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.

To work out the probability of winning the car, we have to know what challenge is set to win the car.

I think mummy knows she's lost this argument and rather than admit we were right (and she was wrong) is going to keep on spewing out random scenarios which make less and less sense until we all give up so she can then post, 'see, i was right'.

Anybody who is taking/laying bets on the basis of whether you get a car or not (your theoretical landlord if I recall) cares what you have to do to get that car. Because it changes the likelyhood that it would happen and therefore the chance of them paying out.

If the game is the aforementioned 30 second hand tied rubix cube I will happily write bets on a 1/2 basis
If it's on a coin toss, I won't.

No you don't.
The production team know the show means they order ten cars on a returns basis. The maximum wins on the show is 10 cars.
Each show has an outcome of someone wins a car or someone goes home empty handed.
The odds of handing over 10 cars is 1/1024.

I refer you back to chomalungma's question at 15.59 What do you think the chances are of an event with a 2|3 probability happening 10 times in a row?.

I refer you back to chomalungma's question at 15.59 What do you think the chances are of an event with a 2|3 probability happening 10 times in a row?.

How does this change that the contest either goes home with a car or is empty hand?
So long as you have a chance to not win that still remains an end result.

If played by people who understand maths, the odds of handing over 10 cars is (2/3)10 = 1024/59049 = 0.017 not (1/2)10.

Of course if played by people who don't understand and switch boxes 50% of the time, then the odds go to

(2/3x1/2 + 1/3x1/2)10 = (1/2)10 = 1/1024 = 0.000976

If played by people who always stick to what they said first (even when it's not right) and never switch boxes the odds go to

(1/3)^10 = 1/59046 = 0.0000169

Yes your answer gives the possibilities of winning, and is right for that question.
But that was not my question.
Mine was win or no win. 10 times in a row.

But the winning can be either weighted or not weighted and this affects the outcomes (see teenplus sums).

No one in the pub cares about weighted doors.
They care about a brand new car driving upto the pub.

Mine was win or no win. 10 times in a row.

What are the odds of winning?

It depends on the chances of winning.
If the chances of winning are 2/3rds you get one answer
If the chances of winning are 1/2 you get another
If the chances of winning are 1/3 you get another still.

There are two possible outcomes, but not equally likely (if we're still talking the boxes game). The probability of winning 10 times in a row if they are independent events is p^10 where p is the probability of winning once.

Try with a dice. You win if you throw 1,2,3 or 4. You lose if you throw 5 or 6 (so 2/3 chance of winning)

Throw just 3 times. Do this 50 times. How many times do you win 3 times in a row?

Now swap it round, you 'win' if actually you scores 5 or 6. How many times did you win 3 times in a row now?

In your question your implying the weighted door will always win . There is still a 1/3 chance the other door wins.
But it does not change the fact the show only needs 10 cars to cover 10 wins.

I think mummy knows she's lost this argument and rather than admit we were right (and she was wrong) is going to keep on spewing out random scenarios which make less and less sense until we all give up so she can then post, 'see, i was right'

It seems to be their SOP

Well of course in 10 shows the most number of cars given away could be 10. That's kind of obvious.

And no I am not saying the weighted door will always win! We are calculating the probability the person has the winning door.

If every contestant choose to switch then every contestant has a probability of 2/3 of having the right door.
To then get the probability of 10 wins in a row you go to (2/3)^10 (because that's the maths).
Of course actual events can buck the probabilities (which is why they are probabilities not certainties ).

Just because there are 2 possible outcomes, doesn't mean that the two are equally likely.

Well, i agree with you mummy, that if the game is played 10 times, you won't need more than 10 cars. But you do keep changing the rules of play. In some examples you have a 50/50 chance of winning, and in others there is a 33/66% chance. This will affect the odds of 10 wins in a row,

My odds are based on once the game has been played .
When you know if they got a car, or didn't.
Simply that. Not the calculations of how likely you were to win the car.
It is a simply probability problem.
Used in the pub to bet.

It seems that, in mummy2017's head, the odds of winning a coin toss ten times in a row are the same as drawing the Ace of Diamonds out of a full pack of cards 10 times in a row.

Cards one question.
Coins another question.
You can combine the results to create a third question.

@TeenPlusTwenties
If you flip a coin 10 times, what is the possibility of 10 Heads?

My odds are based on once the game has been played .

Then the odds are 1:1, because you already know the result.
The odds apply before you know the result, so that you increase your chances of winning or know how much to bet.

What you mean is frequency, which can be used to calculate odds.
Even so, you'd find out that the frequency that you'd observe is not what you seem to think it is.

If you flip a coin 10 times, what is the possibility of 10 Heads?
It's possible. Possibilities aren't shown in numbers. It's either possible or its not.

@Lweji
True , wrong word.
But since the landlord did the betting before the shows were recorded, he had no idea of the result.

If you flip a coin 10 times, what is the possibility of 10 Heads

It's 1 in 1024 because the odds of getting a head is 1 in 2.

Going back to the producers.

If there were 9 shows in a week, there is a 2/3 chance that 6 cars would be won.
There are lots of permutations so you can get 6 cars.

There is only 1 way you can get 9 wins.
There is only 1 way you would never have to give a car away.

But there are lots of other permutations so you can get 2, 3,4,5,6,7 or 8

And the fact that people know the strategy give a win of 2/3 and a loss of 1/3

So as a producer, it would be a rare event to give away 9 cars in a week
It would be even rarer to give away no cars.
The most likely event is 6 cars - a slightly skewed distribution.

So, why do you want "odds" after the game is played?
It makes no sense.

If someone is betting on a contestant winning or not, they'll have to factor in how likely it is that the contestant will swap or not in addition to the odds of where the winning box is.