Card Question!

[ryan_kidz]

A bag holds 5 cards identical except for color. Two are red on both sides, two are black on both sides, and one is red on one side and black on the other. If you pick a card at random and see that the only side you can see is red, what is the probability that the other side is also red?

Thnx!

 

[Denis]

A bag holds 5 cards identical except for color. Teo are red on both sides, two are balck on both sides, and one is red on one side and black on the other. If you pick a card at random and see that the only side you can see is red, what is the probability that the other side is also red? Thnx!

Remove the 2 blacks on both side; only there to confuse you.

So you have a 1/3 chance of having picked the red:black card;
leaves a 2/3 chance of having picked a red:red card.

 

[Gene]

The possibilities are you see red-front, red-back, red-front, red-back, red-only
4/5 the other side is red.

 

[pka]

Lets set up some machinery.
R means the card is red on both sides: P(R)=2/5.
M means the card is mixed on both sides: P(M)=1/5.
B means the card is black on both sides: P(B)=2/5.
r means the face seen is red:
P(r)=P(r^R)+P(r^M)+P(r^B)
>>=P(r|R)(P(R)+P(r|M)P(M)+P(r|B)P(B)
>>=(1)(2/5)+(1/2)(1/5)+0.

Now here is what you want to know: given that the seen face is red, what is the probability the card has two red faces?
Or, P(R|r). But that is P(R|r)=P(r^R)/P(R).
You finish it off.

 

[Gene]

Denis: Apply your reaasoning to a simpler game.
3 cards RR,RB,BB
Probability of RR is 1/2
(same reasoning if you see black)
Probability of BB is 1/2
It is never RB?????????????

PKA: I'm not familiar with P(r^R) but from

P(r)=P(r^R)+P(r^M)+P(r^B)
>>=P(r|R)(P(R)+P(r|M)P(M)+P(r|B)P(B)
>>=(1)(2/5)+(1/2)(1/5)+0.

it seems to be 2/5 as is P(R). That makes P(r^R)/P(R) = 1???????

I gotta stick with my 4/5

 

[pka]

Gene, Sorry I should have been clear on notation.
Not knowing if the general browser can read intersections, I wrote P(r^R) to mean P(r&R). In the history of probability the notation would have been P(rR).
Lets use P(AB) to be the P(A and B).
If the face is red then P(r)=P(rR)+P(rM)+P(rB)= P(r|R)P(R)+P(r|M)P(M)+P(r|B)P(B).
That is because, P(AB)=P(A|B)P(B)=P(B|A)P(A).

Of course you are correct that answer is 4/5.
P(R|r)=P(rR)/P(r)
........ = P(r|R)P(R)/[ P(r|R)P(R)+P(r|M)P(M)+P(r|B)P(B)]
........ = (1)(2/5)/[ (1)(2/5)+(1/2)(1/5)+0]
........ = (4/5)

 

[soroban]

Hello, ryan_kidz!

A bag holds 5 cards identical except for color.
Two are red on both sides, two are black on both sides,
and one is red on one side and black on the other.
If you pick a card at random and see that the only side you can see is red,
what is the probability that the other side is also red?

I agree with Denis . . . discard the two double-black cards,
. . but I have a different answer.

Represent the three cards like this:
. . . \(\displaystyle [R_1,R_2}\;\;\;[R_3,R_4]\;\;[R_5,B]\)

Since you see a red face, there are five possible equally likely cases:

. . [1] You see $\displaystyle R_1$; the other side is $\displaystyle R_2.$
. . [2] You see $\displaystyle R_2$; the other side is $\displaystyle R_1.$
. . [3] You see $\displaystyle R_3$; the other side is $\displaystyle R_4.$
. . [4] You see $\displaystyle R_4$; the other side is $\displaystyle R_3.$
. . [5] You see $\displaystyle R_5$; the other side is $\displaystyle B.$

In four out of the five cases, the other side is red.

~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

This is the basis for a classic "Sucker Bet".

There are three cards:
. . one is red on both sides,
. . one is black on both sides,
. . one is red on one side, black on the other.

You draw a card at random and see a red face.
What is the probability that the other side is black?
. . Would you bet "even money" on it?

Here's the bait:
Since you see a red face, it is not the double-black card.
. . Hence, it is either the double-red or the red-black.
You have a 50-50 chance of winning the bet . . . obviously.

WRONG! . . . Your probability of winning is only one-third.

The two cards are: . $\displaystyle [R_1,R_2]\quad[R_3,B]$

Since you see a red face, there are three cases:
. . [1] You see $\displaystyle R_1$; the other side is $\displaystyle R_2.$
. . [2] You see $\displaystyle R_2$; the other side is $\displaystyle R_1.$
. . [3] You see $\displaystyle R_3$; the other side is $\displaystyle B.$

In only one of the three cases will you win the bet.

. . . . . ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

Let's make the situation crystal-clear.

You are betting that the other side of your card is the opposite color.
I am betting that the colors match.

You're betting that you draw the red-black card . . . get it?

 

[Gene]

PKA: Thanks for clearing that up. I now see that there was a minor typo, at the end, which you corrected in the explanation.
P(R|r)=P(r^R)/P(R)
should read
P(R|r)=P(r^R)/P(r)
It works much better now.
------------------------
Gene

 

[Denis]

I said, that is, I typed(!):
"Remove the 2 blacks on both side; only there to confuse you.
So you have a 1/3 chance of having picked the red:black card;
leaves a 2/3 chance of having picked a red:red card."

Praytell WHY that's wrong; remember it's not the solution, but a hint.

 

[Gene]

Denis,
Hmmmmm, if its a hint I missed it. I took it as your solution.
If you pick an RR card the other side is red.
You pick an RR card 2/3 of the time.
Therefor: the other side is red 2/3 of the time.
QED

Sic semper modus ponens.
--------------------
Gene