# Singed Random Numbers

Upon my arrival in Hell, the Devil predictably gave me a chance to win my freedom:

• The Devil's printer would print 101 random real numbers, picked independently and uniformly between -1 and +1, in invisible ink on 101 cards (each with infinitely many decimal places of precision).
• I would be allowed to tear up one of the cards.
• The Devil's card sorter, which can read invisible ink, would order the remaining cards from lowest to highest.
• I would pick one of the sorted cards.
• The Devil would lightly singe all the sorted cards, causing the numbers to be revealed.
• If the one I picked was the closest of the sorted cards to 0, I would be released.

Should I tear up a card?

• Probability zero that two cards have exactly the same absolute value so this case can be ignored, right? Commented Aug 3 at 0:35
• @Lucenaposition that's right. Commented Aug 3 at 7:24

You should choose...

...not to worry about it, because it doesn't matter whether you tear up a card or not.

Why?

There are either 101 cards or 100 cards; that's the only difference tearing up a card makes.

In choosing the numbers, we can model the choice of absolute value and the choice of sign as being independent.

Let's say there are $$n$$ cards. Imagine that first, the absolute values of the numbers are chosen. Then, the sign of the number of smallest absolute value doesn't matter -- it will be sorted to the same place either way, between the positives and negatives; we can just allow the signs of the remaining $$n-1$$ cards to be chosen, and if they are correctly balanced around our chosen index, we win. Thus, if we index from $$1$$ and choose index $$k$$, the chance of winning is $$\frac{1}{2^{n-1}} \binom{n-1}{k-1}$$ -- in other words, the chance (from the binomial distribution) of having exactly $$k-1$$ negative numbers from among $$n-1$$.

Knowing how the binomial distribution behaves, this verifies the intuitive fact that it is better to choose an index nearer the middle.

The result follows from the easily-verified equality $$\frac{1}{2^{100}} \binom{100}{50} = \frac{1}{2^{99}} \binom{99}{49}$$ -- the chance of winning by choosing index 51 from 101 coins on the left, and choosing index 50 from 100 coins on the right. (I'm sure there a million ways to show this with a nice combinatorial argument, so choose your favorite).

• But then by induction, you will always choose the correct card? Let P(n) be the probability of choosing the correct card among n cards. Then P(101)=P(100)=...=P(1)=1? Commented Aug 3 at 1:04
• @BenjaminWang No. I think it's an even/odd thing. P(99) would be $\binom{98}{49} \frac{1}{2^{98}}$ which is a different number. P(2k)=P(2k+1) Commented Aug 3 at 1:14
• rot13(Bar bs gubfr zvyyvba jnlf hfrf gur snpg gung avargl-avar pubbfr sbegl-avar rdhnyf avargl-avar pubbfr svsgl.) Commented Aug 3 at 2:00
• @chux-ReinstateMonica If the number of cards is bigger, the chance for any particular one to be the closest to zero gets smaller. So more cards reduce your chance of guessing correctly. Your obsevation 1) is correct and in this example it just so happens that the two effects exactly cancel out and the probability is the same. Commented Aug 4 at 0:55
• @Suthiro: The fact that one could choose either of two cards that would each have a probability 0.0796 of winning, doesn't offer any advantage over a situations where one could identify a single card that has such a probability of winning. Commented Aug 4 at 18:30