Here are two variations on the Monty Hall question - one where the host knows and one where the host doesn't. There is consensus on the former, but I'm interested in the latter.(adsbygoogle = window.adsbygoogle || []).push({});

I'm using playing cards here, but the principle is identical.

THE HOST KNOWS

The object is to find the Queens.

There are 6 rows of 3 cards, all face down. In each row there is a Queen and two Jacks.

From each row you select a card and keep it face down.

From each remaining pair of cards, a host who knows what they are turns over one of them and reveals a Jack. (So there is no misunderstanding, the host will always do this)

So now there are the six cards you chose and six cards not eliminated.

To acquire the most Queens, should you stick or swap? Or are the odds 50/50?

Monty Hall Answer: Swap - advantage 2/3.

Reasoning: probability was always that you originally chose 2 Queens and 4 Jacks. You know the other six cards are the opposite of your choice. Therefore, on average, the other cards will most likely have 4 Queens and 2 Jacks.

Essentially this is standard Monty Hall. I'm assuming we're already agreed on this answer. Now consider the following scenario, which is the same in all respects except for the host's state of mind.

THE HOST DOES NOT KNOW

The object is to find the Queens.

There are 6 rows of 3 cards, all face down. In each row there is a Queen and two Jacks.

From each row you select a card and keep it face down.

From each remaining pair of cards, a host randomly eliminates a Jack. (The host's random elimination is defined as follows: if a Queen is discovered, you simply start again; all the cards in each row are shuffled and you make a new selection; keep doing this until you get the desired outcome from the host - 6 eliminated Jacks.)

So now there are are the six cards you chose and six cards not eliminated.

Is the solution the same?

Simon

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# Monty Hall - a conundrum within a conundrum

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