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Maths from the Monty Hall of Fame

July14

montyhallMonty Hall (left) was a game show host on “Let’s Make a Deal” who offered contestants a brand new car. All they had to do was choose the lucky door out of 3 possibilities. There was, clearly a 33.3% chance of choosing the car, right? Yes…right! However, after you chose a door, Monty Hall would, in a rather surprising move, open one of the other doors and, surprise… behind it was a booby-prize (not a goat as these would be too messy). He would then say, “Now do you want to change your mind and choose the other door?”

Of course, this new option would mess with the contestant’s mind and make them wonder whether they were being “set up”. What should they do – stay with their original choice or switch? This became known as “The Monty Hall Problem“. Mathematically, there is a strategy for this kind of “choice” which will improve your chance of winning the car (or some other prize). Check out how this works:

Now, this dramatic use of the Monty Hall problem is quite good, but if you didn’t fully understand the Mathematics behind it, might prefer this explanation (click on the image to link to the movie):
game show goatH3 Note: Should the Monty Hall Problem apply to us in life’s choices? For example, you get accepted for a job and turned down for another. Then, a third option opens up. Now, that is something to think about! (note to Editor: “Don’t be such a goat!”)

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y(y-8) = 9 –> y.y – 8y – 9 =0
–> (y-9)(y+1) = 0, therefore y = 9 (can’t have a distance of – 1 for the other solution for y)
Using the top and bottom of the rectangle,
x = (y-8)(y+2) = (9-8)(9+2) = 11
but, the left side = (x-4) = 11-4 = 7, but rhs = y+? = 9+?, which is greater than the value of the opp. side??
[I think that the left had side was a mistake and should have read (x+4)?]

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