Vagueness and the Aftermath - A sporadic diary

[EvilPie]

Lovely.

I found this fact quite interesting:

"Even if every one of the 7  billion men, women, and children on Earth played one billion card games every year for a billion years, they would not even make a dent in the number of possible shuffled decks of cards"

Well not so much interesting as incredible!!

[RED-DOG]

well 52! is a pretty big number



that's the number of different possible shuffles, approximately 80 unvigintillion in the US apparently or 80 undecillion in the UK. the difference being the disagreement over how many are in a billion or a trillion

this is a nice thing to read

http://thebigblogtheory.wordpress.com/2011/03/13/s04e18-the-prestidigitation-approximation/

because it makes me think of this

Wow! That's interesting.

I never knew it was prestidigitation approximation that caused me to keep thinking about scantily clad girls.

[gatso]

it's not pretty girls Tom, it's the prettiest girl

that blog is the science behind the big bang theory, one entry for each episode explaining all the stuff they were talking about but everyone was too busy looking at Kaley Cuoco to bother trying to understand at the time

[mulhuzz]

You'll like this one Tom. It's a maths fact.

If you took a pack of just 12 cards and shuffled them 1000 times, there's only a 0.1% chance two shuffles will be the same.

The human brain struggles to comprehend large numbers and probability, and I think I read something on blonde previously that said it's unlikely that two (proper) shuffles of a deck of cards have ever resulted in the same order of the 52 cards.

What do you think to those 'facts' Gatters?

that said, take two 52 card decks and shuffle accordingly. now turn over one card from the top from each deck in turn, making your way through the pack.

what do you think the chances of having at least one matching card (suit and rank) in those decks are?

50/50. and not in that jokey 'either it happens or it doesn't' way

I think it's less, something like 37% I seem to remember an old maths teacher telling me...i could well be wrong though.

[gatso]

I think it's less, something like 37% I seem to remember an old maths teacher telling me...i could well be wrong though.

if I take a 51 to 1 shot 52 times my expectation is break even surely?

[gatso]

that's the number of different possible shuffles, approximately 80 unvigintillion in the US apparently or 80 undecillion in the UK. the difference being the disagreement over how many are in a billion or a trillion

hmm, there seems to be some dispute over the english naming conventions here

[mulhuzz]

I think it's less, something like 37% I seem to remember an old maths teacher telling me...i could well be wrong though.

if I take a 51 to 1 shot 52 times my expectation is break even surely?

no because the probabilities aren't independent totally --> like if your deck rolls the Ace of spades the first card and it's in the second position in my deck, try as I might, I can't get there.

[doubleup]

http://sportsillustrated.cnn.com/vault/article/magazine/MAG1135210/index.htm

http://www.dailymail.co.uk/sciencetech/article-2065728/Whist-players-dealt-complete-suit-opening-hand.html

[gatso]

nah, before we look at any of the cards we have 104 unknown cards giving us 52 pairs, each of which is 51 to 1 to be a match

to simplify your objection lets take the 2 decks and strip out 50 of the cards leaving us with just the    s and the  s

shuffle them up and half the time both pairs will match, half the time neither so on average one pair matches from each 2, it's 50/50

your objection

if your deck rolls the Ace of spades the first card and it's in the second position in my deck, try as I might, I can't get there

while completely true has no bearing on the odds. we're dealing with unseen cards, not peeping at some of them first

at least I think that's right

[Redsgirl]

2 of the birds we took pictures of on our weddingmoon: The one with the spike we called Jedward. No idea of species.

The first is an orange weaver, ( I think) the second is a red-whiskered bulbul.

Tom, did you already know that - if so, very impressive! - or did/do you use a website to identify them?

I have tried a few sites which help identify birds, several books, too, but few really work well for me.

Left my house @ 5am this morning, I could hear the birdsong, loads of different birds singing away, & I know as sure as hell they were not pigeons, but that is all I see in my mini-garden so far. I've spread some niger seed around the ground, too, now, some grated cheese, & bread crumbs. Guess them plump pigeons will just get plumper.

The web, (speshly google images) is great for confirming your identification, but useless if you don't have a starting place.

When I was young, I used to go to the annual 'National' bird show in London with my dad. There were hundreds of different spices there and, (as you do when you're young) I must have absorbed information without trying or even realising I was doing it.

I knew the orange weaver because I remember we spent some time talking to a man who was exhibiting them. (We were actually trying to sell him an escapee Zosterop that we had caught, but that's another story).

I also had a vague recollection of a crested bird called a bulbul. The name stuck in my mind because of a Val Doonican song which was popular at the time.

"We would like you to know, that you trod on the toe, of Mr Abdul Abul Bul Amir"

I put Bulbul Mauritius into google images and there it was.

Your missing an E.

[mulhuzz]

nah, before we look at any of the cards we have 104 unknown cards giving us 52 pairs, each of which is 51 to 1 to be a match

to simplify your objection lets take the 2 decks and strip out 50 of the cards leaving us with just the    s and the  s

shuffle them up and half the time both pairs will match, half the time neither so on average one pair matches from each 2, it's 50/50

your objection

if your deck rolls the Ace of spades the first card and it's in the second position in my deck, try as I might, I can't get there

while completely true has no bearing on the odds. we're dealing with unseen cards, not peeping at some of them first

at least I think that's right

right, I got the pen and paper out and had a bash.... I'm pretty sure this is right.

we need to basically solve (for n=52)

p = 1 - [1/2! - 1/3! + 1/4! - 1/5! + . . .1/(n-1)!]

that series (after the 1-) converges really quite quickly and approximates to 1/e (it's a McLauren (sp?) series expansion for e^-1, as I remember from school) and so the chance is 37% that there is no pair (1/e) and 63% that there is a pair. (approx)

The reason for this is that, since my deck (A) is shuffled, whether yours (B) is or not is wholly independent and you could just have all the cards in order -> so we can label them 1 to 52.

So basically, for there to be no match, we're looking for the probability for all n 1-52 that A(n) =/ n -- this turns out to be a regular nCk problem (how many ways can we deal two decks) which summises to:

n! − nC1 · (n−1)! + nC2 · (n−2)! + ...  nCn · (n−n)!

or

n! − n!/1! + n!/2! + ... (−1)n

which for n=52 gives the first 53 terms of the MacLaurin series for e^-1 as above.

that make sense?

[gatso]

that make sense?

not to me, no

it may well be right in which case it's very impressive but I'm lost

but

P A(n)=n is still 1 in 52 so I still contend that we have 52x 51 to 1 shots giving us an overall even money bet

[mulhuzz]

P A(n)=n is still 1 in 52 so I still contend that we have 52x 51 to 1 shots giving us an overall even money bet

some googling has told me that this is known as the Montfort problem:

http://www.math.uah.edu/stat/urn/Matching.html

[tikay]

[gatso]

ffs, had hoped that was going to be really quick to read as I have to go out, I'll read it later

assume it means I'm wrong. were you right?

it actually makes quite a lot of sense empirically that it should be less than 50% when you start to think of examples

like I always wondered why secret santa draws don't constantly end with someone buying for themselves