Re: Why is Deal or No Deal not the Monte Hall problem?
I got to the same place as you did, Shane, up to here.
[ QUOTE ]
The difference between that is the cards in this case are picked at random, without knowledge. In the Monte Hall problem, if we always have to show 8 of the 9 cards (and none can be the ace) then the last 1/9 probability becomes a probability of 1 (since we can always flip over 8 cards without an ace turning over if we know which one is the ace). Thus if the 'host' flips over 8 of the 9 cards and tells us he isn't flipping an ace, then the probability the unflipped card is an ace is (9/10) * 1 = 9/10, and we should switch.
[/ QUOTE ]
This is where I'm running into problems trying to translate for the person. The person's argument was, if it was 90% for the Ace to be available in the big pile... why isn't it 90% for the 5 to be in the big pile instead?
I know we're not lowering the percentage chance of a card being left from the big pile, because we're intentionally not lowering the initial probability (we're only exposing bad cards). My friend sort of understands that.
But their argument is that I am arbitrarily picking the Ace to be the 90% card, rather than the 5, to say that you should always switch in a Monty Hall version of Deal or No Deal.
I can't figure out a way to explain this sticking point.... especially when you compare end results of the Monty Hall version (8 unwanted cards exposed, down to choosing two, when we want the higher value one) and the DoND version (8 unwanted cards exposed, leaving 2 of which we want to pick the higher value one).
It's hard to look at that and say "Oh yeah, one should be 90% and the other 50%"... and even harder to think of a way to explain it.
Any more help, Shane? Or anyone else?
|