Tres Amigas!

The order of the heights can be

• $E \to B \to F$
• $E \to F \to B$
• $F \to E \to B$

each with equal probability.

Therefore, the chance that Ella is taller than Frabduzella is $\boxed{\dfrac{2}{3}}$

For those of you still convinced that 1/2 must be the right answer, I think @Varsha Dani explains it best in the comments below, copied here:

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"Does this probability change if I then state c, d , and e are taller and f, g , and h are shorter? Or if I state that everyone else in the world is taller?"

Yes, it could. It depends on what new information you get and what question you are asking.

I believe I already replied to the person who brought up the gambler's paradox -- Those coin flips are independent. Here, the heights themselves are independent, it is true. But the comparisons of the heights are not independent.

So, if I have three girls, E B and F, then it is true that comparing any two of them is like a coin flip. But take note of the fact that there are three possible comparisons. E vs B, B vs F, and E vs F. This would be three coin flips, but if they were independent there would be eight possible outcomes. In reality there are only six possible orderings of the heights. Two of the outcomes of the coin flips lead to inconsistent orderings (E >B, B> F, F> E and E<B, B<F, F<E). Thus you cannot make the analogy to the gambler's paradox, and you have to think of this problem in terms of the distribution of all the heights together. Now, being told some information can affect the probability, and in this instance it does.

Note that if you had four girls, E, B, F, and S, then being told that E is taller than B affects the probability that E is taller than F or that B is taller than S, but it does not affect the probability that F is taller than S, which is still 1/2.

Hope this helps.

• @Varsha Dani

The three friends are Ella $(\color{#20A900}E)$ , Bella, $(\color{#3D99F6}B)$ , Frabduzella $(\color{#E81990}F)$ . Total there are three are three possibilities keeping one condition constant : $\color{#20A900}E > \color{#3D99F6}B$ $\begin{cases} \color{#20A900}E > \color{#3D99F6}B > \color{#E81990}F \qquad {\color{#D61F06}(√)} \\ \color{#20A900}E > \color{#E81990}F > \color{#3D99F6}B \qquad {\color{#D61F06}(√)} \\ \color{#E81990}F > \color{#20A900}E > \color{#3D99F6}B \qquad {\color{#D61F06}(×)} \\ \end{cases}$

$\therefore$ The probability that that Ella is taller than Frabduzella is $\dfrac{\color{teal}2}{\color{#69047E}3}$ .

I sort of made an educated guess. I thought that since Ella was taller than Bella, on average Ella will have a height greater than the median. So on average Ella will be taller than Frabduzella. So that left 2/3 and 3/4 as they are both greater than a half. I guessed 2/3 as it seemed more reasonable. I really need to brush up on some probability/statistics :)

link text

This a conditional probability problem. Our initial possible events set is:

$B<E<F$

$B<F<E$

$E<B<F$

$E<F<B$

$F<B<E$

$F<E<B$

We have the information that Ella is taller than Bella, so we can remove all possibilities where Bella is taller than Ella. Now we have:

$B<E<F$

$B<F<E$

$F<B<E$

In only 2 of these 3 cases Ella is taller than Frabduzella. So the probability must be $\boxed{\frac{2}{3}}$

For anyone that's interested, this is a variation on a classic statistical problem called the "Monty Hall Dilemma". Famously, it featured on an American game show, and the "smartest woman in the world", Marilyn vos Savant, was the first one to solve it correctly.

To start with, the probability for any of the girls being tallest is 1/3.

Knowing E is taller than B adds 1/3 chance for E being tallest, making it 2/3, while F remains at 1/3.

I was first thinking about the odds that Ella would be taller than both Bella and Frabduzella, not caring about that I already have half that information.

The odds that Ella is tallest is $\frac{1}{3}$ .

The odds that Ella is taller than Bella is $\frac{1}{2}$ . The odds that Ella is taller than Frabduzella is $\frac{1}{2}$ . That mean that the odds that Ella is taller than both Bella and Frabduzella is $\frac{1}{2} * \frac{1}{2} = \frac{1}{4}$ which I clearly see is wrong otherwise the odds that any girl would be tallest would be $\frac{3}{4}$ instead of 1.

This mean that the odds that one girl is taller than both the other girls are less than the odds that she individually is taller than both girls, one at a time. Since we now that the odds that Ella is taller than Bella is $\frac{1}{2}$ we call the odds that she is taller than Frabduzella x. This lead to that the odds that Ella is taller than Bella ( $\frac{1}{2}$ ) multiplied with the odds that Ella is taller than Frabduzella ( $x$ ) should be $\frac{1}{3}$ .

That gave me $\frac{1}{2} * x = \frac{1}{3} \implies x = \frac{2}{3}$

Same of Monty Hall problem

At the start each has equal probability of being the tallest of 1/3 each. Now two girls are decided so it's now 2/3 vs 1/3