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.

One door is not stronger or older than the other.

The difference is that one door was chosen randomly by the contestant, while the other was one of two, and chosen by someone (the game host) who knows what they are and opened a door that didn't have the prize.
That's why the odds are not 50-50.

Everyone in your pub is excited your on a quite show.
You will either win a car or come home with no car.
If you win a car you do the next show to win another car.
What are the odds on winning 10 cars.
There are no doors to open .

You need to explain how to win the car in your latest pub example. Is it truly random (coin toss) or a less random choice? Because that determines the odds.

There are no doors to open .

How do you get in the car then?

It does not matter as the odds are calculated on car, no car.

you know exactly what I mean by swap given the length of this thread - take the door you didn't select originally. But thank you for just proving once and for all that you are just trolling.

But I'm more likely to win the car each time if the challenge is weighted in my favour.

By this logic, annuities rates should sky rocket

"at the end of each year you are either alive or not. 50/50. By 10 years the odds that you are still alive is 1/1024....Please calculate my annuity on that basis"

StatisticallyChallenged
But once you swap doors you can only win or lose.
While the swap gives you a better chance to win, there is no guarantee it wins everytime.

In your own words While the swap gives you a better chance to win, there is no guarantee it wins everytime.. Can you not see how having a better chance, changes the odds?
Otherwise at the start of a horse race, they'd all have the same odds, wouldn't they?

"While the swap gives you a better chance to win, there is no guarantee it wins everytime."

This is true. No-one has said otherwise.
I think this thread is so beguiling because Mummy keeps mixing absolute truths with absolute lunacy.

So she is either a moron or a genius. I give it a 50/50 shot.

Unlike the probability of winning the car.

I think this thread is so beguiling because Mummy keeps mixing absolute truths with absolute lunacy

When you place a bet it is before the event is known.
The odds are worked out on what each coin toss could, so if you bet on Toss 10 only , that is very different to before Toss 1 has happened.

Mummmy

What do you think the chances are of an event with a 2|3 probability happening 10 times in a row?

The landlords bet is do you have 10 new cars on your drive .
Not about the door odds.
How hard is it too see , two different questions.

@chomalungma your answer to your question is right....
My answer to my question is right.

Jesus Christ. I’ve genuinely had more intelligent debates with my 8-year-old than this.

@mummy2017 - you’re just embarrassing yourself. You’re not as nearly smart as you clearly think you are. You don’t actually seem to know what you’re asking, and your ability to string together a coherent question is about on par with your ability to understand quite basic maths.

I’m out.

That answer, which you keep posting, is only valid if the two options (win/lose, car/no car) have equal probabilities. In a coin toss they do. Unless you are winning the car based on a coin toss it's not the right answer. It is not 1/2 because there are two possible outcomes, it is 1/2 because the two possible outcomes are exactly equally as likely.

Let's say it's "you can win the car if you throw a 1 or a 6" on a normal 6 sided dice.

What is the probability of 10 cars?

I think Mummmy just wants to argue.

*not nearly as smart

Oh god this thread was half hilarious, half painful to read. Can anyone tell me the odds of answering a probability question correctly when the question has been phrased in an ambiguous way?

My question is based upon only on two outcomes. Car or no car.
The other bits your adding in change the question, which means your answers are not for my question.

The outcome is still car or no car; but the chance of that happening depends on what you have to do to get the car and the probability of you doing that thing.

Number of outcomes tells you nothing about probability of achieving outcome, unless the number of possible outcomes is 1!

It does not matter as the odds are calculated on car, no car.

Of course it matters.
If the game is "turn up on the show, get given a car" you have a 100% chance of winning a car every time.
If the game is "toss a heads, win a car" you have a 50% chance of winning a car each time.
If the game is "solve a rubiks cube in 30 seconds with your hands tied behind your back and being chased by a snake, win a car" then your odds of winning the car are very small.

What the game is depends how likely it is you'll win a car ten times in a row. Or however many times you play the game.

The pub does not care about what the show is about.
It could be coin with car one side, no car the other.
All they want to know is are you arriving at the door with a new car each time.