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I ask because I have just found out about Bayesian networks and have just started to pay attention to probability.
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Please pardon me, I did not explain what was meant by 'again'.
1/4 + (3/4 * 1/3)
= 1/4 + 1/4
= 1/2
Thinking of it another way, you can set it out as: "You have these 4 cards, number them 1-4 and place them in front of you. What are the odds that the red one is #1 or #2?" Asked that way it becomes obvious that it's 50/50
EDIT: my top method is essentially a tree-diagram without writing it down:
Chance of getting Green Card = [chance of green on first pick] + [chance of getting to second pick] * [chance of green on second pick]
It's that middle part that's easy to miss.
(I think)
Despite having two attempts from four cards I don’t think the answer is 50%. This is because the first-picked card doesn’t go back into the bag thus giving the second attempt (if required) better odds.
If the bag belongs to me, then you'll get the right card when I am damn well ready for you to have it!
https://speakerimpedance.co.uk/?act=two_parallel&page=calculator
We have two events and the chance of success is additive. The probability of success on the first draw is .25. The probability of success on the second draw is .33. So the overall probability of success is .58, or roughly six out of ten.
1/4 + 1/3 + 1/2 = 13 in 12 = 1.08333333333
which is better that a dead cert. I thought that the probability runs from 0 to 1
This is going to highlight my misunderstanding, doesn't 1.083 mean that there is no chance of pulling three red cards in a row? Am I wrong to assume that is there is still a red card in play, that it can be drawn?
Which trap have I fallen in to here?
Another way of looking at this is to work out the chance of losing: you'd have to pick one of the red cards on the first go (3 in 4 chance) and also one of the red cards on the second go (now a 2 in 3 chance, since one of the red cards has gone), so the chance of losing is (3/4)x(2/3) = 2/4 = 1/2.
The chance of winning is then 1 minus this, i.e., 1 - (1/2) = 1/2.
Not quite - the chance of success on the second attempt is not .33 - you need to take into account the fact that in order to have a second go, you have to pull one of the 3 reds out first. Hence it becomes 3/4 * 1/3 = 3/12 = 0.25, as stickyfiddle says earlier.
That's why, if you only take into account the draw in isolation, you end up with a >1 probability when drawing 3 cards.
So you have 12 ways to win, and 12 to lose. 50%.
Based on that @stickyfiddle is correct.
if one were to pull out all the cards they could come out in these combinations:
Green, Red, Red , Red - Win
Red, Green, Red , Red - Win
Red , Red, Green, Red - Lose
Red, Red , Red, Green - Lose
So that way it is a chance of winning 50% or a possibility of .5
The not losing way of looking at may be:
All possibilities add up to one
The possibility of not drawing a green on the first card is 3/4
However
that not winning leads to a second opportunity, with one less card against the candidate, who then has a 2/3 chance of not winning.
For the candidate to not win, in total they would have to select red both draws i.e. to qualify for a second draw you have to not win the first draw.
Therefore the candidate has 2/3 chance of losing, but only if they lose the 3/4 draw first.
i.e the it's a 2/3rds of the 3/4s chance of losing , which equals 2/4 = 1/2
Take that 0.5 chance of losing away from 1 and 0.5 is left.
BAZIIIINGA!
I await my roasting for getting it wrong,
just because you do, doesn't mean you should.