

Trading two for one - A Monty Hall variant. - RiderOfGiraffes

Suppose I have ten cards, one of which represents a prize of great value. I shuffle them, spread them out face down, and let you choose any two. You take them, but you don't look at them.<p>I take the remaining eight cards and look at them. One by one, I show you six of them. I want to keep the tension high, so I definitely don't reveal the card of great value. After all, I might not have it.<p>After revealing those six of the cards, I have two, you have two, and I make you this offer:<p><pre><code>    You can choose and keep one
    (but only one) of my cards,
    if you give me both of yours. 
</code></pre>
Assuming this isn't a card trick, and that everything is as I've described - should you accept?
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po
From the outset when you take your cards you have a 2 in 10 (20%) chance of
having the prize. The dealer has a 8 in 10 (80%) chance of having it. After
the dealer looks at and discards six cards it is still a 8 in 10 (80%) chance.

I would trade my 20% chance for a 50/50 shot on the 80% chance which is 40%.

It's a good variation on the problem though. The thing is that the Monty Hall
problem gets less intuitive when you deal with smaller numbers. When you're
dealing with 10 cards you _feel_ like you probably don't have it to begin
with. When it's 1 in 3, you feel like you're closer.

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RiderOfGiraffes
It's a great variant for teaching about probability, because switching is the
right answer, even if you don't then get the prize. Sometimes you switch, but
choose the wrong one of the two.

So for bonus points, what values of 10, 2 and 6 make this a 50-50 proposition?
More particularly, what does the relationship between 10, 2 and 6 have to
become?

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mooism2
I'd accept, and expect to double my chances of winning the prize by doing so
(from 20% to 40%).

