to ask if you can answer a question re probability (Maths question)

[Fainne]

So, say I have 20 cards in a pack.

I pick one. It's the Ace of Diamonds let's say for argument's sake.

I then pick another one out of the same pack of 20 cards the following day.

Am I correct in saying that the odds of me picking the same card is a multiple of the single odds?

So 1/20 x 1/20 = 1/400

?

Because I've someone telling me the odds are still 1/20 that I'll pull the same card.

*nailing

Sorry lord, you’re wrong too.

Before you draw any card, the chance of you getting the same card twice is 1 in 400.

That is only true if you need the same specific card to come out twice, otherwise there are twenty different ways of getting two matching cards making it 20/400 or 1/20.

A maths troll. That's new

Worryingly little discussion about cards being shuffled between day one and day two for 153 posts.

I'm outta here.

There are basically three different scenarios flying around i think. Excluding dice...

1. At outset, what is the probability of me picking a specific pre defined card on two consecutive days, assuming card replaced and a 20 card deck. 1/400

1. Having achieved my specified particular card on day 1, what is the probability of getting that card on day 2. 1/20

1. Without specifying what card I want, what are the odds of getting the same card on two consecutive days. 1/20. It doesn't matter what you pull on day 1, only that you pull it again on day 2, which makes it the same probability as option 2.

Exactly.

If the first card doesn't matter then the only purpose of the first round of drawing the card is doing is selecting the card for the second round.

There's no way of the "challenge" failing at this point because there is no possibility of drawing a "wrong" card, they're so correct. So 20 out of 20 cards are sufficient to fulfill the first round criteria. So 20/20 * 1/20=1/20

If you need the 7 of spades or King of clubs specifically then there are 19 cards which will fail the first round and only one which will be successful. Same again in the second round. 1/20*1/20=1/400

It's 1/20 of getting the same card both times.
It's 1/400 that it will be the ace of diamonds both times.

There are basically three different scenarios flying around i think.

I agree, and the op has moved seamlessly between them depending on which poster she fancies insulting at the time.

I wonder whether they’ll be back.

Anyone else questioning the timing?
Why draw cards in different days? Why wait 24 h?

I am questioning many things...

The odds of picking THE ACE OF DIAMONDS twice is 1/400.

The odds of picking THE SAME CARD TWICE NO MATTER WHAT IT IS is 1/20 because there are 20 different ways that could happen - 1/400 odds for each of the 20 cards.

(I thought @PurpleDaisies was right but I asked my dad who has a masters in maths from Oxford)

SerendipityJane

Hey, I mentioned shuffling. [smug]

@PurpleDaisies is correct. I have a PhD in statistics, if it matters OP

I'd love to see the OPs take on the Monty Hall problem.

The highlight (lowlight) of the thread has got to be that the probability of getting matching numbers on two dice is less than one in 60 million. That ones going in my “ridiculous common sense test fails” folder.

I'd love to see the OPs take on the Monty Hall problem.

Don't. I have stuff to do this morning. I'd have to turn my wi-fi off if she came back and you asked that!

I am reminded of people who felt the "need" to choose new numbers when the lottery draw changed ....

It's a shame that OP is a troll, as it's easy to see how someone could be confused by this, but there are lots of great explanations on the thread with different ways of looking at it that should help someone who was really struggling.

Having said that, I did find it quite entertaining in the middle of the night.

@lotsofoysters I love the Monty Hall problem!

@PurpleDaisies I hope you got some sleep! xkcd.com/386/

Sod it. Just turn all the bloody cards over and choose the one you want. Problem solved.

Ha yes, Serendipity.

I also remember trying and failing to explain to DM that the combination 1 2 3 4 5 6 is as equally likely to come out as any other combination of numbers.

It was also widely reported when the lottery started that about 10 000 people had chosen that combination, so if it does come out, they will have to share the jackpot with all the other people with the same numbers, so will win a few hundred pounds.

Purpledaisies - yes you are right, and I thought of that as I was typing, I was distracted by the Ace of Diamonds as a specific example.

Before drawing anything, chance of getting Ace of Diamonds twice = 1 in 400.

After drawing Ace of Diamonds once, chance of getting Ace of Diamonds a second time in a row = 1 in 20.

If the specific card doesn't matter, then 1 in 20 chance of drawing same card twice in a row, because the first card can be any card so doesn't have a probability attached to it. (Or to put it another way - if you draw one card from a pack of 20, you have a 20 in 20 probability that the card you draw will be one of those 20 cards).

My point though, was - the way you express the problem, is what changes the solution. As the thread shows, not being specific about what you're calculating is what leads to confusion...

What is interesting is that if someone knows the cards and shows you one that isn't the matching card, the odds are still the same of you picking the matching card, apparently. Is that correct, stats wizards?

Agree PurpleDaisies, change the question and insult those who answer...then claim the dice answer. 😂

Just found this thread and read the whole thing Very impressed by your patience, @PurpleDaisies!

I think most people who've read and responded do understand that the people OP has been throwing insults at are the ones who are mathematically correct; so I sincerely hope OP is a troll, because otherwise their behaviour is pretty embarrassing!

I have to thank those who have been up over night for this most entertaining thread. It has made me want to smash my keyboard a couple of times, but it's been exceptionally cringey in a way that could match The Office.

There is a difference between P[A and B] and P[A|B] which people always fail to understand. It's like when flipping a coin and it's come out heads 5 times, the probability of it being heads on the 6th time is still 1/2.

One of the Sunday tabloids used to (and probably still does) publish a short analysis of how many times each individual number had been rolled in the National Lottery. Think it was the Mail on Sunday, but it might have been the Express.

My mother used to buy that paper partially for those stats. She used them to 'work out' which balls were 'most likely' to come up next, which is what it was intended for. At one point she was doing five different lottery tickets on Wenesday and on Saturday, trying to cover all the bases between her own personal lucky numbers and the numbers that allegedly were most likely to come up on a Wednesday and so on.