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.

Try the Monty Hall problem with 100 doors Mummy. You pick a door and then Monty does you a HUGE favour by removing 98 goat doors. Are you still sticking with your lucky red door?

OK

Imagine a bag with 20 balls in
19 are red
1 is black

It is very possible to pull out 10 reds in a row.
Because there are more reds in the bag. So the more likely event is going to happen every time.

Someone has pulled out
RRRRRRRRR

Do they have a 50% chance of pulling out a red ball next time?

Your changing the number of door so the original odds would be 1/100 or 99/100.
Of course you swap.
But with the original show the advantage just tell you there two doors in front of you one has more chance of a win, not that that it will definitely win. I want to stick.
Even on the 100 door , the door you choose could still win.

I was also told in the race it was 1/2 at the last. Which I agreed with. The 1/1024 were for the start

Yes - if you flip a coin, there is a 1 in 2 chance of it being heads.
There will always be a 1 in 2 chance of it being heads if you spin it 100 times, a 1000 times, a million times.

There is a 1 in 1024 chance of getting 10 heads in a row.
Just as there is a 1 in 1024 chance of any combination of Heads and Tails in a row happening.

Because there are 1024 possible outcomes that could happen if you flip a coin 10 times in a row.

Look at the coin in your hand.
Imagine how many times since it was made that it has been flipped.
Flip it.
What are the chances of it being heads?

@chomalungma
Your talking about the odds for the last game. One game only. Do you agree?
I want the odds of before you start the game. When you had no red balls.

Mummy, yes, it could win, but you're more likely to win if you swap - first game or tenth game.

ut with the original show the advantage just tell you there two doors in front of you one has more chance of a win, not that that it will definitely win. I want to stick

Of course it won't definitely win.
There is no way you can guarantee a win in this situation because there is a possibility that you did pick the correct door at the start.

But the chances that you picked the correct door was 1 in 3.
It will always be 1 in 3.

So the other door left must have a 2 in 3 chance of being right.

Which is better odds than 1 in 3.

AND IT DOES NOT MATTER WHAT HAPPENED ON THE PREVIOUS SHOW.

Sure, the odds are much better the more doors there are. Take it back to the three doors and you MIGHT win by sticking, but you’re more likely to win by switching. People want to win so most would choose the option that is more likely to result in a win. Unless they believe in ‘runs of luck’ or favourite colours!

I want the odds of before you start the game. When you had no red balls

What odds are you talking about?
The odds of what happening?

The odds of getting 10 reds in a row is better than the odds of getting 9 reds followed by a black, when there are 19 reds and 1 black ball in a bag.

If you had to gamble between the two choices (and only those 2 outcomes), which one is more likely and why?

If you came along and told me that I had a 50% chance of getting a red ball out of a bag of 19 reds and 1 black ball, then I would suggest that you don't understand probability.

Chomalungma hubby is on the show, if he wins he gets to stay on the show
Chomalungma places a bet at 1/1024 that hubby wins ten shows.
hubby wins ten shows the company say here have £2000 .

Oh you bet £1000 to start with.

Chomalungma places a bet at 1/1024 that hubby wins ten shows

I am not betting on 10 shows though. Winning 10 shows is very hard.
But you stand a much better chance of winning 10 shows in a row if you choose the door with the higher probability than if you decide on the 10th show to pick the door with the lower probabilty because it hasn't come up yet.

Which I agreed with to win many shows you swap.
And in the bet you placed you bet for 10 wins.

hubby wins ten shows the company say here have £2000

Ok - so let's say that I place a bet at 1024 to 1 on someone getting 10 heads in a row.

Someone would take me up on that because that is very hard to get.
Yet I have been lucky...

and I have got 9 heads in a row. The chances of that happening are 1 in 512. So very unlikely.

But it could happen.

I have a 50% chance now of getting a heads on my next throw.
Or a tails.

I always have had that chance.
If I win, then I would expect my original bet to be honoured at 1024 to 1.
But the chances of the entire run happening is very rare.

But in the Monty Hall problem, it's not 50 / 50. Its 1/3 to 2/3.

So you should always go for the higher probability if you want to win.

Mummy, f you agree Which I agreed with to win many shows you swap.
Why would you not apply that logic to show 1, or show 10. You clearly agree you have more chance of winning if you swap, regardless of what's happened before.

Which I agreed with to win many shows you swap

You swap every time.
On day 1
On day 100

Every time.
The day you do it on makes no difference.

The doors don't care about you. The doors aren't lucky. You aren't guaranteed to win but you are more likely to win.

Simple question:

You have a bag with 9 red balls and 1 black ball in.
You pull a ball out, look at it and out it back in.
There have been 9 red balls in a row.
You have £1000.
Do you put it on
a) Reds - because reds have a 90% chance of being drawn
b) Black because blacks are due
c) It makes no odds because there must be 50% chance of it being red or black

Yes the 1/1024 odds are rare but that is the bet you placed, if your hubby won you would want your million because it was what you bet.
This is the bet I placed on ten wins.
The fact a single toss has a 1/2 chance means nothing when your asking for odds on 10 wins.
And yes I have never said that you shouldn't take door 3 if your playing more games.

This thread is a metaphor

And yes I have never said that you shouldn't take door 3 if your playing more games

But you have said that you would stick with your original choice at the start rather than swap

And that's what people are struggling to understand
You are not guaranteed a win. But you are twice as likely to win than if you don't swap at the start.

If you were going on this game show, and you wanted to win, and then if you won you would go on the next show as well, what strategy would you approach?

@OffToTheMoon
On show one you end up looking at two doors.
One door may have a higher probability of a win but if your only playing one game then you can just choose the door you like.
You open the door, you win or you lose.
50/50 chance
Yes there is a third more chance with one door, but your door could have the prize.
The more games you play the more the swap advantage kicks in.

One door may have a higher probability of a win but if your only playing one game then you can just choose the door you like

You open the door, you win or you lose

The more games you play the more the swap advantage kicks in

I really really have to go and do something more useful now

There is an element of luck to the doors.
Your taking a gamble, you have no way of knowing where the prize is.
I just watched the show, you do know there is no swap. Right.

You open the door, you win or you lose.
50/50 chance
Yes there is a third more chance with one door, but your door could have the prize

Mummy, can you see that it isn’t possible for the doors to simultaneously have 1/2 odds and 1/3 odds?

And of course you can just ‘choose the door you like’, if you’re happy to accept that you’re less likely to win 😆