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To: emrs1 who wrote (13131)	6/11/2006 12:26:25 AM
From: E. Charters	Respond to of 78419

The reason as most people do not see is that the game changes once you block out an unknown door, which has only a 1/3 chance of being a cadillac and THEN a door which has 0 chance of being a Cadillac after it is opened for you. That means that the chance if 100% that the Cadillac is behind the door you selected -AND- the door that was not opened. It is now 33.33 per cent chance that the Caddie it behind your door as it always was, and 66.67% chance NOW THAT YOU kNOW WHERE ONE OF THE GOATS IS, that is NOT behind the door you selected. It ALWAYS is that SAME chance -- that does NOT CHANGE ... but now you know WHICH DOOR to change TO!!!!!!!!!! So we know that the chance it is behind the two doors, yours and the unselected door is 100%. Which is 33% for your "picked" door, and what must it be for the unselected door? 33.33+66.67 =100% So it is 66.67% chance to be behind the unselected door. Think of it this way. What chance is there that given any door, you have a goat behind it? 66.67% right? So when a door is opened and it shows a goat, the chances for the remaining doors have changed, right? The doors remaining must be Cadillac-goat, but they can be g-c or c-g. But remember the chance you landed on a goat is NOT 50%! It is 66.67% right?! So the chance you see are not g-c, c-g but G66-C33, C33-G66. The remaining doors, i.e. the one you selected and the one that was not opened that you did not select, can only be two ways, g-c or reverse, but which one did you land on most often, the g or the c? It is NOT even!!!! No matter where the Caddie is of the other two doors that are not opened, you landed on the goat most often. It sounds crazy, but what it means is that it is twice as likely that you landed on a goat as a Cadillac from the get-go, so the distribution you imagine for the remaining two doors is never 50-50. It nore often puts you on the goat, it is never even. The key is the first selection. It is 66% goat so to speak. Then when you add a goat to that duple you have one goat + a 66% chance for a second goat. So 66% chance you have two goats in that "fixed" duple. A goat plus one other door, IF the unknown door is first selected but not opened.. order matters... IS 66% goat-goat.. So it follows that it is 33% chance the remaining door is NOT a goat door. If I select a known goat-door, and then bid you pick, it goes back to 50-50. You select first, it is 33-66. EC<:-}

To: emrs1 who wrote (13131)	6/11/2006 2:27:47 AM
From: Gib Bogle	Respond to of 78419

There is an explanation along the lines of EC's post at mathforum.org I'd like to invite Mr Bayes into the discussion. First, some terminology. P(A \| B), where A and B are events, means: "The probability of event A given that event B occurs" You can convince yourself that the probability of both A and B occurring, written P(AB), is P(A\|B).P(B) and also that P(AB) = P(B\|A).P(A). This leads directly to the Bayes rule: P(A \| B) = P(B \| A).P(A)/P(B) Now I reckon I can use the Bayes rule in this situation. We are interested to know, in the case that I had guessed door #1, and door #2 opens, what is the probability that the Caddie is behind door #1? In Bayes rule, let A = The Caddie is behind door #1 B = Door #2 opens then we know that P(Caddie is door #1 \| Door #2 opens) = P(Door #2 opens \| Caddie is door #1)P(Caddie is door #1)/P(Door #2 opens) We can evaluate these 3 terms. P(Door #2 opens \| Caddie is door #1) = 1/2 (I'm assuming here that Monty Hall is equally likely to open doors 2 and 3 in this situation). P(Caddie is door #1) = 1/3 (this is the a priori probability) P(Door #2 opens) = ? This is a bit harder. We want the probability that door #2 opens, given only the information that I guessed #1. We have to look at the 3 possibilities for where the Caddie is. Caddie is door #1 --> Door #2 opens with probability = 1/2 Caddie is door #2 --> Door #2 doesn't open (prob = 0) Caddie is door #3 --> Door #2 opens with probability = 1 Combining these 3 probs, each weighted equally by 1/3, gives (1/2 + 0 + 1)/3 = 1/2 Now we can determine P(Caddie is door #1 \| door #2 opens) = 1/2 1/3 / (1/2) = 1/3 In other words, the probability that the Caddie isn't door #1 (i.e. that it is door #3) = 2/3. Luckily this is the same number that EC came up with. Was this helpful? Probably not.

To: emrs1 who wrote (13131)	6/11/2006 2:58:42 AM
From: E. Charters	Read Replies (1) \| Respond to of 78419

based on new information it now seems the Caddy has a 50-50 chance of being behind 1 and 3 (and zero of being behind door 2)? You mean the Cadillac has a 50-50 chance of being behind doors one or three. Well not exactly. It has a 100% chance of being behind one of the doors remaining, but the chance that it is behind the door that the contestant picked is always the same as from the first time he picked. 33%. The chance does not change from the contestant's point of view because someone else knows the position of a goat. He picks from 3, he gets 33%. He has first choice. From the remaining doors, the one not selected and the one with the goat shown, the field of the probability of the Cadillac being in that duple is still 67%. (If one selected is 33% the remaining two is 67%.) Always the same. If we see the goat door, the the remaining door has the only possible position for the 67% chance of finding the Cadillac. So we switch to a 67% probability from a 33%. It is like switching to a duple from a single. The duple always has the better chance. Why do you not stick with the duple of the one you picked originally and the goat shown? It is not the same. "Adding a goat" to one selected first at random makes that duple a 33% chance no matter what.