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.

No-one is trying to get to 10 T in a row. The odds of doing that, or indeed any pre-declared combination of heads and tails, is 1 in 1024. But that is unrelated to what you are saying.
What you are saying is that the odds of you picking the right door increase the longer you play, which makes no sense whatsoever. And neither does not picking the most likely outcome because "you feel lucky".

Let's play 1000 games. I am going to use my strategy of picking the most likely option every single time. So my expected number of wins at the end of 1000 turns is 667.
What is the expected value of your strategy?

This is exactly how casinos make money because people think their number must come up soon.

If you keep betting £1 on a 6 coming up, then after say 600 goes, you should have £100.

But if you had had a run of say 20 numbers that weren't 6, you wouldn't bet £20 on the next one being 6. Because the chances are that you would lose that money.

Here’s Derren Brown flipping 10 heads in a row: m.youtube.com/watch?v=XzYLHOX50Bc

Whenever I teach my classes probability I show them the video and ask them how he did it.

I also ask them how to guarantee a win on the lottery.

No I am saying each game is an event.
At the end of the day I win it or lose it.
But the more times you win the less chance the next will be another win due to the doubling of the possibility of the odds.
The baby thing again, each mum has a chance of a boy or girl.
But the hospital's chances of a row of boys is indepedant of each birth. But there are still calculated odds on having the 11 boys.

I really do have 50% chance of picking the right door.
With the knowledge my door has a true value of 33% of winning.
So I am only gaining a 16ish %

Wrong. In the MH problem, by switching you go from 33% to 66% chance of winning. Which is not a 33% increase in chance, but a 100% increase!

I say again anyone got 10 T in a row

The fact that few people will does nothing to prove your point. Nobody is saying it is any more than a miniscule probability.

An experiment to prove your theory correct would involve taking people who have already flipped tails nine times, and collecting data on the result of their 10th try. If significantly more of them flipped heads, you'd be onto something.

I also ask them how to guarantee a win on the lottery

There was a case of some students who rigged the lottery.

www.theatlantic.com/business/archive/2016/02/how-mit-students-gamed-the-lottery/470349/

But the more times you win the less chance the next will be another win due to the doubling of the possibility of the odds.

Nope, not true. This weeks lotto winners have exactly the same chance of winning next week as you or I (assuming we buy the same number of tickets).

But this afternoon, before the draw, the chances of tonight's winners winning 2 weeks on the trot were infinitesimally low.

But the hospital's chances of a row of boys is indepedant of each birth. But there are still calculated odds on having the 11 boys

Yes - but given a hospital has had 10 boys, what are the chances of the next baby being a boy?

Would you give me £10 if it was a boy and I give you just £1 if it was a girl?

Or am I cheating you with those odds?

"But the more times you win the less chance the next will be another win due to the doubling of the possibility of the odds."

As has been pointed out so many, many, times - NO.

I say again anyone got 10 T in a row

No, because I haven't given in 1024 tries. because I have better things to do. But if I did, it would probably happen once during that sequence of goes. It might not, or it might happen more than once. But if I did that every day, on average, it would happen once a day.

What event is more likely?

BB
BG

Or are both events just as likely?

@Fainne, the reason the person you were trying to explain this in real life "didn't get it" is not because of the problem in itself (regardless whether you were right or wrong). The reason is you attitude: you come across as arrogant, dismissive, aggressive and having a superiority complex, which will not help trying to explain anything to another person.

But the more times you win the less chance the next will be another win due to the doubling of the possibility of the odds.

This is utterly wrong. What has come before does nothing to change the chance that the next go will be another win/tails/6/boy.

Let's go with 5 tails, for ease.

At the start, the probability of 5 tails in a row =1/32. Probability first throw is a tail =1/2. It's a tail

2nd throw. Probability of 5 tails in a row given we have one already =1/16. Probability second throw is a tail =1/2. It's a tail

3rd throw. Probability of 5 tails in a row given we have two already =1/8. Probability third throw is a tail =1/2. It's a tail

4th throw. Probability of 5 tails in a row given we have three already =1/4. Probability 4th throw is a tail =1/2. It's a tail

5th throw. Probability of 5 tails in a row given we have four already =1/2. Probability 5th throw is a tail =1/2.

This happened in Monte Carlo in 1913

Monte Carlo Casino
Perhaps the most famous example of the gambler's fallacy occurred in a game of roulette at the Monte Carlo Casino on August 18, 1913, when the ball fell in black 26 times in a row.

This was an extremely uncommon occurrence: the probability of a sequence of either red or black occurring 26 times in a row is (18/37)26-1 or around 1 in 66.6 million, assuming the mechanism is unbiased. Gamblers lost millions of francs betting against black, reasoning incorrectly that the streak was causing an imbalance in the randomness of the wheel, and that it had to be followed by a long streak of red.

But your not talking about the same thing as me.
It took Derren Brown 9 hours to get a row of ten Heads.
It took him less than a min to get a single heads .
It's like your dismissing how hard it is to get ,9 wins in a row to claim the 10th win.

I don't see anyone dismissing that it is hard to get 9 heads in a row. But it still does not change the probability on the 10th.

@chomalungma
Say you have a million casinos in the world, they each have 200 spins a night.
300 days a year and your event happened once in a 100 years.
6,000,000 Millon to 1 chance.

But your not talking about the same thing as me

It's like your dismissing how hard it is to get ,9 wins in a row to claim the 10th win

It is hard to get 10 wins in a row.

But it is easier to get 10 wins in a row than to get 9 wins and a loss.

It's easier to get 2 wins in a row than a win followed by a loss.

Because the chances of winning are twice as high as the chances of losing.

And that will never change.

@iseetodaywithanewsprintfray
But your total forgetting just how hard it would be to win 9 times in a row.
To say how easy it is to win on the 10th.

Unlikely events do happen.

But if you have 25 blacks in row in roulette, and assuming it's completely fair, you would be stupid to be your life savings on red.

Because the odds are the same for red or black.

But your total forgetting just how hard it would be to win 9 times in a row

But it's happened.

It's easier to win 9 times in a row than to win 8 times in a row and then lose.

It's always easier to win than to lose.

@mummmy2017 - I am not forgetting anything. It is hard to win 9 times in a row. But that does not change the probability of winning the 10th.
And as @chomalungma so beautifully puts it, it is harder to win 9 times and then lose, because you are more likely to win. Every. Time.

@mummmy2017

Your original statement was that given the chance to play the game once you would not choose to switch doors.

I think this could be one of the key points you are missing in understanding why swapping doors always gives you a better chance of winning (this is copied from the Wiki article on the Monty Hall problem as it explains it better than I could):

"Your chance of winning by switching doors is directly related to your chance of choosing the winning door in the first place: if you choose the correct door on your first try, then switching loses; if you choose a wrong door on your first try, then switching wins; your chance of choosing the correct door on your first try is 1/3, and the chance of choosing a wrong door is 2/3."

Does that help? No matter if it's the first time or the millionth time you've played the game the above statement will always be true and therefore switching doors will always give you a better chance of winning.