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[Job]

Yes the slow creep towards proving youre not comitting a crime.

It actually started with SORN on your vehicle.

 

[Scouse]

It actually started with SORN on your vehicle.

You really don't like having to tax and insure your car properly do you.

 

[Job]

Not a problem, I just think its very concerning having to tell them Im not breaking the law, and outside of your tired joke, you know thats true.

 

[Moriath]

China proposes controversial HK security law

Thats a load of shite China rule is something i dont agree with

 

[Lamp]

Here's one for you.

5 people each have a standard deck of 52 cards. They each shuffle their pack randomly. Each person draws one card, places it face down, and after everyone has a card, they all turn them over.

What is the probability that everyone has the same card?

(eg 5 x 6 of clubs; 5x jack of spades, etc)

 

[Scouse]

Here's one for you.

5 people each have a standard deck of 52 cards. They each shuffle their pack randomly. Each person draws one card, places it face down, and after everyone has a card, they all turn them over.

What is the probability that everyone has the same card?

(eg 5 x 6 of clubs; 5x jack of spades, etc)

One in 380,204,032?

 

[Wij]

One in 380,204,032?

52 times too high:

7,311,616

 

[Rubber Bullets]

Why is that Wij? I got the same as Scouse, 52^5 is that wrong?

 

[Moriath]

Why is that Wij? I got the same as Scouse, 52^5 is that wrong?

Maybe the first guy doesnt have to match with anyone so his draw could be any card? Its the only reason i could think of to not x the last 52

 

[Moriath]

480 Rainier Dr, Pittsburgh, PA 15239 | MLS #1445093

A mad house but price isnt much

 

[Lamp]

All 5 cards have to match. The order is irrelevant

 

[caLLous]

Why is that Wij? I got the same as Scouse, 52^5 is that wrong?

Without putting too much thought into it, I'd say 52 ^ <number of packs> as well. If there were only 2 packs there wouldn't be 52 possibilities, there'd be 52^2.

 

[Rubber Bullets]

Maybe the first guy doesnt have to match with anyone so his draw could be any card? Its the only reason i could think of to not x the last 52

Yeah, that's it isn't it. The first guy picks any card, so the second guy has a 1 in 52 chance to match it. The third guy 1 in 52^2, and so the 5th 1 in 52^4.

 

[Lamp]

I got what @Wij had. It was a question on QI (series M).

(1/52)^4

but according to the "QI elves" the odds are one in 2 billion. Eh?

I'll ask my sister. She has a degree in mathematics.

 

[Job]

Here's one for you.

5 people each have a standard deck of 52 cards. They each shuffle their pack randomly. Each person draws one card, places it face down, and after everyone has a card, they all turn them over.

What is the probability that everyone has the same card?

(eg 5 x 6 of clubs; 5x jack of spades, etc)

Thats where probability blows my skull.
If its 1/52 say to get the 5 clubs..then the next day I say whats the prob of the 5 of clubs, its the same 1/52.
But I got a 5 clubs yesterday...doesnt matter
But if I turned two seperate packs at the same time it wouldnt be 1/52 for them both being 5 clubs.
That would be joint probability
How small a time gap can you leave..how does this work?

What's the Difference Between Probability and Cumulative Probability?

 

[caLLous]

Pakistan passenger plane crashes in Karachi

Carrying ~90 passengers (the Indy article says 107 on board), it crashed into a "residential area" in Karachi.

 

[Scouse]

Pakistan passenger plane crashes in Karachi

Carrying ~90 passengers (the Indy article says 107 on board), it crashed into a "residential area" in Karachi.

That's horrid.

Not to belittle in any way - it's interesting to make the point that we were at over 8 times that (at 854) on daily deaths from coronavirus.

 

[Zarjazz]

Why is that Wij? I got the same as Scouse, 52^5 is that wrong?

Because while 52^5 is the odds of one particular identical combination (say all Ace of Spades), since there are 52 possible combinations where all cards match the odds are then 52^5 / 52 = 52^4.

I'm almost certain the QI Elves are wrong about this one unless there was some other detail in the original question that changes the described scenario.

 

[Scouse]

Because while 52^5 is the odds of one particular identical combination (say all Ace of Spades), since there are 52 possible combinations where all cards match the odds are then 52^5 / 52 = 52^4.

I don't follow that. Pls explain.

If one card combo is 52^5 the existence of other combos is independent and the chances of one of those other combos is the same, independent, probability. No?

 

[caLLous]

Because if you have 2 packs of cards and you pick one card from the first pack, there's a 1 in 52 chance that you pick the same card from the second pack, so it's 52 ^ <number of packs - 1>.

 

[Scouse]

Because if you have 2 packs of cards and you pick one card from the first pack, there's a 1 in 52 chance that you pick the same card from the second pack, so it's 52 ^ <number of packs - 1>.

That's unrelated to @Zarjazz' explanation which is about cards, not packs?

Anyway, because the draws are unrelated to each other the probability should be calculated multiplicatively like this:

1/52 x 1/52 - for 5 packs.

The draw of a card from any pack has no effect on any subsequent draws.

 

[Scouse]

It's been a while since I had to look up basic rules of probability. But here goes.

The pertinent bit:

If A and B are independent events, the probability of this event happening can be calculated as shown below:

So, multiply the probabilities of separate independent events.

Edit. Gah, barely readable, but that's what it says

 

[Scouse]

Because if you have 2 packs of cards and you pick one card from the first pack, there's a 1 in 52 chance that you pick the same card from the second pack, so it's 52 ^ <number of packs - 1>.

I do get your point about packs. But yoir thinking seems to imply conditionality.

Think of it as either one pack of cards with the selected card being replaced and the cards shuffled, or, five packs of cards all having a card picked at the same time - each deck has a probability of 1/52 for a specific card, independently...

 

[Wij]

The chance of drawing the 'right' card from the first deck is 1. It's then multiplied by 52 for each of the subsequent decks.

Shall we do the Monty Hall problem next?

 

[Job]

The universe isnt interested in our attatchment to individual cards.

Thats what starts to eat in to me, Im sure the key to consciousness is hidden in probability.

 

[Job]

That's horrid.

Not to belittle in any way - it's interesting to make the point that we were at over 8 times that (at 854) on daily deaths from coronavirus.

Celebrity deaths.

 

[Zarjazz]

I don't follow that. Pls explain.

If one card combo is 52^5 the existence of other combos is independent and the chances of one of those other combos is the same, independent, probability. No?

We are furiously agreeing. I just flipped the odds so it may look confusing. 52 x probability of each unique combination (1 / 52^5)

 

[Scouse]

We are furiously agreeing.

My favourite type of agreeing

In fact my favourite type of a lot of things you do in front of the internet

 

[Scouse]

The chance of drawing the 'right' card from the first deck is 1. It's then multiplied by 52 for each of the subsequent decks.

Nope. It's 1 in 52.

You're creating conditionality (bad english I know) if your looking at it after you've drawn your first card - but we're looking at five untouched decks and saying "what's the chance of drawing five the same?"

Five blokes, 3... 2... 1... draw!

 

[Rubber Bullets]

1 in 52 is the chance of drawing a specific card from the first deck. In this case we don't care what that card is, so the first deck just acts as a random card generator, the only requirement is that the card from the second deck needs to match it, with the odds of 1 in 52. for the third deck then to match is 1 in 52^2 etc.

I've no idea what you mean by conditionality in this case, but I don't think it's relevant, all three cards can be drawn at the same time or in order, the odds don't change.

And I don't care what anyone else says, the fucking plane will take off!!