Please Explain (mathematics question)

[J_Cubed]

@RadicalPi

I don’t know if anyone’s mentioned it, but ! represents the factorial function.

0! time, in 6.5238!, for 3.878! minutes.

[SeldomSeen]

Thanks to all who responded, explaing a difficult (for me) problem in layman’s terms. I believe I understand it better now.
SS

[Spectre_of_Pithecanthropus]

@Malacandra

[sup]n[/sup]C[sub]r[/sub].

Written as [noparse][sup]n[/sup]C[sub]r[/sub][/noparse].

Thanks for replying.

The symbol I’ve seen is different, but it was in a very old textbook so I wouldn’t be surprised if it’s obsolescent. This was a tall set of parentheses with containing n and r, one positioned over the other. Or is that something else?

[Malacandra]

No, writing n over r in vector parentheses means the same thing, but defeats my coding skills!

[Chronos]

Quoth Wendell:

You seem to think that it would be necessary to produce about (10^68) different orderings to expect to see even one case where two of those orderings match. That’s not true. It’s only necessary to produce about (10^34) different ordering to expect to see about one case where two of those orderings match.

But that’s not what brocks said. He didn’t say that if all those people shuffled for all that time that they’d never match each other; what he said was that they’d never match that one specific shuffling you did.

[Wendell_Wagner]

Now that I reread what brocks wrote, I’m not clear what he meant. If he means that one starts with a given ordering and waits until that ordering is repeated, then his calculations are correct that about random (10^68) orderings have to be produced. If he means that one starts producing orderings and waits until there is some repetition, then my calculations that one only needs to produce (10^34) random orderings to see the first repetition are correct.

[DHMO]

@Wendell_Wagner

Now that I reread what brocks wrote, I’m not clear what he meant. If he means that one starts with a given ordering and waits until that ordering is repeated, then his calculations are correct that about random (10^68) orderings have to be produced. If he means that one starts producing orderings and waits until there is some repetition, then my calculations that one only needs to produce (10^34) random orderings to see the first repetition are correct.

I think it is very clear what brocks meant:

@brocks

If you shuffle a deck thoroughly, then you have produced something that the world has never seen before, and if left only to chance, will never see again.
… the chance that one of them would have produced ***the same order of cards you did ***would be about one in a trillion trillion.

[emphasis mine —DHMO]

[Wendell_Wagner]

Incidentally, even if you assume that’s what he meant, his arithmetic still isn’t quite correct. As I showed in my calculations, he’s talking about (10^53) different orderings being produced. He claims that there’s only one chance in a trillion trillion that this will match a given ordering. But there are only about (10^68) different orderings, so that’s one in a thousand trillion, not one in a trillion trillion.

And I still don’t find it clear what he meant, but this is pointless to argue about.

[don_t_ask]

@J_Cubed

It’s six of one thing, 3! of the other.

Just thought I should say this is a pretty clever joke but somehow seems to have caused the thread to turn to crap.

[brocks]

@Wendell_Wagner

Incidentally, even if you assume that’s what he meant, his arithmetic still isn’t quite correct. As I showed in my calculations, he’s talking about (10^53) different orderings being produced. He claims that there’s only one chance in a trillion trillion that this will match a given ordering. But there are only about (10^68) different orderings, so that’s one in a thousand trillion, not one in a trillion trillion.
And I still don’t find it clear what he meant, but this is pointless to argue about.

You’re right, I somehow managed to add 54 to 24 and get 68. But I’m danged if I can see how you find it unclear that I’m specifying a particular deck to match.