin{align*} u_{new}-u_{initial} &= \left(u(A)p_A (1-b) + u(B)b + u(C)(1-p_A)(1-b)p_C \right)\\ &- \left( u(A)p_A + u(B)(1-p_A)p_B + u(C) (1-p_A)(1-p_B)p_C\right)\\ \\ &= -p_A b u(A) + ( b- (1-p_A)p_B) u(B) \color{red}{-(1-p_A)(b-p_B)p_C } u(C) \end{align*}.$$

The coefficient of $u(C)$ above is strictly negative. Therefore, the more worth $C$ has (i.e. the higher $u(C)$ is), then all else equal, the lower your tendency to prefer the new proposition. If you want to maximise your expected utility, you will take the new proposition iff $u_{new} - u_{initial} > 0$.

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The new proposition effectively increases the probability of getting $B$ while decreasing the probabilities of each of $A$ and $C$, so your final decision is influenced by the weights of all three.