How long is an Eternity?

I once heard about an old Arabic definition of eternity. The definition was:

“Imagine a pyramid in the desert, a thousand meters high, a thousand meters long and a thousand meters wide, made of pure diamond. Every thousand years, a small bird comes by and pecks at the pyramid [once]. When the pyramid is gone, an Eternity has passed.”

OK… that seems like a pretty long time, right? But just how long is it? And how does it relate to other vast time periods like the age of the Earth, the age of the Universe or something else?

*Figure 1: The figure shows the great pyramid of Giza. Its size is about* $2.6\times 10^6 m^{3}$ *or approximately 130 times smaller than the pyramid of diamond.*

*Figure 1: The figure shows the great pyramid of Giza. Its size is about* $2.6\times 10^6 m^{3}$ *or approximately 130 times smaller than the pyramid of diamond.*

Before we start calculating, let’s assume some basic things. We assume that the pyramid is only affected by the bird and not by, say the weather. Also, assume that diamond is perfectly stable and will not be destroyed by “time” itself (e.g. quantum mechanical effects going on in the nucleus etc.).

Now, let’s start calculating. We start with the size of the pyramid. The volume of a pyramid is

$\begin{aligned} V = \frac{1}{3} \times w\times l \times h \end{aligned}$

where $w = width$ , $l = lenght$ and $h = height$ , so in our case that would be $1/3\times 1,000\times 1,000\times 1,000$ or $0.33*10^9 m^{3}$ . Diamond has a density of $3.520 kg/m^3$ , so the pyramid would weigh about $1.2*10^{12} kg$ . That’s a big and heavy pyramid!

So how much of this gigantic pyramid can be removed by a pecking bird? I have no idea, but let’s take a very conservative approach and say it removes one single atom. It can’t really remove less than one atom, because if it did, then the pyramid would never decrease in size and the Arabic definition of eternity would just be a long version of saying “forever”. An atom is extremely small, so maybe the bird would remove a lot more, but assuming one atom per peck gives us an upper limit.

So how many atoms are there in the pyramid? As it happens, the definition of a kg is “the weight of $6.022*10^{23}$ carbon atoms”, also called Avogadro’s number, so in other words, the number of atoms in the pyramid is $7.1*10^{35}$ . So, a bird (probably different birds!) showing up every 1,000 years to remove one atom at a time would need $7.1*10^{38}$ years to finish the job! In other words, the Arabic definition of an eternity is at most 710 000 000 000 000 000 000 000 000 000 000 000 000 years.

How does this number relate to other big numbers? Let’s have a look at the table below, before we move on to my own definition of eternity.

7.07\times 10^{38} — *Table 1: The table shows the age of Earth, the Universe and the Arabic eternity. The Arabic eternity is much longer than the others are! About* $5.1\times 10^{29}$ *times longer than the age of the Universe. I’m not even sure there’s a word for a number of that size.*

4.543\times 10^{9} — *Table 1: The table shows the age of Earth, the Universe and the Arabic eternity. The Arabic eternity is much longer than the others are! About* $5.1\times 10^{29}$ *times longer than the age of the Universe. I’m not even sure there’s a word for a number of that size.*

Something that you can do for a long time is play cards. One of my favorite card games is Casino, which is quite common in the Nordic countries. It is a so-called fishing game. The complete rules can be found at https://www.pagat.com/fishing/nordic_casino.html, but in short, this is how it works. A complete game consists of a number of “mini-rounds”, in which four cards are dealt to each player. The aim is to match the value of one of your cards to the value of one or more cards on the table. If you can’t match any card, you add a card from your hand to the table. When all the cards of the first mini-round have been played, four new cards per player are dealt. Whatever cards were left on the table at the end of a mini-round are left there for the next. The game then continues with mini-rounds, until all cards in the deck have been played. At the end, the cards you matched during the game are scored and whoever ends up with most points win.

Figure 2: The figure illustrates an example of a round of the card game Casino. With two players, it is played in 12 mini-rounds. In each mini-round, each player is dealt four new cards from the deck. Any remaining cards on the table at the end of a mini-game stays put to the next round.

So, how long can you play this game, without it repeating itself?

It all depends on what you mean by “repeating”. Let’s say both players are dealt the same hands twice, but they choose to play their hands differently the second time. Would you consider it to be one or two different games? Also, what if the hands are the same, but shifted? So, in game two player A gets the same set of hands as player B had in game one and vice versa. Are those different games? Also, since the game is played in mini-rounds, what if player A gets the same cards as in a previous game, but in different mini-games? For example, if the first four cards were all aces in game one, but in game two all the aces came in the last mini-round, would that be considered as one or two games?

Hrm… It’s not simple, but let’s try this definition: If both players have the same set of hands in the same order as in a previous game, it’s considered “repeating”. So, if player A has the same set of hands as player B had in the previous game, they are considered as two different games. Also, if player A has the same cards as in a previous game, but in different mini-games, they are viewed as different game. Using this definition, the number of combination of hands is $9.2\times 10^{49}$ .

When you play cards, however, you shuffle the deck and deal. You do not keep track of which set of hands have already been dealt. So, except for the first hand, there is always a possibility – although very slim – that you may end up with the exact same set of hands as before. This probability grows larger and larger as the number of games played increases (imagine having played all hands except one – it’ll be pretty unlikely that you will be given that specific hand in the next round, right?).

So how many games do you need to play before you’ve played say, half of all possible set of hands? Having played that many hands would mean that you’d go “We’ve already played this game!” typically every two games. At that point, I guess it would be pretty boring.

Calculating this was a bit more complicated than I first thought, and the details can be found below, but it turns out that you need to play $5.4*10^{51}$ games. If we assume that each game takes 15 minutes to complete, this would take $1.5*10^{47}$ years. That’s more than 200 million times longer than the Arabic eternity!

Which leads us to my own definition of eternity:

“Imagine playing your favorite card game Casino with your best friend. When you seem to be playing the same set of hands as before more often than new hands, an Eternity has passed.”

To me, that seems like a nicer image of an eternity than waiting for that bird that shows only up every 1,000 years.

/Anders Björling

How did you calculate that? A short introduction to combinatorics

If you are not familiar with combinatorics and what to know how I arrived at the numbers, please read on. If not: move one there is nothing here to see here.

Dealing cards can be thought of as what in combinatorics goes under the name of random selection without replacement. In other words, you pick a card and you don’t put it back.

In Casino with two players, the first four cards are put on the table. In how many ways can this be done? Well, the first card can be any of 52 so we can pick just one card in 52 ways. When we pick the second one, we only have 51 cards left to choose from because one is already picked. For card three and four, we have 50 and 49 cards to choose from, respectively. One would think that the first four cards could then be selected in $52\times 51\times 50\times 49$ ways, but that is not the case. The reason is that the order is unimportant. It doesn’t matter if you get cards 2, 3, 4 and 5 of clubs or if you get 5, 4, 3 and 2 of clubs, they are the same! To account for this we need to divide by the number of combinations four cards can be selected, which is $4\times 3\times 2\times 1$ or 4!. So the total number of combinations for selecting 13 hands of four cards is then

$\begin{aligned} \frac{52!}{(52-4)!\times 4!} \times \frac{(52-1\times 4)! }{(52-2\times 4)! \times 4!} \times \frac{(52-2\times 4)!}{(52-3\times 4)! \times 4!}\times ....\times \frac{(52-11\times 4)!}{(52-12\times 4)! \times 4!} = \frac{52!}{(4!)^{13}} = 9.2\times 10^{49} \end{aligned}$

When you deal a set of hands, there is always a possibility that this exact set of hands have been dealt before. In daily life, the probability of this is extremely small and more or less unimportant but in this theoretical example, we must keep track of it.

If N is the total number of possible set of hands, and $(n-1)$ is the number of already played set of hands, the probability that we in game n get a previously unplayed set of hands is $p(n) = (N-n+1)/N$ . This probability has a geometric distribution with a expected value of $t(n)=1/p(n) = N/(N-n+1)$ . This is true for all values of n, so the expected value of drawing half of all possible hands is then

$\begin{aligned} T \end{aligned}$ = $\begin{aligned} E[t(1) + t(2) +...+ t(\frac{N}{2}-1) + t(\frac{N}{2})] = \end{aligned}$

$\begin{aligned} E[t(1)] + E[t(2)] + ... + E[t(\frac{N}{2}-1)] + E[t(\frac{N}{2})] = \end{aligned}$

$\begin{aligned}N\times (\frac{1}{N} + \frac{1}{(N-1) \, } +...+ \frac{1}{2} + 1) -N\times (\frac{1}{(\frac{N}{2})} + \frac{1}{(\frac{N}{2}-1)} +...+ \frac{1}{2} + 1) = \end{aligned}$

$\begin{aligned}N\times (H(N)-H(\frac{N}{2})) \, \end{aligned}$ , where $H(x)$ denotes the so-called x:th harmonic number.

For large numbers of $x$ , then $H(x) \approx log(x) + \gamma + \frac{1}{2x}$ . This means that $T = \frac{N}{2} \times (log(N) + \gamma + log(2))$ , where $\gamma \approx 0.5772156649$ .