It’s not a Monty Hall thread without someone bringing up the third possibility: Monty knows where the car is and will reveal it if possible.
Under this scenario, if Monty reveals a goat, you can reason that he must have been choosing between two goats. You will always lose if you switch and will always win if you do not switch.
Note that the above is 100% compatible with the Vos Savant phrasing of the problem.
Oh, and the OP is a trick question: As you only remove balls in pairs, you never remove the last ball. The final ball removed will be one of a pair, and the final pair must be one red ball and one blue ball. As you remove them two at a time, neither can be fairly said to be the ‘last’ one.
Agreed that is consistent with various ambiguous descriptions including Vos Savant’s. What it’s totally inconsistent with is the actual performance on the show.
Monte did not routinely reveal the car. Which under that proposed decision rule would happen about 2/3rds of the time after the contestant randomly picked his/her first choice.
Hmm, interesting point, Don. You and Tim are both right about the fact that there will be two balls in the last step and we know exactly what color those two balls are. Then we are instructed to “remove” them as a pair, which means…
the last pair we remove will be one red and one blue (because the number of red balls can’t change from odd to even). So at that point you’ve got a red ball and a blue ball in your hand. But look at the instruction which says “add in one extra red ball”. If we interpret that to mean put the same red ball which is already in your hand back into the bag, before you take your hand out of the bag, then it would be fair to say the “final ball removed” is in fact blue. The single ball left in the bag is red, but it never gets removed.
The version which I saw on reddit goes like this: I have a bag of 17 red balls and 17 blue balls. I (blindly) remove two balls. If they are both the same colour I put a blue ball back in, if they are different colours I put a red ball back in. I repeat this until only one ball remains. What colour is it? Notice the reddit version doesn’t ask what color is the final ball removed, but rather it asks what color is the one ball which remains. I like this version much better, except it still doesn’t address the problem of how you can “put a blue ball back in” if the first two you grab are both red. Perhaps this could be cleaned up by saying “if they are both red, paint one of them blue and put the freshly painted one back in the bag”.