Wine/water paradox

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The wine/water paradox is an apparent paradox in probability theory. It is stated by Michael Deakin as follows:

A mixture is known to contain a mix of wine and water in proportions such that the amount of wine divided by the amount of water is a ratio ${\displaystyle x}$ lying in the interval ${\displaystyle 1/3\leq x\leq 3}$ (i.e. 25-75% wine). We seek the probability, ${\displaystyle P^{\ast }}$ say, that ${\displaystyle x\leq 2}$. (i.e. less than or equal to 66%.)

The core of the paradox is in finding consistent and justifiable simultaneous prior distributions for ${\displaystyle x}$ and ${\displaystyle {\frac {1}{x}}}$.^[1]

Calculation[]

We do not know ${\displaystyle x}$, the wine to water ratio. We only know that it lies in an interval between the minimum of one quarter wine over three quarters water on one end (i.e. 25%), to the maximum of three quarters wine over one quarter water on the other (i.e. 75%).

Now, making use of the principle of indifference, we may assume that ${\displaystyle x}$ is uniformly distributed. Therefore, the chance of finding the ratio below ${\displaystyle a}$ is

Prob${\displaystyle \{x\leq a\}={\frac {a-x_{\mathrm {min} }}{x_{\mathrm {max} }-x_{\mathrm {min} }}}={\frac {1}{8}}(3a-1).}$

This is the linearly growing function which is ${\displaystyle 0}$ resp. ${\displaystyle 1}$ at the end points ${\textstyle {\frac {1/4}{3/4}}={\frac {1}{3}}}$ resp. ${\textstyle {\frac {3/4}{1/4}}=3}$.

With ${\displaystyle a=2}$, as in the example of the original formulation above, we conclude that

Prob${\displaystyle \{x\leq 2\}={\frac {1}{8}}(3\cdot 2-1)={\frac {5}{8}}}$.

Now consider ${\displaystyle y={\frac {1}{x}}}$, the inverted ratio of water to wine. It lies between the inverted bounds. Again using the principle of indifference, we get

Prob${\displaystyle \{y\geq b\}={\frac {x_{\mathrm {max} }(1-x_{\mathrm {min} }\,b)}{x_{\mathrm {max} }-x_{\mathrm {min} }}}={\frac {3}{8}}(3-b)}$.

This is the function which is ${\displaystyle 0}$ resp. ${\displaystyle 1}$ at the end points ${\displaystyle 3}$ resp. ${\textstyle {\frac {1}{3}}}$.

Now taking ${\textstyle b={\frac {1}{a}}={\frac {1}{2}}}$, we conclude that

Prob${\displaystyle \left\{y\geq {\tfrac {1}{2}}\right\}={\frac {3}{8}}{\frac {3\cdot 2-1}{2}}={\frac {15}{16}}={\frac {3}{2}}{\frac {5}{8}}}$.

Paradoxical conclusion[]

The second probability exceeds the first by a factor of ${\textstyle {\frac {x_{\mathrm {max} }}{a}}\geq 1}$, which for our example numbers is ${\textstyle {\frac {3}{2}}}$.

Finally, since ${\textstyle y={\frac {1}{x}}}$, we get

${\displaystyle {\frac {5}{8}}=\operatorname {Prob} \{x\leq 2\}=P^{*}=\operatorname {Prob} \left\{y\geq {\frac {1}{2}}\right\}={\frac {15}{16}}>{\frac {5}{8}}}$,

a contradiction.

References[]