Alice in Wonderland -- Red Queen at tea party

Probability Level 4

Alice, Mad Hatter, March Hare and Dormouse were talking happy at the tea party when Red Queen came to look for Alice and Mad Hatter (Of course, to cut off their heads).

In order to evade Red Queen, Alice and Mad Hatter managed to shrink themselves and randomly hid into five of the teapots. Since the room is relatively small, Alice and Mad Hatter cannot be in the same teapot.

When Red Queen came, she saw nothing except five teapots (March Hare and Dormouse had run away). She suspected that Alice and Mad Hatter must be in two of the teapots, so she decided to check some of them until she found both Alice and Mad Hatter.

Red Queen is very clever and good at logic. She knows that if she can make the conclusion that Alice or Mad Hatter must be at remaining teapots from the previous results, she doesn't have to check them. She will directly take them to her palace. Also, if she finds both Alice and Mad Hatter in previous teapots, she doesn't have to check rest of them.

Let $X$ be the number of teapots Red Queen has checked . What is $E(X)$ ?

\dfrac{17}{5}

\dfrac{7}{2}

\dfrac{16}{5}

\dfrac{18}{5}

2 solutions

Δrchish Ray
Jun 22, 2019

2 Teapots:

If Red Queen picks up 2 teapots, the only way she can do it is:

$\textcolor{#20A900}{c}$ $\textcolor{#20A900}{c}$

Note: $\textit{c}$ means Red Queen guessed correct and $\textit{w}$ means Red Queen guessed wrong

The probability of this happening is $\frac{2}{5} \times \frac{1}{4} = \frac{1}{10}$

3 Teapots:

If Red Queen picks up 3 teapots, there are 3 ways she can do this. The first two are:

$\textcolor{#20A900}{c}$ $\textcolor{#D61F06}{w}$ $\textcolor{#20A900}{c}$

$\textcolor{#D61F06}{w}$ $\textcolor{#20A900}{c}$ $\textcolor{#20A900}{c}$

The probability of either of these happening is: $2 \times \frac{2}{5} \times \frac{3}{4} \times \frac{1}{3} = \frac{1}{5}$

The other way she can do this is by guessing the wrong teapot 3 times, and deducing that the remaining two have Alice or Mad Hatter in each of them. The probability of this happening is:

$\frac{3}{5} \times \frac{2}{4} \times \frac{1}{3} = \frac{1}{10}$

4 Teapots:

If Red Queen picks up 4 teapots, there are 6 ways she can do this. The first three are:

$\textcolor{#20A900}{c}$ $\textcolor{#D61F06}{w}$ $\textcolor{#D61F06}{w}$ $\textcolor{#20A900}{c}$

$\textcolor{#D61F06}{w}$ $\textcolor{#20A900}{c}$ $\textcolor{#D61F06}{w}$ $\textcolor{#20A900}{c}$

$\textcolor{#D61F06}{w}$ $\textcolor{#D61F06}{w}$ $\textcolor{#20A900}{c}$ $\textcolor{#20A900}{c}$

The probability either of these happen are $3 \times \frac{2}{5} \times \frac{3}{4} \times \frac{2}{3} \times \frac{1}{2} = \frac{3}{10}$

The other three can be done if she picks up one correct and three incorrect, deducing the final one has someone under it:

$\textcolor{#20A900}{c}$ $\textcolor{#D61F06}{w}$ $\textcolor{#D61F06}{w}$ $\textcolor{#D61F06}{w}$

$\textcolor{#D61F06}{w}$ $\textcolor{#20A900}{c}$ $\textcolor{#D61F06}{w}$ $\textcolor{#D61F06}{w}$

$\textcolor{#D61F06}{w}$ $\textcolor{#D61F06}{w}$ $\textcolor{#20A900}{c}$ $\textcolor{#D61F06}{w}$

The probability of either of these happening are $3 \times \frac{2}{5} \times \frac{3}{4} \times \frac{2}{3} \times \frac{1}{2} = \frac{3}{10}$

Finally, we can calculate the expected number of cups Red Queen will pick up.

$2 \times \frac{1}{10} + 3 \times \frac{1}{5} + 3 \times \frac{1}{10} + 4 \times \frac{3}{10} + 4 \times \frac{3}{10} = \boxed{\frac{7}{2}}$

David Vreken
Jun 22, 2019

There are ${5 \choose 2} = 10$ ways the Red Queen can pick the pots, which are organized in the chart below. $\text{Y}$ means the Red Queen picked a pot with either Alice or the Mad Hatter in it, and $\text{N}$ means the Red Queen picked an empty pot. The strike-through text means the Red Queen did not have to check those pots, because either both Alice and the Mad Hatter have been found (already has $2$ $\text{Y}$ 's) or the Red Queen concluded that they were in the remaining teapots (already has $3$ $\text{N}$ 's).

Sequence	Teapots Checked
$\text{YY}$ $\sout{\text{NNN}}$	$2$
$\text{YNY}$ $\sout{\text{NN}}$	$3$
$\text{YNNY}$ $\sout{\text{N}}$	$4$
$\text{YNNN}$ $\sout{\text{Y}}$	$4$
$\text{NYY}$ $\sout{\text{NN}}$	$3$
$\text{NYNY}$ $\sout{\text{N}}$	$4$
$\text{NYNN}$ $\sout{\text{Y}}$	$4$
$\text{NNYY}$ $\sout{\text{N}}$	$4$
$\text{NNYN}$ $\sout{\text{Y}}$	$4$
$\text{NNN}$ $\sout{\text{YY}}$	$3$
Total:	$35$

Therefore $E(X) = \frac{1}{10} \cdot 35 = \boxed{\frac{7}{2}}$ .

Yup, same approach. Any ideas about how it might be extended? ( $n$ people hiding in $t$ teapots)

Chris Lewis - 1 year, 11 months ago

For $n$ people and $t$ teapots I'm getting

$\frac{1}{t \choose k}\bigg(\displaystyle \sum_{k=n-1}^{t-2} {k \choose {n-1}} \cdot (k + 1) + \sum_{k=t-n-1}^{t-2} {k \choose {t-n-1}} \cdot (k + 1)\bigg)$

where the first sum deals with already finding the $n$ people (already has $n$ $\text{Y}$ 's) and the second sum deals with the conclusion that the remaining people are in the rest of the teapots (already has $t - n$ $\text{N}$ 's). I'm not sure if this can be simplified more or expressed in an easier way, though.

David Vreken - 1 year, 11 months ago

Alice in Wonderland -- Red Queen at tea party

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