Yes, it applies. Assuming you switch to the other box, you lose if and only if you originally pick the correct box, which has probability 1/22. Turns out the probability of winning £250,000 is 21/22.
If you are playing deal or no deal and you pick a box at the start of the game that you think contains £250000(top prize) out of 22 boxes and while you are playing most of the boxes are eliminated, except from your box, and another box that you know only contains £1.00.
In this specific case does the monty hall paradox principle apply? I know the host has no prior knowledge of what is in the boxes but everyone knows that there is now only two boxes, one that contains £1 and one that contains £25000.
So you are now left with a box that contains £250000 and a box that contains £1.00. If you are given the option to change your mind about which box you want would it make sense to do so?
I disagree. This is NOT the Monty Hall problem, because in that sitation the host specifically opens a door that he knows does not containg the grand prize - in other words his openng of a door is not random. But in Deal or No Deal all cases are selected randomly. So if you get down to just two cases remaining then the odds of the big priize being in either one is the same, and you don't improve your odds by changing your selected case. You can apply Bayes theorem here:
P(my original case is the winner given no winner in the first 20 = P(my original case is the winner & no winner in the first 20)/P(no winner in the first 20) = (1/22)/(2/22) = 1/2.
P(the 21st case is the winner | no winner in the first 20) = P(the 21st caseis the winner & no winner in the first 20)/P(no winner in the first 20) = (1/22)/(2/22) = 1/2.
So your odds of winning are 50:50 either way.
But still, 20 of the boxes are eliminated; it doesn't matter if the host knew whether they contained the prize or not. For the Monty Hall problem, assuming you switch, you lose if and only if you originally pick the correct door. Therefore you win if and only if you originally pick an incorrect door, which occurs with probability 2/3. Can it be generalized to more doors?
Agreed that for Deal or No Deal it doesn't matter what the host knows, because he makes no decisions in the game. But for the Monty Hall problem the host DOES make the decision as to which doors to open, and he specifically selects doors to open that he knows do not have the prize. If you generalize to N doors and assume that the host opens N-2 doors that he knows do not contain the prize, you chance of winning if you don't change to the last remaining closed door is 1/N, and you odds if you do change are (N-1)/N. If the game had 100 doors with 1 prize, and the host opens 98 for you, then switching from your original choice to the last remaining closed door improves your odds to 99/100.
In the TV game show called "Let's Make a Deal" an audience member is presented with three closed doors, and is told that behind one of the doors is a fabulous prize while behind the other two was worthless junk. The contestant has to pick a door, either 1, 2, or 3. The Monty Hall Problem supposes that the host (named Monty Hall) knows which door contains the prize, and he opens up one of the doors that he knows does not have the prize, then asks the contestant if he'd like to switch his choice to the remaining closed door. So for example if the contestant picks Door 1 and Monty knows the prize is behind Door 2 he would open door 3 to reveal it's empty, then ask if the contestant would like to change his choice from Door 1 to Door 2. The question is - should the contestant do so? The answer is yes, as it improves his odds of winning from 1/3 to 2/3. As shown in the attached table there are nine possible cases to consider and in 6 of the 9 situations the player wins by changing.
But in the Deal or No Deal game it's different because no one is opening cases that they known ahead of time are not the winner. If you were to write out all the possible occurrence in a 3-case game (to keep it simple), limiting it to situations where the player has already selected a case to open that was not the winner, there are twelve possible combinations to consider (leaving out the situations where the first case opened is the winner, shown as "N/A" in the attached table). Six of the twelve are helped by changing and six are hurt by changing, so the odds of winning are not affected by whether you change your choice or not.
Wait a minute, so in "Deal or No Deal" the host can randomly open the winning door? That's what you implied in the past attachment.
For the Monty Hall three-door case, we can assume that, without loss of generality, that the contestant originally picks door number 1. If door 3 is removed, then the player has a 2/3 chance by switching to door 2. If door 2 is removed, the player has a 2/3 chance by switching to door 3.
If door 2 or 3 is eliminated and it doesn't contain the prize, then it doesn't matter whether the host previously knew whether it contained the prize. All that matters is that the door was eliminated, and that it doesn't contain the prize.
So if Monty truly doesn't know where the prize is the player has a 6 out of 18 = 1/3 chance of winning regardless of whether he changes his choice or not. This is a different result than if Monty knows where the prize is and specifically avoids opening that door. Under this scenario the number of possibilities collapses to just 9, reason being that all the "N/A" are eliminated, and the cases where the player has selected the correct door are reduced from 2 to 1 as it doesn't matter which door Monty opens:
There are no "doors" in deal or no deal - there are brief cases. And the host doesn't open anything. Perhaps a refresher is in order (I had assumed that everyone commenting on this thread was familiar with both games - shame on me).
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