5?

Basically, everyone on the Earth is not going to drop everything and shuffle cards all day. We couldn't even survive that for very long. So, in taking the total poulation of the Earth as my number, I'm making an impossibly generous concession, in order to imagine this actually happening. The 40-hour week is something familiar to us; I'm trying to put it in real terms.

If you like, I could go back to having each person shuffle an exponential amount of decks every second...

that's less than 1/3 of the hours in a week....what do you do with all of that free time? ya know....it sounds appealing....

This site disagrees

So on the ninth time, the shuffle will be the same as the first shuffle.

He expects a "perfect riffle shuffle", though.

:lol:

Why should you start injecting reality and common sense into this discussion at this late date?? You might as well start thinking about whether those monkeys with the typewriters have had touch typing classes or not.

Okay, I hear what people are saying. You want more shuffles getting done, okay. Let’s look at that.

I’m going by HLJ’s calculation (I’m too lazy to do it myself) that in 9.e+33 permutations you would have a 50% chance of repeating the same shuffle.

I’m using the entire population of planet Earth (6,648,429,413), and I’ll have them shuffle one deck of cards per second, year-round. It would take 42,925,646,062,489,266.972805985537223 years to get through 9.e+33 permutations.

How long is that? Well, the Earth is estimated to be 4.54 billion years old. It’s 9,454,988.1194910279675784109112819 times longer than that.

So, if every living person on the Earth shuffled one deck of cards per second, for 9 million times longer than the Earth has existed, there would be a 50% chance of the same shuffle coming up.

THE SAME EXACT SHUFFLE HAS NEVER HAPPENED, AND NEVER WILL HAPPEN. EVER.

unless it does...

It totally happened to me the other day. Take my word for it.

Would you settle for 1% chance?

Flint, are you OCD?

If you're offering to calculate how many permutations of 52! you would have to go through to have a 1% probability of repeating a permutation...
I would love to know that.

I'd better present my logic to see if anyone disagrees.

(n*(n-1))/(2*52!) = P; solve for P, where P = probability of 2 or more matching shuffled decks in n shuffles.


(n-1)^2 < n*(n-1) < n^2, so

(n-1)^2/(2*52!) < n*(n-1)/(2*52!) < n^2/(2*52!),
so the first equation can be simplified (for large values of n) as

n^2 ~= P*2*52!

n = square_root(P*2*52!)

If P = 0.5 (50%), n = 8.98e33
If P = 0.1 (10%), n = 4.0e33
If P = 0.01 (1%), n = 1.27e33

I just noticed a flaw in this logic. Can you see it?

lol - I can't even follow that!

pull my finger!