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.

@mummmy2017 - and it is your decision to be completely illogical, if you choose. But doubling the probability of winning really should be enough to sway you. You can't compare your 33/67 with the 50/50 - the 50% is not an option to you, so you either have 33% or 67%.

Not worth swapping to double your chances of winning?

Is Mummy2017 the OP with a name change fail?

But your blind chances in a door choice are 50/50.
Therefore your knowledge is only worth 16 ish present.

Not really enough to sway me to swap from my door. I feel lucky.

That's not a logical decision, it's an emotional one.

If I have 99 goes then yes you swap, the more goes you have the more the probability kicks in and shows results.

At least you got that right. The more goes you have at something, the closer to the expected outcome you will be. Toss a coin 10 times, and you might get 5 heads 5 tails. You probably won't. You'll likely get 6/4 or 7/3. Maybe even 8/2, 9/1 or 10/0
Probably no more than 65/35. Toss it a thousand times and you won't be too far off 500/500. Maybe 550/450. A million times and you'll be even closer.

Insurance for teenage drivers is expensive. Now one teenage driver might go claim free for a year, whilst his mum has 4 claims. But insure 10K teenagers, and 10k of their mums, and you know exactly which group is going to cost the insurance company more.

Therefore your knowledge is only worth 16 ish present

But it's not a blind choice.

Your knowledge is worth more to you.

@Purpledaisies and other clever maths folk...

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?
(I don't care what babies I have, just literally what is the probability for each?)

I really don't get maths so I'm probably wrong but that's what my brain thinks lol

Or am I mixing up probability and outcome

@mummmy2017 - yes, if you are comparing to a blind chance. But you're not. You're comparing not switching (33%) with switching (67%).
And even if you were comparing with 50%, you should still switch, because 67% is larger than 50%.

You need to watch Rosenkrantz and Guildenstern Are Dead by Tom Stoppard: the odds remain the same no matter how many times you perform the trick. They will always be 1/20. Always.

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

There are 3 outcomes.
2 of them have a 25% chance of happening
1 of them (2 of the same sex) have a 50% chance of happening.

Please remember you are gambling.
You do not know 100% where the prize is.
So on your first play while your knowledge does increase your chance of a win, it is no guarantee.
They are saying it should on probability be behind the door, not that it will be, you could still lose.
Only after many tries does the probability mean your knowledge equals more wins.

@BrownStripePJ - the same outcome in different orders are different outcomes. Probability of a boy and a girl = 50%. Probability of a boy then a girl = 25%

But are there not only 3 actual outcomes rather than 4?

@BrownStripePJ - but the point is that different outcomes can have different probability. For example, if I play the Lottery, there are two outcomes: win the jackpot v don't win the jackpot, but it doesn't make it 50:50.

Your first explanation where you list the 4 possibilities is more helpful, because you've listed 4 equally probable outcome: a girl and a boy is twice as likely as 2 girls, because of the two ways it can happen as your list shows.

@mummmy2017 - but when gambling, you should still pick the option that is most likely, no?

Only after many tries does the probability mean your knowledge equals more wins

No - the probability is always the same.

It's just that after many goes, you will see that you win 1/3 of the time.

But the probability is always the same.

If I wanted to pick the Ace of Spades, I have a 1 in 52 chance of picking it.

I always have a 1 in 52 chance of picking it.

I might get it the first time, I might not get it after 52 goes.

But do it enough times, the probability will be 1 in 52.

Because that's what the probability is.

They are saying it should on probability be behind the door, not that it will be, you could still lose

In a 2 horse race, if you know that a horse is much faster than the other one, would you pick that horse or does it not matter?

You could still lose. But it's very likely you will win.

Here’s Marcus du Sautoy and Alan Davies actually doing the Monty Hall game lots of times and showing you win more times if you switch www.interactive-maths.com/monty-hall-problem-video.html

People gamble on horses, not everyone backs the fav .
People back what they like.
This is why people don't always swap on the show, on one game.
Because when there are two doors you can choose and win on the one with lower odds.
So by not using the info your taking a 50/50 gamble of winning.

What if the faster horse has a bad day.
Oh it lost.

So by not using the info your taking a 50/50 gamble of winning.

Nope. It’s only 50:50 if you’re presented with two doors and don’t know which one the presenter chose not to open. Not using that info by switching is irrelevant.

It would only become 50:50 if after the presenter opened one door and then the remaining two doors were muddled and it wasn’t ‘stick or switch’ but ‘pick entirely from scratch because you don’t know any more which you chose originally’

Not using that info by switching is irrelevant.
I meant not using the info and sticking with the original

@mummmy2017 by not using the info you are taking a 33/67 chance on winning.
And that's fine, but it's completely illogical.

People back what they like

And lose.

People with knowledge win more.

What if the faster horse has a bad day

Yes - it happens.

But you are more likely to win if you back the faster horse.

I would love to gamble with you.

Ok so the ODDS of picking the right door when faced with 2 doors is 1/2.
The probability is 2/3 due to prior knowledge.