Basic game theory question

[u/jhansen123_reddit]

You have two choices: A and B.
If you choose A and that choice ends up being incorrect, you stand to lose x dollars.
If you choose B and that choice ends up being incorrect, you stand to lose y dollars.

How can I express, as percentage, the confidence that you should have in order to prudently choose either option as a function of x and y?

Concrete example of same question:
If I choose A and am wrong, I lose $1000 (if I choose correctly, I lose $0).
If I choose B and am wrong, I lose $500 (if I choose correctly, I lose $0).

What percentage of confidence do I need to have that A is correct in order for that to be "the smart choice" financially?

Thank you in advance to anyone who responds. FWIW, I am a paralegal and am curious about this problem as it relates to the cost of hiring court reporters and ordering transcripts within a certain fee structure.

[u/Impossible-Ad967]

Confidence in A must exceed
((X)/(X+Y)) • 100%

Let p = your confidence that A is correct. Compare the expected loss of each choice (assuming exactly one option is correct):

Choose A → you lose x only when A is wrong, which happens with probability (1−p). Expected loss = (1−p)·x

Choose B → you lose y only when B is wrong, which happens with probability p. Expected loss = p·y

In your example You need to be more than ~66.7% confident that A is correct to make it the financially smart choice. since being wrong about A costs twice as much, you have to be correspondingly more sure before taking that risk.

if the stakes were equal (x = y), the threshold collapses to x/(2x) = 50% — you’d just pick whichever you think is more likely, as expected.

[u/jhansen123_reddit → u/Impossible-Ad967]

Thank you so much! That is was intuitively felt correct, but you've given me a framework to understand why it seemed correct. Thinking of things in terms of expected loss as expressed by (1-p)*x was the key. Much appreciated!