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.

It's still 1/20. If the question was what are the odds of pulling same card twice in a row (assuming you shuffled etc) it's 1/400. I think!

So you're saying it's what I want that determines probability. Fascinating.

Wanting a specific outcome and for it to happen a specific number of times is a different situation to just wanting the same thing to happen a specified number of times. You're effectively introducing an additional selection criteria if you want a specific thing, rather than just two (or more) the same

Let us hope OP is not as terribly unpleasant in real life as he/she comes across in this thread.

Funny that people mention the Monty Hall maths problem, that drives me nuts every time I watch the film 21 (great film).

Fainne, you are describing two different scenarios.

Scenario 1) you are preselecting the first card: eg ace of spades. That means you have a 1 in 20 chance of getting that card. Then the chance of getting another ace of spades in the next drawing is 1/20. So multiply together to get 1/400

Scenario 2) not preselecting the first card. This means there is 1/1 chance in your first drawing as you don't care what it is and 1/20 of matching that random card in your second drawing. Multiply to get 1/20.

As I also have higher maths from leaving cert and a degree in maths I just wanted to put you right and try to stop you from embarrassing yourself and the nation any further.

How's the weather under the bride, OP?

Thank you for a hilarious thread. Even if it's trolling I am truly amused. P. S. I am rubbish at maths and even I understand this

Bridge* not bride 😳

Another way of looking at scenario 2 is that the cases that would satisfy the criteria are:
Ace of spades and ace of spades (1/20 x 1/20)
Two of spades and two of spades (1/20x1/20)
Three of spades and three of spades...
For all of your 20 cards.
Since you can have any one of these options to satisfy the requirement that you have 2 of the same card, (eg 2 ace of spades OR 2 two of spades OR...) this makes 1/20 x 1/20 + 1/20 x 1/20 + ... (adding up the probabilities of the 20 possible pairs). This makes:
20 x 1/20 x 1/20 = 20/20 x 1/20 = 1/20

@Fainne

I understand where you are coming from.

But, to help you understand...

The outcome is the outcome.
That will never change.

The probability DOES change based on what you want (or don't want. Or don't care)

I have had great fun trying to explain the Monty Hall problem.

I have done it with 100 cards.

Told someone that I know where the prize card is.
Asked someone to see if they can guess it. They pick a card.
I then turn over ALL the other cards except their card and the one that I know. Assuming they haven't picked the actual card.

Then I ask them if they want to stick with their choice or do they think that they might want to go with the other card, given that I know where it is and I have turned over all the cards that I know are definitely not the correct answer,

Some people have still stuck with their original choice, thinking it's the same probability.

People really struggle with the difference between the probability of the outcome and the probability of them making a choice. If you pick at random between two choices then you have a 50/50 chance of picking the winning thing, but that doesn't mean the thing you picked has a 50/50 chance of being the winner.

Eg if there are two horses in a race and I choose one randomly, I have a 50/50 chance of picking the winner. But once I've chosen, the horse I've picked doesn't have a 50/50 chance of winning. One horse might be a prize stallion and the other might be lame.

The Monty Hall problem is similar. Monty fixes the choices so that one door is more likely to win, but people think because they have a 1 in 2 choice, each door has an equal probability of winning.

Have been enjoying this thread and I am not a mathematician. The last few pages are enlightening in terms of the crux of the difficulty in the OP's understanding (if indeed not a troll).

OP, you seem to think that what people are telling you is that what you want to happen, in secret in your head, somehow influences the likelihood of that thing happening. It doesn't change how likely it is for the secret thing to happen, but what you want to happen changes how likely you are to get what you want.

So, if I am hoping for sunny weather today, that has a certain likelihood of happening. If I change my secret wish such that I am now hoping for a sunny day with a rainbow appearing at 5pm, that doesn't make it any more or less likely that it will be a sunny day with a rainbow appearing at 5pm, but it does make it significantly less likely that I get my secret wish than when my secret wish was simply for a bit of sunshine. Me wishing for something doesn't make it more or less likely to happen, but me wishing for something that is less likely to happen means I am less likely to get my wish. That may sound a bit circular but I don't think it is.

Flipping a coin is good.
A good way to make some money from someone who doesn't get maths.

I offer to flip a coin twice.
If I get a double, I win.
Chances of getting a double is 50%
If I want double heads, it's 25%

That is GCSE maths Draw a tree diagram and add up the probabilities.

After reading the first few pages of this thread I was hoping the OP wouldn't turn out to be Irish..... We're not all as rude and ignorant as that I promise!

@PurpleDaisies there's a vacancy at the school my children go to. Please will you apply for the job??!!!

The Monty Hall works long term.
But for the contestant on that day the odds change for them as well.
They had a 1/3 chance of winning.
Take a door away and they are looking at a 1/2 chance of winning.
So from 33.3% chance to now 50% chance.

Take a door away and they are looking at a 1/2 chance of winning

No it isn't.

Imagine if I had 1 million doors.
I know the door the prize is behind.
You pick a door.
You have a 1 in a million chance of being right.
I then open all the doors except 2. I don't open them at random because I know the correct door.
You are left with 2 doors.
Do you have a 1 in 2 chance of having picked the correct door?

The chances of you having picked the correct door is 1 in a million.
That probability hasn't changed.

I admire @PurpleDaisies for her patience, persistence and clear explanations. I'd have given up after the second failure to understand.

@chomalungma
Yes it has.
Forget all the taken away doors.
What your saying to the person is here are two door choose one.

What your saying to the person is here are two door choose one

So you are saying that if you had 1 million doors and I knew the correct door but you didn't.

If you pick a door at the start. At random.
I then open all the other doors except 2.
There is a chance you could have picked the correct door at the start.
1 in a million.
I deliberately ensure that one of the 2 doors left is the correct one.

Do you think that if you never changed door, you would win 50% of the time because there is now a 50% chance of being right?

Try it with 10 cards - you will need someone to know where the card is that is 'the correct one'.

See if you win half the time by not changing.

No, mummy the host isn’t allowed to open a door with a prize. They have avoided opening the other door when given the opportunity. In the million door case, they opened door 1, 2, 3, 4 then skipped door 5 then opened door 6, 7, 8.....skipped the one you picked, then opened all the other doors up to 1,000,000. Is it more likely that you correctly guessed the door in a million on your first go and they randomly skipped door 5, or is it more likely that you picked the wrong door out of a million and they skipped door 5 because it’s the one with the prize behind it?

"By similar logic, let's say the game was that you had to pull the Ace of Diamonds five days in a row. At the outset, the odds are 1 in 3,200,000.

The OP seems to think that, even if you had already successfully drawn the Ace of Diamonds 4 days in a row, that your odds of then drawing it on the 5th day remain 1 in 3,200,000, despite there only being one choice left to make with 20 cards to choose from."

Yes the events "draw an ace of diamonds two times in a row" and "draw an ace of diamonds given that you have previously drawn one" are different events, with different probabilities.

To be fair to the OP I think they are considering the event "draw any card twice" but failing to appreciate that it is a different event than "draw a pre-decided specifically named card twice".

@chomalungma understands the Monty Hall problem, mummy2017 doesn't.

Yes on the million to one door the odds of the door left increase on it being right and yours being wrong.
But what if your door was the right one.
On stage on the night when they show you two doors sat which will you open it is 1/2 chance of being the right door.
As you only have two doors left to open.
The odds maybe stacked on all the other doors holding the answer the more doors you include in the puzzle, which would change how you play the game.
But on the 3 door game.
You had a 1 in 3 chance of winning a prize.
Remove one door and your odds change.
You now have a new game with a 1 in 2 chance.

On stage on the night when they show you two doors sat which will you open it is 1/2 chance of being the right door
As you only have two doors left to open

No - one of the doors has a 1 in 3 chance of being right. As that was the odds you had at the start.
The other door has a 2 in 3 chance.

The best thing to do is to do lots of trials. Experiment .

Get someone to pick a card. Put it in a pile of 3.
You guess which card it is.
Tell the other person to show the card that it is not.
That bit is crucial.

Don't swap.

Do you win 50% of the time or a third of the time?

You should do enough trials to see a pattern. Not just 3!!

@mummmy2017 that's not how it works.

Three doors, A, B, C. Say I pick door A. 1/3 chance of the prize behind A, 2/3 chance it's behind one of the other two. Monty essentially offers me the choice of A, or the choice of B and C together, since if the prize is behind B or C, I get it if I switch doors.

You're confusing the probability of picking the right door with the probability of the prize being behind each door. They're two different things.

Another interesting probability that is a bit counter intuitive is that if you spin a roulette wheel 8 times, there is around (to decimal places) 50 percent chance a number will come up twice.