Goat and the car
Tuition love rekindled! The satisfaction of getting a student understand the concepts is orgasmic! After a couple of idiotic students (who simply refuse to help themselves), I've got a new hardworking student. Coincidentally, she is from VJC. I've yet to meet a student who is non-victorian yet works as hard. Perhaps i teach in the east...
You are on a television game show called “Let’s Make A Deal”. The show’s host gives you the choice of three doors to open. Behind one door is the grand prize, a car. Behind each of the other two doors is a goat. You choose door number 1. Before you open this door to see what’s behind it, the host tells you that he will open a goat door (he knows what is behind each of these doors). He opens door number 3 and a goat appears behind it. The host now asks you: “Do you want to stick with your original choice of door number 1, or switch to door number 2?” How would you answer? (Hint: Compare the probability that the car lies behind door 1 with the probability that it is behind door 2, given that door 3 houses a goat). - My econometrics tutorial, answer is pasted in comment box.
3 Comments:
Death of Flipflop
Anyway, the answer...
Let Ci represent the event that “the car lies behind door i” and G3 the event that “the host shows you a goat behind door 3”. First note that: P(G3) = P(G3 | C1) × P(C1) + P(G3 | C2) × P(C2) + P(G3 | C3) × P(C3) = (1/2)(1/3) +(1)(1/3) + (0)(1/3) = 1/2 The probability that the car is behind door 1 before the goat door was opened is 1/3. The opening of door 3 does not change this probability (imagine that you were not given the chance to change your choice). In other words, P(C1 | G3) = P(G3 | C1) × P(C1) / P(G3) = (1/2 × 1/3) / 1/2 = 1/3 If the probability that door 1 is the correct door is still 1/3, then the probability that it’s the wrong door is 1 – 1/3 = 2/3. But if door 1 is the wrong door, then door 2 must be the right door, since door 3 houses a goat. So the probability that door 2 is the right door is also 2/3. Formally, P(C2 | G3) = P(G3 | C2) × P(C2) / P(G3) = (1 × 1/3) / 1/2 = 2/3 Hence, you should switch your choice to door 2 because your chances of winning the car will be doubled by doing so.
jc got learn before!
wah cheem i'm not cut out to be a mathmatician or take part in such game shows
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