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I am now pretending to be a newbie student. I write the following sample space for the Monty Hall problem (It is a famous brain teaser, I assume you know it).

$$ S=\{ (C,G1),(C,G2), (G1,G2), (G2,G1) \} $$

where the first tuple represents the first choice taken by the guest and the second tuple represents the optional switch offered by the host.

As a teacher I have difficulty to explain that $\{(C,G1),(C,G2)\}$ must be simplified to just one $\{(C,G)\}$ and $n(S)=3$. $C$ is the car, $G1$ and $G2$ are the goats.

Do you have any idea to explain it?

Edit

More information:

  • for $(C,G1)$, if you switch you will lose
  • for $(C,G2)$, if you switch you will lose
  • for $(G1,G2)$, if you switch you will win
  • for $(G2,G1)$, if you switch you will win
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  • $\begingroup$ [Not quite an answer, but] I think of the initial sample space as being: goat, goat, car. (Picking here gives you a 1 in 3 chance of finding the car: 3 choices, random.) Once the contestant picks one of these, there is now more information; specifically, a goat is revealed. So the modified sample space is: goat, car. (Picking here - i.e., swiching - gives you a 1 in 2 chance of finding the car: 2 choices, random.) The parenthetical remarks here are how I, personally, understand that switching is wise: for winning probabilities, 1/2 > 1/3. $\endgroup$
    – Benjamin Dickman
    Commented Oct 9, 2018 at 20:26
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    $\begingroup$ @BenjaminDickman: But you win 2/3 of the time by switching! $\endgroup$
    – Davis Herring
    Commented Oct 10, 2018 at 0:57

1 Answer 1

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I think the key here is that $(C, G_1)$ and $(C, G_2)$ are each only half as likely as each of the other two cases - and the standard "counting" approach to probability only works if all the cases are equally likely. To fix it, you need a third event - think of it as Monty flipping a coin. If it comes up heads, he picks the leftmost goat that's still hidden; if tails, the rightmost. If the player picks a goat, the coin flip doesn't matter, but it happens anyway. This gives us six outcomes: $(C, H, G_1), (C, T, G_2), (G_1, H, G_2), (G_1, T, G_2), (G_2, H, G_1), (G_2, T, G_1)$.

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    $\begingroup$ The fact that the coin flip doesn’t matter in the initial-goat cases corresponds to the crucial part of the setup that Monty knows where the car is and actively avoids it. This avoidance conveys information, which is how you can win so often. A Monty with no information can reveal none, but must sometimes ruin the game by revealing the car. $\endgroup$
    – Davis Herring
    Commented Oct 10, 2018 at 3:04

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