Pick a Door (Math)

Three Hundred Eighty Eighth Post: Pick a Door (Math)

There is a classic puzzle dealing with 3 doors and one prize. The player chooses and then 1 of the 2 remaining doors is taken away. This leaves 2 doors. The question is should the player switch or stay with the original answer.

Well this problem is explored in the book “Statistics and Probability with the TI-89,” by Brendan Kelly. The theme behind this book is presenting the problem with just enough information to allow the reader to reach their own conclusions.

My opinion on this puzzle is indecision. I ran a random number program on the calculator that the book supplies. I tested the door problem by both switching my answer and staying with the original. I didn’t believe it but those who say to switch are correct. Their claim that the player should switch because once one door is removed the probability is ½ . The first answer was chosen with a 1/3 chance.

This didn’t make sense to me. Isn’t the original choice also ½ once one is eliminated. I can follow the logic that by switching the answer is now ½ , but even after running the program I still don’t agree with it. Unfortunately I see no easy way to prove it wrong.

If one was to take 3 lottery balls. One ball would be marked with a red dot (that the player couldn’t see.) If the user chooses 1 of the 3 and then one is taken away. So 2 lottery balls remain.

Would it be correct to say that each ball has a ½ chance? And then check the answer with the red dot.

This needs to be explored further. I know the calculator still says you win more often by switching your answer, but what is the reason for this? Is there mathematic evidence? I have a “hunch” that there is still much to debate about this problem.

References: “Statistics and Probability with the TI-89,” Brendan Kelly, Brendan Kelly Publishing Inc. 2000

So until the calculate gives the answer we’re looking for: May the Creative Force be with You