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SPOILER ALERT!!!
This is the official Monty Hall Discussion thread.....which CONTAINS THE ANSWER to the Monty Hall Puzzle found on the ChessandPoker.com homepage. I STRONGLY urge you to read the original text before proceeding: Quote:
Am I correct that it's always to my advantage to switch based on the facts presented or Is my antagonist from the audience correct that it doesn't matter either way since each door has an equal probability of hiding the car? Please try to reason out the answer for yourself BEFORE you check the solution. What fun is a puzzle if you don't try to solve it yourself? And besides...you might be surprised by both the result and the delicious logic behind the reasoning! THE REVEAL (scroll down to reveal the solution and discussion) ...... ...... ...... ...... ...... ...... ...... ...... ...... THE ANSWER Heh.... Of course I'm right!! It's always to my advantage to switch. But why? THE LOGIC ![]() When we made our primary decision to select Door #1 each door had an equal probability based on the facts provided by the puzzle. As the graphic above shows, each door held a 1/3 chance (33.3%) of hiding the car. However, once I made my selection, Monty Hall proceeded to add additional parameters to the equation. He opened up Door #2 and revealed a goat behind it. That made a HUGE difference in our resulting dilemma, should we stay with our original selection of Door #1 or switch to Door #3. Mathematically...we should always switch, and here's why! ![]() We can view our selections as two subsets of groups as detailed in the second graphic above. Our original selection is now segregated into its own section, and the two remaining doors in their own respective section. As we can now see, our decision is in fact made quite a bit easier. We selected Door #1 to start the puzzle off, which still holds a 1/3 chance of hiding the car since we selected it from 3 doors all of which held the same probability of winning for us. However, now that Monty was forced to open one of the doors to reveal a goat based on a few specific parameters (he cannot reveal the car, and if both hid goats he would choose one of them randomly) we have SIGNIFICANT information to assist our decision. Actually, it makes it a lock! The two doors outside of our original Door #1 selection are now the focus of attention. And mind-bogglingly enough, since we know Door #2 hides a goat, Door #3 now actually has a 2/3 (66.6%) chance of hiding the car! Why is that? Because the 1/3 probability Door #2 originally had cannot just disappear. In fact, Door #3 effectively absorbs its chance since both doors are grouped together now as my alternative to keeping Door #1. ![]() Not convinced yet? Here's a further example. In the third graphic above we'll utilize playing cards to illustrate the point further. We now see 5 playing cards face down. One of the cards is an Ace, and the other four are the lowly deuces. We of course want to find the mighty Ace! We'll once again be forced to select one card out of the five without any advanced knowledge, so what the heck let's go with Card #1. As the graphic shows, each card originally has its appropriate 1/5 (20%) of being the Ace. Once again Monty does his thing but this time he must turn over THREE cards and of course cannot reveal the Ace if it's out there. ![]() As we see in the fourth and final graphic here, while our original selection of Card #1 still has its initial 1/5 chance of being the Ace, since we now have three of the Deuces revealed, the lone face down card in the remaining card group now has a whopping 4/5 (80%) chance of being the Ace. As you can imagine, it is clearly still mathematically in our best interest to always switch. THE DEBATE This solution is often debated as being inaccurate, with each side defending their beliefs quite passionately.....and I expect nothing less from our inquisitive readers! Please use this thread as a platform for the discussion. Debunk it, Debate it....Let us hear what you think! -James Yates |
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