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.

What do you mean “if you’re wrong?”
Surely now you’ve sobered your and had some sleep you can see it.

The comments section on mysticbanana.com does not count as a credible source of mathematical information to me.

Yes, you're not the only one who is wrong.

Fainne
If I'm wrong, I ain't the only one who's wrong!

I once watch You've Been Framed, where a soldier marched in the opposite direction to the whole troupe. Everyone laughed at the video, then the presenter said 'What was really funny about that?'. The soldier said "I was the only one marching in the right direction'.

Being in large number doesn't always make you right...

Any comments on the dice issue? I’d love to see if anyone else thinks the odds of throwing two matching numbers is less than 1 in 60 million.

Try “Ineedatutor.co.uk” or “mymathsisbad.com”.

I've got to hand it to the OP, they've managed to get another angle on this. Looking at mathematical probability on a fucking tarot reading site. That is some next level shit.

11, 12, 13, 14, 15, 16,
21, 22, 23, 24, 26, 26,
31, 32, 33, 34, 35, 36,
41, 42, 43, 44, 45, 46,
51, 52, 53, 54, 55, 56,
61, 62, 63, 64, 65, 66.

The above is all the ways you can throw a red and blue dice.
There are 36 ways.

Probability throwing a double 6 = 1/36
Probability throwing any double (don't care, just want the blue to be the same as the red) = 6 x 1/36 = 1/6.

Same with playing cards.

And even on said site:

It’s 1 in 78. It works that way because you didn’t specify which card. If you draw a card, all that needs to happen for your condition to be met is for that same card to be drawn again. It would be 1 in 6084 if you had a specific card in mind that you wanted drawn twice in a row.

All you've shown OP, is that you as stupid as the stupid people on a tarot reading site....

Did anyone else think the reason the OP chose 20 cards instead of the more traditional 52 is because they are sitting at home somewhere with their shoes and socks off ?

I didn’t initially think this was genuine, but if I were to start a trolling thread, I don’t think my pride would let me choose a ‘thick as shite’ persona - so I think it must be real.

@Fainne
If I'm wrong, I ain't the only one who's wrong!

Correct, lots of people are as hopeless at maths as you.

With dice: Chances of throwing double 6 = 1/36, but
Chances of throwing any double 1/6

And there we have it. How can the OP not get this.
Chances of picking the ace of diamonds twice, 1/400
Chances of picking the same card twice, 1/20.

If picking the ace of diamonds twice, you pick the first card, it's not the ace of diamonds, it's the 3 of clubs...game over. You've failed.

If picking the same card twice, you pick the first card, it's the 6 of hearts, you're still in the game. All you have to do is get the 6 of hearts next time, and if you have 20 cards, it's a 1/20 chance.

Ffs....this isn't even secondary school stuff, it's primary.

Hi OP!

Let's go back to the dice. You correctly identified that the chance of double sixes is 1/36. Now, what does the denominator (the bottom half of the fraction) represent?

It represents the number of possible outcomes. In this case, there are 36 ways to throw standard dice.

Now what are the chances that I will throw snake eyes or double sixes, if the chance of each on its own is 1/36?

We add them, so the answer is (1/36)+(1/36) =2/36 (cancels down to 1/18)

Next, let's imagine I can win 100 euro on throwing the same number twice with two (six-sided) dice. How many ways are there for me to win 100 euro?

See answers from mathjoe and Bill A in this link. About 5th and 6th replies more or less - so you won't have to scroll too far.

The mathjoe who said the same thing as all of us, you mean?

The probability of pulling any one card is 1 / 78.

Replace the drawn cardback into the deck and shuffle.

The probability of pulling the same card as from the first pull is 1 / 78.

Since the pulls are independent events, we can multiply the probabilities of the seperate pulls to calculate the probability of pulling the same card twice with replacement.

^P(pulling same card twice with replacement)
= (1/ 78)(1 / 78)
= 1 / 6084^

Yes Pothole. He's saying what I'm saying. That it's a multiple of it.

The reason I ended up on a tarot reading site is because it's the only site where the question probability of pulling same card twice was answered on google.

I've seen other sites who state as the majority here are saying.

But logically, I still can't see how the probability of pulling any card is the same as pulling the same card twice.

Mathjoe is wrong (I’m assuming he is answering “what is the probability of pulling two of the same card if the card is replaced”).

What the first card is is irrelevant. The second card only needs to match the first card. The probability of that is 1/78.

The other way of thinking of it is the two matching cards scenarios. There are 78 different ways of pulling two identical cards. 78/6084 which is equal to 1/78.

Mathjoe has got confused and is thinking of the possibility of pulling the same specific card twice.

Btw - there's absolutely no need for being so insulting to someone, even if you think they're thick as shit.

Fainne, I am on page 2 of this thread and your are UNBELIEVABLY rude, especilaly to Purple.

You are being told that you are talking about 2 different scenarios which give 2 different sets of odds, but you refuse to hear that, and are telling everyone else that they are stupid.

Sorry Fainne, but you are the one who is unable to do maths here.

Two words.

Cumulative possibilities.

The reason I ended up on a tarot reading site is because it's the only site where the question probability of pulling same card twice was answered on google.

I will ask just one more time.

Do you mean the probability of pulling a named card twice with replacement, or do you mean the probability of pulling any two matching cards?

I wonder how many people the OP would think need to be in a room to get the odds of two people having the same birthday to be greater than evens ?

I'm hiding this thread now, as it's just a pile on. So knock yourselves out castigating me and insulting me. I won't be reading.

Well, I’m sure everyone is gutted that you’re going.

How unsatisfactory in the end.

So why are you choosing to agree with Bill off the Tarot site?

What makes you think that he is right and everyone else is wrong?

But logically, I still can't see how the probability of pulling any card is the same as pulling the same card twice

I'll try again. The first draw, your objective is to pull any card. You pull a card and get 'the 6 of clubs' (card chosen at random, obviously you could get any card out of the 20).

You put the card back in the pack of 20 cards. You now have 20 cards and are looking for the 6 of clubs, which is 1 out of the 20 cards.

You pull the card - the odds of getting the six of clubs is 1 in 20. Agree. The first draw was just to identify which card was going to be 'your' card and the chances of getting a card is 100% or 1 in 1.

Therefore 1/1 x 1/20 = 1/20.

what is the probability of pulling two of the same card if the card is replaced”).

Yup. But, even so, he specified that the chance of getting the same card as you did before is always 1/78, and doesn't turn into 1/6084.

The odds of getting the ace of diamonds on day one is 1/20.
The odds of getting the ace of diamonds on day two is 1/20.
The odds of getting the ace of diamonds on day one AND day two is 1/400.

Yep

pothole he’s talked about multiplying the probabilities together without considering the fact there are multiple ways that two matching cards can occur. His final answer and reasoning here was wrong...

Since the pulls are independent events, we can multiply the probabilities of the seperate pulls to calculate the probability of pulling the same card twice with replacement.

P(pulling same card twice with replacement)
= (1/ 78)(1 / 78)
= 1 / 6084

That’s not what we were all saying. Hold the line! It’s 1/78 for (any) pair of the same card.