You are asked to play a classic shell game with two shells.

Each round, a ball is placed under one shell, the shells are rapidly, continuously rearranged for 15 seconds with distraction*, and then you are asked to pick which shell contains the ball.

*By "with distraction," all I mean is that the guy shuffling the shells tries to distract you, and sometimes he is successful, so at the end of each round, you tend to have some knowledge of where the ball is, but you are rarely 100% confident.

If you are correct, you get 4 points.

If you are incorrect, you get -1 points.

If you aren't sure which shell has the ball, you are allowed to opt out of the choice, and receive 2 points.

In order to perform the task optimally, how confident do you need to be each round (in the form of a percentage) to guess (i.e., to forgo the automatic 2 points and shoot for 4)?

Please show your work.

(Yeah, this is simple as hell, but my answer is different from a friend's, and I want to be sure which of us is right.)

  • 2
    $\begingroup$ How is this not just a math exercise in probability? $\endgroup$
    – Rubio
    Jul 1 '17 at 19:00
  • $\begingroup$ When you say that the shells are rearranged with distraction, is that equivalent to saying that the ball position is random? $\endgroup$
    – MikeQ
    Jul 1 '17 at 19:05
  • $\begingroup$ @MikeQ no, i was just trying to imply that the position isn't obvious at the end of each round. your level of confidence in the ball position varies from trial to trial, depending on how distracted you are in each round. $\endgroup$
    – dbliss
    Jul 1 '17 at 19:06
  • $\begingroup$ @MikeQ and if you're asking whether the guy moving the shells picks the final position of each shell randomly, then i guess you can assume the answer is that you wouldn't know whether he does or not. (he doesn't always put the shell with the ball on the left, for example. he doesn't seem to prefer one side over the other.) $\endgroup$
    – dbliss
    Jul 1 '17 at 19:12

Expected Value = Probability of Correct * Reward if Correct + Probability of Incorrect * Punishment for Incorrect

The question is, if we are at a certain probability or below, we should pass. What is the probability that we should pass at? The break even probability will be the one at which the expected value of passing is equal to the expected value of picking.

P(correct choice) = p  
EVp = p * 4 + (1 - p) * (-1)  
2 = 4p - 1 + p  
5p = 3  
p = 3/5  

If your odds of being correct are 60% or greater, you should pick the shell. Otherwise, it is better in the long run to pass.

  • $\begingroup$ this is what i got. so, at the least, we think alike. hopefully we also have great minds. my friend says the answer is 66%. not only that, he is a professor at a major research university. not only that, he published his answer in a leading experimental psychology journal. it made it through peer review. $\endgroup$
    – dbliss
    Jul 1 '17 at 19:20
  • $\begingroup$ Well if it's published already he'll hear about it I'm sure. To be honest there could be more information in the scenario than what you have posted here. $\endgroup$
    – LeppyR64
    Jul 1 '17 at 19:25

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