Re: Why is Deal or No Deal not the Monte Hall problem?
 

I'm not sure if this is going to help me, but what you and Sheets added made me think of this possible explaination:

- In the "LL's Monte Hall does DoND", I'm going to intentionally not show you the Ace.

- Also, the other card you end up with to choose from (in this case, the 5) is not going to be exposed either 90% of the time, because it's ....

(never mind, that doesn't work- too many conditions).

I have to come up with a simple explaination that makes it clear how unlikely I was to not have to expose the 5 in the big pile, along with intentionally not exposing the Ace in the big pile...

Something along the lines of "you knew I wouldn't expose the Ace, so how likely was it that I wouldn't have exposed the other card (the 5) before I'd gotten down to two cards to choose from?"

Maybe something with the birthday thing that Sheets came up with.... eh, I don't know. [img]/images/graemlins/confused.gif[/img]

Re: Why is Deal or No Deal not the Monte Carlo problem?
 

In the Monte Hall door scenario, if Monte picked the door randomly like you did in the card game then the player would lose the game 2/3 and win 1/3 of the time when switching. Why? The player can lose because of the the first pick or if not, then by swicthing. The player would instantly lose the 1/3 of the time Monte randomly picked the prize door first. When Monte didn't pick it first then the player now has a one of two chances when switching. So overall, 2/3 of the time the player would still have a 1/2 chance of loss=1/3 for a total of 1/3 + 1/3=2/3 total chance of loss.

When Monte specifically picks a non-prize door first (as in the non-random orginal version classic problem) it eliminates the "first pick" loss chance and leaves the player with only the 1/3 chance of loss when switching.

In your card game the player would instantly lose if you happened to pick an Ace. You left the player the chance of the "first pick" loss along with the potential loss from switching so it is equivalent to the Monte Hall random pick scenario.This is NOT the original version of the problem. Under your card game scenario, the chances of loss at the end must be 50%. You could re-configure the card game so there was no immediate player loss from you picking an Ace. So if you found an Ace, you would replace it without showing. Then when it got down to one card, the player would have 90% chance of winning by switching.
http://farm3.static.flickr.com/2418/...0bd68580_t.jpg

Re: Why is Deal or No Deal not the Monte Carlo problem?
 

[ QUOTE ]
What is it about one condition- being forced to exposed losers only- that makes the probabilities different in the end result?

[/ QUOTE ]

Player needs ace to win.

Player picks a card.

You randomly expose all but one.

Just ace and 5 are left.

This is 50/50.

What we are forgetting is 80% of the time the game ends when we expose the ace.

So---of the 20% of the time we randomly expose cards and no ace appears, we are 50/50 by switching.

If we look at the cards, the game is always played out, and the ace is on our side 90%.

So, by not looking at the cards, game ends 80% of the time.

So from this point ---the 20% played out with no ace---10% he has ace, 10% the other card, we are 50/50 (10% of 20%).

If you want to make more conditions, the ace and 5 both play say, then the game ends every time those cards are exposed.

So of the times that only the 5 and ace are left , it is 50/50.