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Best strategy for The Million Pound Drop

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Trog Woolley

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Nov 2, 2010, 1:15:53 PM11/2/10
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I somehow managed to see the TV program The Million Pound Drop.
For non UK readers, here is some info about the program
http://www.ukgameshows.com/ukgs/The_Million_Pound_Drop

Assuming that you are like me and know almost nothing about the
world of celebrity (a subject which comprises a large number of
the questions) but yoy are quite good at general knowledge, what
would be a good strategy to maximise your winnings? Contestants
split their remaining pool evenly if they don't know an answer.
Is this really the best strategy? My gut feeling tells me that
you should divide the pool unevenly to maximise your chances.

--
Trog Woolley | trog at trogwoolley dot com
(A Croweater back residing in Pommie Land with Linux)
Isis Astarte Diana Hecate Demeter Kali Inanna

Mark Brader

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Nov 2, 2010, 3:04:22 PM11/2/10
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> I somehow managed to see the TV program The Million Pound Drop.
> For non UK readers, here is some info about the program
> http://www.ukgameshows.com/ukgs/The_Million_Pound_Drop

Interesting game-show concept. Endemol scores again.



> Assuming that you are like me and know almost nothing about the
> world of celebrity (a subject which comprises a large number of
> the questions) but yoy are quite good at general knowledge, what
> would be a good strategy to maximise your winnings?

If you think one answer is most probably correct, your expected
return is maximized if you bet everything on that one. If you have
no idea, your expected return is identical no matter how you bet.

> Contestants split their remaining pool evenly if they don't know
> an answer. Is this really the best strategy?

You didn't ask for the best strategy, you asked for the strategy to
maximize winnings. Splitting the pool gives up some money in order
to increase the probability of winning *some* money rather than zero.
In short, it's insurance against risk -- and the players are likely to
be concerned about that.
--
Mark Brader, Toronto | "Professor, I think I have a counterexample."
m...@vex.net | "That's all right; I have two proofs."

My text in this article is in the public domain.

David

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Nov 3, 2010, 7:58:11 AM11/3/10
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"Trog Woolley" <tr...@email.fake> wrote in message
news:dHXzo.34593$r35....@newsfe14.ams2...

> what
> would be a good strategy to maximise your winnings?

I would get somebody in the audience that's connected to the internet to
cough me the answers. I would then be caught out for cheating and sell my
story to the papers for £1m.
(It worked on Who wants to be a Millionaire)


James Dow Allen

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Nov 3, 2010, 2:10:45 PM11/3/10
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On Nov 3, 12:15 am, Trog Woolley <t...@email.fake> wrote:
> What would be a good strategy to maximise your winnings?  Contestants

> split their remaining pool evenly if they don't know an answer.
> Is this really the best strategy?  My gut feeling tells me that
> you should divide the pool unevenly to maximise your chances.

Suppose only one question will present any difficulty,
and you assess answer A as 60% likely to be correct,
B as 40% likely to be correct. If you bet it all on A,
your *average* winnings will be $600k, but you might get zero.
If you bet 60% on A and 40% on B, your average winnings
are somewhat less (only $520k), but you're guaranteed
at least $400k. Surely most poor people would prefer the
second strategy.

A convenient way to cope with such problems is to seek
to maximize
Expectation [log (W + B)]
rather than
Expectation [W]
where W is winnings, and B is the rest of your bankroll.

The example strategy (bet K% on the answer you judge K% likely)
was, IIRC, the correct answer to an IBM Ponder This Puzzle
several years ago. That problem was quite similar to this one,
but I am too lazy/stupid/apathetic to confirm
the correctness of the same answer here. :-)

James Dow Allen

James Dow Allen

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Nov 3, 2010, 9:15:08 PM11/3/10
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On Nov 4, 1:10 am, James Dow Allen <jdallen2...@yahoo.com> wrote:
> A convenient way to cope with such problems is to seek
> to maximize
>      Expectation [log (W + B)]

In the case B=0, with two exclusive answers whose probabilities
sum to 1, and where your bets must sum to 1, one seeks
Max (p log x) + ((1-p) log (1-x))
Recalling that log t(x) is maximized when t(x) is maximized, which
is when dt(x)/dx = 0, the above solves (I think!) to give indeed
x = p

In the actual game you have three-answer questions and compound
questions but I think they can be decomposed into binary
decisions of the simple type just "solved".

> The example strategy (bet K% on the answer you judge K% likely)
> was, IIRC, the correct answer to an IBM Ponder This Puzzle
> several years ago.

http://domino.research.ibm.com/Comm/wwwr_ponder.nsf/challenges/January2007.html

That problem may seem quite different. Proving (or disproving)
that it is the same underlying problem is left as exercise.

> but I am too lazy/stupid/apathetic ...

My "apathy" was more a matter of waiting for my hot date to
show up! :-) Instead, we both fell asleep waiting for
the kids to fall asleep. :-(

James

Rob Sharp

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Nov 4, 2010, 7:04:42 AM11/4/10
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One further complexity is that you have to always leave one answer with
no money on.

So, given 4 answers you cannot split your money four ways if you do not
know the answer and cannot completely eliminate one of the possibilities.

Also, as the game goes on the number of available answers decreases from
4 to 3 to eventually 2 on the final question (one of which must still be
left empty)

R

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