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</p>
<p>You should switch. Am I right?</p>
<p>Xiggi, I almost sent my original question to you in PM rather than start a thread, but I couldn’t guess as to your availability and I was in a hurry for a result. At my current level of comprehension, I will be able to figure out your post, and Simba’s response, and your next post, by, say, Christmas. And I certainly cannot figure out why there would be an advantage to switching from door #1 to door #2 if door #3 is eliminated. I mean, how could it matter - as the odds are still 50% that one or the other could be the correct door, right? So how could switching the doors improve the chances of guessing the correct door when there are only two to choose from? It seems to me that it doesn’t matter - chances are equal either way, with either door. I do get that the probabilities would be different when there are three doors, but, what does removing one of them have to do with one of the remaining doors suddenly being better odds? </p>
<p>Can you point me to a source that I can go to read to try to understand, on a remedial level?</p>
<p>I tried to google it before starting this thread, and that got me to sources from Yale, MIT, etc., it was completely hopeless as I couldn’t understand ANY of the material found in the links. Then I tried Wikipedia but that was too much the 50,000 mile view, with no real substance and no links to anything that would give me a visual concept. MIT has some very intriguing open courseware but the material is too advanced and the lecture notes are very barren - although there is some very interesting material on digital anthropology.</p>
<p>OK. So now you are an expert in probabilities. Here is a problem for you. There are three red hats and two black hats. There are three people standing in line - face forward (shown below).</p>
<p>A</p>
<p>B</p>
<p>C</p>
<p>They can not turn their heads. A can see B and C’s heads, B can see C’s head. C just has to stare at the wall. You jumble up the hats and put the hats on their heads (be careful or they will see the color). Then you ask the question what is the color of your hat?</p>
<p>Only one person (A or B or C) will be able to answer that. Who was that person? What color hat was he wearing?</p>
<p>Simba: My S was able to solve this problem in 4th grade!</p>
<p>well I first learned about this problem when I was 40. I think I had tough time. :)</p>
<p>I’ve been trying to offload S’s old math books. Lots of math puzzle books that he liked to take on vacation. Maybe I should send them on to you? :)</p>
<p>I am absolutely certain that I am not wearing a hat.</p>
<p>“If I choose door #1, and Bob then eliminates door #3 as a possibility, should I switch my choice to door #2 or stay with my original choice, door #1?<br>
You should switch. Am I right?”</p>
<p>Yes. You should switch.</p>
<p>You are choosing 1 out of 3. The choice looks like this… (1) or (2 and 3).
The doors left represent 2 out of 3. The fact that they show you 1 door out of 2 doesn’t change the odds, that you only choose right 1 out of 3 times.</p>
<p>Example…
Suppose you choose door number 1. You will be right 1 out of 3 times. 2 out of 3 times it will be door number 2 or 3. So you always switch to the 2 or 3 choice. The fact that they always show you one of the 2 doors you didn’t pick doesn’t change the original odds of 1 out of 3. </p>
<p>In the other example, A, B, or C, it’s C.</p>
<p>If A knows his color hat, you have to be wearing black. If A doesn’t know, C could be wearing black or red.</p>
<p>Then we go to B. If A doesn’t know then B looks at C. If C is wearing black, and A didn’t know what A is wearing then B would have to be wearing red. So if A doesn’t know and B sees black, B now knows he is wearing red. If A doesn’t know and B sees red on C, he doesn’t know what B is wearing. It could be black or red.</p>
<p>So we get this.</p>
<p>If A knows, it has to be B is wearing Black and C is wearing Black.
If A doesn’t know, B looks at C. If he sees black on C, B is wearing red.
If A doesn’t know, B looks at C. If he sees red on C, B doesn’t know if he is wearing red or black.</p>
<p>So if A knows what he is wearing, C knows he wearing black.
If A doesn’t know and B knows what he is wearing, C must be wearing black.
If both A and B don’t know what they are wearing, then C must be wearing red.</p>
<p>My daughter is a math major too. lol</p>
<p>dstark: Now you can go to 5th grade.</p>
<p>latetoschool: repeat pre-K</p>
<p>Thanks. The older I get the more my skills diminish.</p>
<p>Simba, I see no advantage in repeating nine months in the time-out chair. LOL.</p>
<p>Meh. How about the two envelopes paradox?</p>
<p>LTS, posting on the forum was the right thing to do; now, we can all have some fun spending some time on entirely trivial matters. Puzzles are a lot of fun, but can drive anyone crazy when the solution does not come easily. Our family loves to spend part of our Christmas vacation challenging one another with the silliest puzzles and trivia questions. Our neighbors think we are nuts! </p>
<p>Here’s a link to a nice illustration of the 3 hats problems. The early explanations are fun to watch. On the other hand, the last part of the presentation requires more of an acquired taste. </p>
<p><a href=“http://www.wcer.wisc.edu/stellar/algebra/Feb18.06/GreenLake.ppt[/url]”>http://www.wcer.wisc.edu/stellar/algebra/Feb18.06/GreenLake.ppt</a></p>
<p>For the three doors problems, I remember having some discussions about the problem a few years back when we had the old board. I believe that reading about the Vos Savant episode is interesting:</p>
<p><a href=“http://en.wikipedia.org/wiki/Marilyn_vos_Savant[/url]”>http://en.wikipedia.org/wiki/Marilyn_vos_Savant</a></p>
<p>
</p>
<p>Also see <a href=“http://www.montyhallproblem.com/[/url]”>http://www.montyhallproblem.com/</a> for a graphical explanation.</p>
<p>xiggi, thanks for those links. I didn’t understand dstark’s solution to simba’s problem, because ummm I guess simba didn’t pose the question properly, with all the constraints.</p>
<p>As for the Monte Hall problem, I struggled with accepting the correct answer for a long time until I expanded it slightly, as follows:</p>
<p>Instead of 3 doors, assume there are 100 doors! Only one has a great prize behind it, the other 99 doors have goats. You choose 1 door, then Monte eliminates 98 other doors that he knows have goats behind them. So you are left with the door you originally selected and one other door. Now do you switch?? (I sure hope so!!!)</p>
<p>(I see your link gives a similar example, calling it an exagerrated solution and uses one million doors for the example. I figured that out for myself!)</p>
<p>Mathematically, all this is fine.</p>
<p>In reality it is 50-50. You win or you lose.</p>
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</p>
<p>No. Having only 2 possible outcomes does not make the chances 50-50.</p>
<p>The problem/puzzle I like comes from Warren Buffet’s letter to shareholders some years ago (I think, in any case it comes from him) observing that it is a much better than 50/50 bet that if you find yourself in a group of 25-30 people or so at a party, that at least two of them will have the same birthday.</p>
<p>I actuallly tried to work out the odds of this based on the number of people in the room, and I think I came out with a 75% probability (or so) when the number of people got up to 30. </p>
<p>Buffets point was that if you could find someone to bet with, and you made a series of bets in that same situation over time, it would pay off handsomely.</p>
<p>blobof: That was a shortened version of an Aggie Joke.</p>
<p>Damn. Stupid internet screwing up my joke detecting skills.</p>
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<p>This problem can be done with 7 hats as well. The solution becomes quite a bit trickier though.</p>