Alexander's Paradoxes

Alexander's paradoxes

Given the concept of the k-th car, the paradox vanishes. The Gold Train looks like just another trivial problem. For example, a poker hand: A normal hand has 5/4 spades, while the hand that contains the Ace of Spade is richer in spades, having 33/17 spades.
I have checked several Statistics textbooks. The issue is still there: Statistics seems to be the science of “them”. The authors never reserve a flight seat, never complete a multiple choice test and they never buy defective component radios. The authors never get involved!

I think this is why such “paradoxes” are missing from culture; this is why I am cramped at my table while the cafeteria administrator reports huge spaces for happy students  The cafeteria administrator consider also the case I could sit at an empty table.

## Alexander in the gold train

### AR calculus

For the average blocks; a train with k blocks has an average block length of n/k.

By looking at color changes, rather than partitions, number of trains with k blocks is

2

(

n
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1

k
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1

)

{\displaystyle 2 \binom{n-1}{k-1}}

so the expected value is

2

2
n

∑

# k
1...
n

(

n
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k
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n
k

=

1

2

n
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1

∑

# k
1..
n

(

n
k

)

=

2
n

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1

2

n
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=
2
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1

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n
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{\displaystyle \frac{2}{2^n}\sum\_{k=1...n} \binom{n-1}{k-1} \frac {n}{k} = \frac {1}{2^{n-1}} \sum\_{k = 1..n} \binom{n}{k} = \frac {2^n - 1}{2^{n-1}} = 2 - \frac{1}{2^{n-1}}}

The expected length of the block containing car k, 1 <= k <= n, is

3
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1

2

k
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1

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1

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n
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k

{\displaystyle 3 - \frac {1}{2^{k-1}} - \frac {1}{2^{n-k}}}

;

so the average is :

3
−

2
n

(

1

1
2

.
.
.

1

2

n
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1

)

=
3
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2
n

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2
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=
3
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4
n

(

1
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{\displaystyle 3-\frac{2}{n} \left ( 1 + \frac{1}{2} + ... \frac{1}{2^{n-1}} \right ) = 3 -
 \frac{2}{n} \left( 2-\frac{1}{2^{n-1}} \right) = 3 - \frac{4}{n} \left( 1- \frac {1}{2^n} \right)}

By (something like) Jensen's inequality, this should always be larger. Calculations show it to be strictly larger for n >= 3.

### anonym

This is a very basic probability question. I do not think it should be downvoted, because it can be interesting for the beginner.

As was pointed out in the comments, Alexander is more likely to see a longer block because a longer block is more likely to see Alexander. It's interesting to derive the answer analytically, assuming an infinite train. The expected number of blocks of size n+1 or longer is exactly half of that of blocks of size exactly n (because the (n+1)th wagon has equal chances of terminating the block or continuing it). Then, the probability that a block chosen uniformly at random has size exactly 1 is 1/2, the probability that it has size exactly 2 is 1/2⋅1/2, and in general the probability that a block has size exactly n equals 2−n.

Then, the probability p(n) that a specific wagon belongs to a block of length n is proportional to n⋅2−n - because the number of wagons in all blocks of size n is proportional to the frequency of such blocks times the number of wagons in such blocks. The expected number of wagons Alexander will see in his block is then:

(

∑

# i
1

∞

n
⋅
p
(
n
)

)

/

(

∑

# i
1

∞

p
(
n
)

)

=

(

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# i
1

∞

n
⋅
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n

2

−
n

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)

/

(

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# i
1

∞

n

2

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=
3

{\displaystyle \left(\sum\_{i=1}^\infty n \cdot p(n)\right)/ \left(\sum\_{i=1}^\infty p(n)\right) = \left(\sum\_{i=1}^\infty n \cdot (n2^{-n})\right)/ \left(\sum\_{i=1}^\infty n2^{-n}\right)=3}

(can you see why it adds up to 3?)

## Alexander goes cafeteria

In Alexander's program there are 50 girls and 50 boys. They use a cafeteria of 25 4-tables, enough to cover the peak hour.

Alexander would like to talk with girls rather than boys during the pause; so, he assumes on average a table will have 2 girls and 2 boys.

After taking his seat at a table, he discovers the cruel reality : on average there are only 1.5 girls at his table and 2.5 boys.

Why the girls are avoiding Alexander ? Why they are choosing other tables ?

This is a very basic question in probability theory. The fact that Alexander, who is a boy, certainly sits at his own table skewes the distribution -- for example, he will never see 4 girls at his table.

The easiest way to think about it is that there are really 3 places left after Alexander sits, for 50 girls and 49 (N.B. not 50) boys. On average, a little over half of these seats (3×50/99≈1.515) will be taken by girls, and a little less than half (3×49/99≈1.485) will be taken by boys other than Alexander on average. So, Alexander is indeed slightly unlucky: he should see about 1.515 girls and 2.485 boys (including himself) at his table, but maybe he just did not get enough samples!