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The Magic Cafe Forum Index » » Puzzle me this... » » Losing games (0 Likes) Printer Friendly Version

Nir Dahan
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Inner circle
Munich, Germany
1390 Posts

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Suppose you have a betting game A, with losing expectancy - i.e. when playing a lot of games, in the long run you are guaranteed to lose your money.
suppose game B is a different game but also with a losing expectancy.

the nature of the games is irrelevant - they can be dice games, coin tossing games etc.

now here comes the famous mathematician Don Ropar and claims that by randomly playing games A or B he is guaranteed to WIN in the long run ??!!!

can this be ??? or is he just another nut case?
If the claim is true prove it, if it is wrong show why.

N.

Googling for the answer is ok (but not for the hardcore puzzler group of Stan, Tomas, Jack etc..)
Psy-Kosh
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Michigan
135 Posts

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Am I allowed to assume that A and B are completely independant games? ie, that game B does not in any way depend on what happened in game A or how often I played it and vice versa, and can I assume that each play of the game is independant of the previous play? Further, can I assume that each time A is played, the same amount is wagered for A?, and each time B is played, the same amount is wagered for B?

_If_ I can assume those things, then I'm going to say that it's false.

Let A(x) be the probability distribution for winning x money on a single play of A, and B(x) be the same thing for game B

now, <A> < 0 and <B> < 0

which is the same as saying sum(x*A(x)) < 0 and sum(x*B(x)) < 0

So now, we create a third game C in which at each step we have use a random event to decide which game to play. (We're switching between them randomly, right?)

So let p = the probability of playing A, so then 1-p is the probability for playing B.

so now the probability of winning x money on a single play of C is C(x) = p*A(x) + (1-p)*B(x), right?

And <C> = sum(p*A(x) + (1-p)*B(x)) = p*<A> + (1-p)*<B>

Both <A> and <B> are negative, and both p and (1-p) are positive, so both terms in that will be negative, so the expected value will be negative.

(I did this the long way because, by the very fact that you're making this a riddle, made me suspicious that there is some snealy unexpexted way of doing it...)

But I will say no. Under the assumptions I made at the begining, of everything being independant and such, the expected value of the game composed of switching between the games will also be negative.
Nir Dahan
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Inner circle
Munich, Germany
1390 Posts

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Both games are statistically independent!

Thanks for the detailed reply psy-kosh

I will PM you with some questions

Nir
Nir Dahan
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Inner circle
Munich, Germany
1390 Posts

Profile of Nir Dahan
Just to make things clear - A and B are not ANY losing games.
the question is can we design games A and B which are independent and that are both losing that when "combined" together in a game will generate a winning strategy.

N
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