Regarding the infamous Monty Hall problem:
I’ve sketched out all the possible answers on paper, and still can’t see how switching doors would have any effect. Rather than revel in ignorance, I’ve decided to take Cecil’s advice and play the game myself. I propose the following experiment:
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Take three unmarked playing cards from the same deck: two jokers and one king. The cards will represent the doors–jokers are doors with no prize; the king is the door with the prize.
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Shuffle cards and place face-down.
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Roll a single six-sided die. If I roll a 1 or a 4, I choose the first card on the left. If I roll a 2 or a 5, I choose the middle card. If I roll a 3 or a 6, I choose the card on the right.
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I move the chosen card to the side.
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I flip a single fair coin once. This represents Monty. If it lands heads, I turn over the remaining card on the left. If it lands tails, I turn over the remaining card on the right.
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If the turned card reveals the king, it is declared “no game.” Monty wouldn’t do that in real life; but as we are relying on random chance to make our moves, this sequence of events will pop up.
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Assuming Monty turns over a non-prize door, I flip the coin again. Heads, I keep my original card. Tails, I switch for the remaining unturned card.
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I turn over my final chosen card to finish the game.
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Repeat.
Now, before I spend an afternoon flipping and rolling, I’d like some feedback. Does this seem like an accurate representation of the game? I wanted to remove all possible bias and make all the moves purely random. So the die chooses my original “door,” the coin chooses Monty’s door, and then the same coin chooses whether I keep or switch. Is that good, or am I missing a subtlety somewhere?
Second, how many iterations would be required to make a reliable test? I’m thinking at least 100. I figure 1/3rd of the runs will end in a “no game,” and I want enough good runs to have some confidence in the results.
Any suggestions or advice? Thanks in advance,
B.