There's a 5/6 chance someone is put onto the side rail (the one the trolley won't go down without interfering with it)

There's a 1/6 chance someone is put onto the main rail (the one the trolley WILL go down)

You're more likely to be on the side track if you're involved in this scenario, so if you wanna get hit you SHOULD try to flip it (if you're the one on the side track, it guarantees a hit. If you're one of the 10 people on the main, you have a 90% chance of having a dead switch and should try anyway)

Unless being tired at work is making me miss something

Except that if people are chosen randomly there is 2/3 chance that you are on the main track according to Bayes. Let's assume there are 10 people.

The probability to be chosen is 1/6 (all are chosen if 6 is rolled) + (5/6) × (1/10) (only one is chosen to go to the side track if 1-5 is rolled) = 15/60 = 1/4.

The probability that you are on the side track knowing that you have been chosen is the probability that you have been chosen knowing that the side track is selected (1/10) × the probability that the side track is selected (5/6) divided by the probability for you to be selected at all (1/4), so (1/10)×(5/6)/(1/4) = 20/60 = 1/3. So there is a 2/3 chance that you are on the main track.

If you do not flip the switch, (2/3)×10 = 20/3 people die.

If you flip the switch, 1/3 (you if on side track) + 10 × 2/3 × 9 / 10 (switch misfires 9 out of 10 times if on the main track) = 190/30 = 19/3 die. This is slightly better than not flipping the switch, you save 1/3 people more. That's an arm and a leg.

Half the fun of trolley problems is adapting them to puzzles for which they are utterly unsuitable: