Monty Hall variation stemmed from Buzztime bar games.

[Eihli]

This problem entered my mind while watching the Saints last weekend at Buffalo Wild Wings.

The Monty Hall problem comes from a game show where a contestant gets to choose 1 of 3 doors. Behind 2 of the doors are goats and behind one of the doors is a new car. After the contestant chooses a door, the host of the show eliminates one of the other 2 doors that has a goat behind it. Then the contestant gets the option to switch doors or keep the one they originally picked. Most people initially think there is no difference in switching or staying, but your odds actually double if you always switch.

There is a Buzztime (http://www.buzztime.com/platforms/potentialsub/index.htm) Trivia game that has a variation of this problem that I was trying to figure out.

You are given a question and 5 answers to choose from. The faster you choose, the more points you get if your answer is correct. If you answer within 3 seconds you get the maximum 1000 points, then it drops to 0 over the next 10-15 seconds. As the time ticks down, it starts giving you hints that eliminate certain answers. So after 5 seconds you'll get a hint that eliminates answer B, but if you guess after the hint you'll only get 800 points.

So I'm trying to figure out the highest EV way of guessing. If you don't know the answer and guess D, then the game gives you a hint that discounts C, would switching your answer be higher EV than staying with the one you picked?


I would think it wouldn't matter since unlike the Monty Hall problem, where the door you guess can't be discounted, the answer you guessed in the Buzztime game CAN be discounted, but I'm not very good at math/logic.

[vhawk01]

Its not quite the Monty Hall problem though, because the game doesn't choose to eliminate one of the choices you DIDN'T pick, it simply gives the clues it is supposed to give. If you wrongly pick D, and clue 1 was meant to eliminate D, you still get clue 1.

[madnak]

It depends on exactly what the structure is for the hints and decreasing outcomes.

At first you have a 20% chance at 1000, worth 200. If you eliminate one answer and it drops to 800, then you have a 25% chance at 800, also worth 200. If it then eliminates another and drops you to 600, you have a 33% chance at 600, 200 value again. And then a 50% at 400, presumably. But then it drops to 0.

Using this structure, since the EV is always 200, in order to minimize variance you want to choose as late as possible. Unless you feel lucky. But this is an easy problem because there are discrete stages and you can calculate the specific EV for each stage.