someone comments on I’ll take their chances

[–] someone@lemmy.today 9 points 4 months ago* (last edited 4 months ago) (2 children)

The initial track has an expected death rate of 1, but that changes to 0 if we flip. So wouldn't we calculate out of 5 tracks for the denominator, meaning the expected death toll is 1 either way?

I understand the math of an expected death of 1.25 (1/4 x 5) and the concept of independent probabilities, but I am not sure we can disregard the knowledge that we are on a track that will have zero probability if we switch..

This reminds me of The Monty Hall problem. (https://en.wikipedia.org/wiki/Monty_Hall_problem) but that may not apply at all. Does someone with a stat background know which is the correct conceptualization?

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[–] EvilHankVenture@lemmy.world 6 points 4 months ago

When you flip it the original track is no longer possible, so it would not be considered. If it was considered, you would also have to consider the one person on that track. If all 5 tracks were equally likely the expected value would be 1.2 deaths.

This doesn't seem like it is similar to Monty Hall. In Monty Hall you win by switching every time unless you picked the winning door first. That's why you have a 2/3 chance to win by switching. This is just a random chance.

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[–] LodeMike@lemmy.today 4 points 4 months ago* (last edited 4 months ago)

We are told the probabilities directly. Expected deaths as number of trials approaches infinity when switching the lever is 0*¾+5*¼ which equals 1.25. Better odds than gamblers face by a long shot but not good.

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