Don’t know ask me how this discussion got started but it did. The topic at hand was the probability of choosing the right door that contained the prize. You’ve heard of that game show where the host gives the contestant a choice of three doors. They select a door and then the host shows them one of the door that does not contain the prize. The contestant now has the option of sticking with their first selection or changing their mind. This is where the probability factor start. It’s a bit tricky to understand, so I’ll try to explain it the way I understand. Of course, I may be wrong.

The problem goes a little something like this. You are given three choice of door to choose from. Behind one of those doors is a prize, say $1,000,000. Behind the other door is something that you don’t want, say a fruitcake. You are asked to select a door, say you pick Door #3. One of the other doors, Door #1 and Door #2, are opened. Usually, it’s the door with the fruitcake behind it, that is shown to you, say Door #1. Now, you given the option of sticking with your choice, Door #3 or switching to the other door, Door #2. Do you stay with your selection or do you switch?

From what I’ve read, the probability of picking the door with the $1,000,000 is higher if you make the switch rather than staying with your selection. The way I see it is like this. You start off with three doors. All three doors have the same probability of containing the $1,000,000 behind it, 33.3%. After selecting a door, one door is eliminated. Now, you have to choose from two doors. By staying with the door you selected, you odds are still 33.3% but by switching, your odds are higher because you’re selecting from two doors rather than three. So it makes sense to say that by switch you’ve increased your odds. By switching to the other door your odds are now 50%. It is a common intuition to belieive so, but it’s not quite right.

You have to try and understand this without numbers because according to the numbers, it makes sense. But the rule of probability states that the probability of A given B equals the probability of both A and B occurring, divided by the probability of B occurring.

= 1/2 x 1/2 / 1/2

= 1/4 / 1/2

= 1/4 x 2

= 2/4

= 1/2

= 50%

Using the numbers from the three doors, you get:

= Door #3 x (Door #1 + Door #2) / (Door #1 + Door #2)

= 1/3 x 2/3 / 2/3

= 2/9 / 2/3

= 2/9 x 3/2

= 6/18

= 1/3

= 33.3%

The probability of selecting a door with the prize are evenly distributed. So by staying with your selection, the calculations are as follows:

= 1/3 x 1/2 / 1/2

= 1/6 / 1/2

= 1/6 x 2

= 2/6

= 1/3

= 33.3%

If you switch, you get this:

= 2/3 x 1/2 / 1/2

= 2/6

= 2/6 x 2

= 4/6

= 2/3

= 66.6%

Understanding this problem depends on your interpretation of the situation. During the second stage, you’re picking between two doors, so you assume that probability is 50/50 but you have to include the probability of your first selection. The next time you get a offer to pick between three things, it’s best to make the switch after one option is eliminated. Your chances of making the right selection is higher. It can go both ways. After making the switch, you may have been right from the start.

But the idea is saying that if you switch, you will get it right more often because your initial odds were 33.3%, so by making the switch, you’re switching over to the 66.6% side because now you’re getting another chance to pick. The 33.3% from your first pick plus the 33.3% chance of your second pick equals 66.6%. If you don’t pick, you’re losing that extra 33.3% chance. It makes sense to choose the one that will give you a higher chance of winning, hence you switch.

My interpretation of the situation is different.

The total probability of selecting one of the three door with the 1,000,000$ with 2 chances is as follows:

WIN % = You have a 33% chance at winning right off the bat [1/3]

Should the first selection fail, your chances at making the right choice is the probability at choosing the wrong selection first: [2/3] multiplied by the remaining chance at selecting the correct box (which is 1 out of 2 boxes left) [1/2] which is 33%.

So in total you have a 66% chance in selecting the correct box!

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LOSE % = There is only one way to lose and that is to make the wrong selection first which you have a [2/3] chance in doing. Doing so leads you to a 1 in 2 chance at making the wrong selection again [1/2]then. You multiply these both to get a 33% total chance at selecting the wrong box.

lol actually what i just said is the same thing as what u expressed mathetmatically, but is more wordy hehe :P.

hahha… thanks for clearing that up.. now everyone will know what to do if they are given this scenario.. I think I’ll stick to the numbers… people like numbers..