# A paradox in probability theory and statistics

1. Dec 18, 2017

### Pouyan

1. The problem statement, all variables and given/known data
In a vessel is a 5 cent coin and two 1-cent coins. If someone takes up two randomly chosen of these coins, and we let X be the total value of the coins taken, what is the probability function for X?

2. Relevant equations
I know that X has a value {2,6}

3. The attempt at a solution
I get two solutions that I do not understand which one is right:

Solution 1:
Let A1 be first coin is 1cent, A2 second coin is 1 cent
we know P(A1) = 2/3
If we want to get 2 cent : P(A1∩A2) = P(A2|A1) * P(A1)
P(A2|A1) = Conditional probability = First coin is 1 cent , second coin is 1 cent = 1/2
then P(A1∩A2) = 1/3 = P(X=2 cent)
And P(X=6 cent) = 1-P(X=2 cent) = 2/3

Solution 2 :

If I assume that the first coin A1 is 5 cent and the other is A2, 1 cent :
P(A2|A1)= 1
P(A1∩A2) = P(A2|A1) * P(A1) = 1* 1/3 = 1/3 = P(X=6) !

Solution 3:
A1 is 1 cent and A2 is 5 cent:
P(A2|A1)=1/2
P(X=6) = P(A2|A1) * P(A1) = 1/2 * 2/3 = 1/3 !

Now I get dizziness!
What is the real value of P(X=6) and P(X=2) ?!

2. Dec 18, 2017

### Orodruin

Staff Emeritus
The conclusion that P(A1∩A2) = P(X=6) is not correct. A1∩A2 is just one of two ways of getting X = 6. The other is covered by your Solution 3.

Edit: In other words, X=6 can be achieved in two ways, first coin 5 and second 1 or first coin 1 and second 5. The probability for X=6 is therefore the sum of the probabilities for these.

3. Dec 18, 2017

### PeroK

The first solution looks right although it is too complicated.

$P(1+1) =( \frac23)(\frac12)$

is all you need.

The other solutions I'm not sure I understand. The 5c coin could be the first or second coin chosen. Perhaps you have calculated tge probability that it is in only one place.

4. Dec 18, 2017

### Pouyan

Thank you ! Now I've got it !
P(X=6) = P(A1∩A5) + P(A5∩A1) = 2/3 !

5. Dec 18, 2017

### PeroK

Except what is event A5?

It's good sometimes to do a simple problem with full formality, but if you do you have to be careful not to get lost in your notation.

I would prefer the notation in post #3 because it doesn't introduce any unnecessary variables.

Otherwise, I would have used A1 for the first coin being 1 cent and B1 for the second coin etc. That would avoid the confusion about which coin is which.

6. Dec 18, 2017

### PeroK

PS note that in any case $A1\cap A5 = A5\cap A1$, so your notation has gone wrong on that respect as well.

You are taking these to be two different sequences of events, but the order of set intersection doesn't matter, so they must represent the same event.

7. Dec 18, 2017

### Orodruin

Staff Emeritus
I agree with @PeroK, the notation is very confusing (not to say wrong). It is much better to state Ax as the event selecting an x coin in the first draw and Bx as the event of selecting an x coin in the second draw. Then your $X = 6$ would be (A1∩B5)∪(A5∩B1).

8. Dec 18, 2017

### mjc123

In this case, as there are only 3 coins, when you have chosen 2 there is just one left. The easiest way to solve this problem is to consider the probability of the left-over coin. There is a 2/3 probability that this will be 1 cent, so P(X=6) = 2/3; and a 1/3 probability that the coin left over will be 5 cents, so P(X=2) = 1/3.

9. Dec 18, 2017

### Ray Vickson

The first way is very obviously correct: either $X=2$ or $X=6$, and $P(X=2) = (2/3)(1/2)= 1/3$ is correct. Therefore, $P(X=6) = 1 - 1/3 = 2/3$ must be correct.

Others have already told you what you did wrong in your second way. However, I am concerned that you lacked confidence in your first way: if you carried out a series of true steps, and did each step correctly, your final answer must be correct.

10. Dec 18, 2017

### Orodruin

Staff Emeritus
Well, I believe this is why the OP used the word ”paradox” in the title. He likely believed that this was true for both the first solution as well as 2 and 3.