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.

I knew you were wrong about the doors.
The odd do change the more you win the less likely you are to win next time....q

The odd do change the more you win the less likely you are to win next time

No. This is why casinos make so much money,

Because people think they are due a win. Or there is a lucky table. Or a machine is due a payout.

Why should the odds change?

If I have 52 cards, I always have exactly the same chance of picking out the same card every time.

It does not matter what happened in the past.

In the Monty Hall problem, even if the host's door has won all week, you should still swap.

Same as on the lottery. Each number has exactly the same chance of coming up as the others. Every single time.

mummy No. The doors have no memory.

Just like if you toss a coin and get 99 heads, it is still 50-50 whether you get heads or tails on the 100th throw. (assuming the coin is unbiased)

If I had a biased coin that came up heads 99% of the time and tails 1% of the time, you should always bet on heads.

Because that's the one most likely to win.
There is a very small chance that tails will come up.
But a very good chance that heads will win.

Which is why bookmakers are prepared to offer people 500 /1 on Italy winning the 6 Nations. Because it's easy money for them.

It's not the doors it's the possibility of getting the run of winning the car...
Car
No car.

Car, car
Car , no car
No car, car .
No car , no car.
So at the second time off playing you get a
1 in 4 chance of a car.

The probability of
HHHHHHHHHH
is exactly the same as the probability of
HHHHHHHHHT
which is 0.5^10

Once you have already got HHHHHHHHH then a H or a T on the 10th throw is equally likely 0.5 each.

So at the second time off playing you get a 1 in 4 chance of a car

How does the door know what happened before?

Mummy2017, by your logic, is you toss a coin ten times in a row, and get heads every time, then on the 11th toss there is a greater than 50% chance of getting tails. There isn't. The odds of the result of the 11th toss are not altered by the previous 10 results.

Seriously.
Bet on Italy. They haven't ever won the 6 Nations so logically they must be due to win soon.

Actually, @TeenPlusTwenties (why do we always meet on the threads with "maths" in the title? ) and @Spamantha , if a coin did produce Heads 10 times in a row, such a 1 in 1000 event might make me question whether the coin was biased in some way, making an 11th head more likely!

But I agree, if it's been established that the coin is fair, then it has no memory of those 10 tosses when you make the 11th toss, and another head is as likely as a tail.

The odds in any random trial do not change based on what has gone before. Each event is independent

The only way the odds change is if you are eliminating options, i.e. they're not independent.

Imagine you have a sack of balls. 5 black, 5 white. Put your hand in, pull one out. 50/50 black or white.
Replace the ball.
Pull one out
still 50/50

If you don't replace it, then the odds change - becoming 4/9 vs 5/9, and so on.

Randomness has no memory

From what I recall, the odds of winning the new National Lottery with 59 numbers are 1 in 45 million.

So there are 45 million possible combinations and only 1 of them is correct.

Yet they weren't selling 45 million different tickets every week. So they started to get rollovers. And no big winners.

But hey,,,if you buy 2 tickets, the chances of you winning the lottery have doubled!!

However, using logic, you can increase your chances of winning more money in the lottery.

And not getting as much money as you expected.

www.telegraph.co.uk/business/2016/03/24/national-lottery-players-in-uproar-over-five-number-15-win/

"Punters with five winning numbers in Wednesday's Lotto draw ended up with just £15. This compared to £25 for people that matched three balls, and £51 for people that matched four.

The winning numbers were 14, 21, 42, 35, 07, 41, and the bonus ball was 43.

Lotto operator Camelot said it was an "extremely rare" set of winning numbers, many of which were multiples of seven. This set of numbers is a popular choice for players and pushed up the number of winners."

"So at the second time off playing you get a
1 in 4 chance of a car."

No. On many levels, no.

The probability of winning the car, assuming you are playing the Monty Hall problem logically, is 66%. That is true every time you play it, regardless of what has happened before.

But you won't go on a winning streak for ever. Noone ever does.
The longer you win the car by swapping the doors the more chance you will lose.
See how many times you can toss a heads in a row.

No, the odds are the same at the point of each individual attempt. What has gone before is irrelevant.

But you won't go on a winning streak for ever. Noone ever does

That's true. If people never changed doors, then the results in the long run would be 1/3 the contestant wins, 2/3 they lose.

But each new game is a new game. It's completely independent of the first.

This thread was a wild ride.

The longer you win the car by swapping the doors the more chance you will lose.
See how many times you can toss a heads in a row.

This is rubbish.

Let's say that there have been 10 deliveries in a hospital and they are all boys.

A new mum gives birth. What are the odds of her giving birth to a boy?

She could have gone to a hospital where there were 10 girls born in a row. Have her odds of having a boy changed?

If I did 10 coins in a row and they were all heads, would you take odds of 10 to 1 on the next coin being a tail?

So if it was a tail, I would give you £1 but if it was a heads, you would give me £10?

"The longer you win the car by swapping the doors the more chance you will lose."

No. They are all independent. You can't go on a winning streak forever because the probability of winning the car is less than 1. But the odds don't change based on how often you play, so you will always have a better chance of winning if you switch.

No-one is saying anything about a winning streak, and we know it's gambling. We are saying that, faced with something that has a 67% chance of happening, and one that has a 33% chance of happening, you should pick the 67%.
I don't fully understand how that is not the obvious thing to do.

Can you help me with the baby example (first pregnancy. Then a 2nd pregnancy)

Possible Outcomes are

1. Boy boy
2. Girl girl
3. Boy girl
4. Girl boy

I understand the probability being 1/4 for having 2 boys. (Or 2 girls). Because 1/2 X 1/2

But are there not only 3 actual outcomes rather than 4? because option 3 and 4 are the same outcome (just in a different order). So does that change to every option being a 1/3 probability from the outset?

No, there are 4 outcomes. Someone may want one of each, with the girl being the eldest, or one of each with the boy being the eldest. Just as someone might want 2 girls or 2 boys. Every one of those has a 1 in 4 chance of happening.

Of course, some people might want 2 of the same sex, not caring if they are boys or girls, and some may want one of each, not caring which comes first. Those people have a 1 in 2 chance of getting their wish. (assuming of course they are able to have 2 children).

No the longer you win the more likely it is that the next time you will lose.
Because you are playing the game over and over.
You have a chance of winning or losing each time. Yes that stays the same, but the odds on you winning games in a row is a different thing to just winning a game.

Ok the mum has a 1/2 chance of a girl.
This is different to the hospital's row of boys.

1 x 1/2 to the power of the births.
1/ 2048.