Probability questions?

Handsome_Chris

A theorecitical problem:  There is a bag with 4 cards in it, 1 green and 3 red.  To win the game a candidate has to draw the green card from the bag.  If they do not succeed on the first go, the card they have draw remains out of the bag, and they get to draw again.

What is the probability of the candidate drawing the green card in two draws or less?

My answer

I believe that it works out at a 50/50 of drawing the green card in two goes or less

1-(the chance of winning on first attempt x the chance of winning on the second attempt)

I ask because I have just found out about Bayesian networks and have just started to pay attention to probability.

thecolourbox

Well presumably it's 100% probability if they just keep having goes until they pick it out?

Handsome_Chris

thecolourbox said:

Well presumably it's 100% probability if they just keep having goes until they pick it out?

They get to draw a card, and if that is red, they get to draw once again.  After that, the game is over.

Please pardon me, I did not explain what was meant by 'again'.

stickyfiddle

Agreed

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.

HAL9000

1 in 4 for the first attempt, then 1 in 3 for the second. Add them together = 7 in 12 or 0.58333...
(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.

fretmeister

Speaking as a magician.

If the bag belongs to me, then you'll get the right card when I am damn well ready for you to have it!

Tannin

Nope. HAL is entirely correct. 

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.

Strangefan

fretmeister said:

Speaking as a magician.

If the bag belongs to me, then you'll get the right card when I am damn well ready for you to have it!

Now that's a skill, the hand and misdirection skills  you guys have is amazing, I have a small hobby of workng out magic tricks, the psychology part of it always intregues me. 

Handsome_Chris

HAL9000 said:

1 in 4 for the first attempt, then 1 in 3 for the second. Add them together = 7 in 12 or 0.58333...
(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.

Therefore., if the second card was red, and a third draw allowed, which would be a 1 in two chance, the maths would look like this 

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?

mart

I'd agree with @Handsome_Chris ;

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.

Garold86

Tannin said:

Nope. HAL is entirely correct. 

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.

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.

UnclePsychosis

For the frequentists among you, if you have four unique cards then there are 24 possible ways to order them. Six of those orderings have the green card first. Six have it second. Six have it third, and six have it fourth. 

So you have 12 ways to win, and 12 to lose. 50%. 

UnclePsychosis

HAL9000 said:

1 in 4 for the first attempt, then 1 in 3 for the second. Add them together = 7 in 12 or 0.58333...
(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.

Ask yourself what your answer would be if the player got to go for a third attempt. 

Musicwolf

UnclePsychosis said:

For the frequentists among you, if you have four unique cards then there are 24 possible ways to order them. Six of those orderings have the green card first. Six have it second. Six have it third, and six have it fourth. 

So you have 12 ways to win, and 12 to lose. 50%. 

This is correct as is the 'what are the chances of not winning?' approach i.e. (3/4)x(2/3)

olafgarten

I like to use tree diagrams for this, it makes it easy to see all the possible outcomes and their probabilities.

Based on that @stickyfiddle is correct.

EricTheWeary

UnclePsychosis said:

HAL9000 said:

1 in 4 for the first attempt, then 1 in 3 for the second. Add them together = 7 in 12 or 0.58333...
(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.

Ask yourself what your answer would be if the player got to go for a third attempt. 

So 7/12ths plus 6/12th which is 13/12. And obviously wrong. 

Interesting as I’d got 7/12 as well. 

Not winning as per Musicwolf is 3/4 x 2/3 = 6/12. Which seems to be the correct answer although I’m not entirely sure I understand why. 

Handsome_Chris

@EricTheWeary, having thought about what @stickyfiddle and @UnclePsychosis have said:

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,

Philly_Q

I've got a degree in Statistics but I honestly can't remember any of it.

bertie

Philly_Q said:

I've got a degree in Statistics but I honestly can't remember any of it.

I was 50% sure you'd say that

Handsome_Chris

bertie said:

Philly_Q said:

I've got a degree in Statistics but I honestly can't remember any of it.

I was 50% sure you'd say that

I was 50% sure that @Philly_Q would say that, and if that were to occur,  50% sure that you'd say that in reply. This which means that I sure there was a 25% chance of you saying that.

Brio

Are the cards shuffled between draws?