<s>[INST] Probability of someone picking the same toy For someone to pick the same toy:
Jon picks car1 and car2
Given 3 people

Hans 
Thomas 
Jon

And 6 toys: 

Car1, Car2, Doll1, Doll2, Ball1, Ball2

Each person is given 2 toys, calculate the probability of someone getting the same toy. (there are two instances of each toy).

The way I tried to solve it is this way:


*

*Only Jon gets the same toys

*Only Thomas gets the same toys

*Only Hans gets the same toys


For 1-3, the probability is as follows:
Hans picks up any toy (5 toys are left)
Jon has to pick any toy that is not what Hans picked (4 options out of 5)
Thomas has to pick the toy which Hans picked (1 options out of 4)
Hans has to pick the same kind of toy he has (1 out of 3)

So the probability for scenario 1 is:
(4/5)*(3/4)*(1/3) = (1/5) 
So for 1-3 the probability is 1/5

Now I have the probability of only 2 getting the same toy, which is not possible as the 3rd person by default also gets the same toy, so the probability is 0.
And the last option is that all 3 get the same instance of a toy.
(4/5)*(2/4)*(1/3)*(1/2) = (1/15)
Overall, the probability of at least one getting two instances of the same toy is: 1/15 + 3*(1/5)
But it seems I am wrong?
Any help will be appreciated.
 [/INST] Someone picking a toy:
One in left hand one in right.
Left hand first, will chose between 6 options. 
Then the right hand will have 5 options.
$5\cdot 6=30$ options
Do we care about what is in which hand? No... So:
In any case all the ways for the first "player" are $\frac{6\cdot 5}{2}=15$.
Second person has to chose between 4 toys anyway.. So, same logic:
All the ways: $\frac{4\cdot 3}{2}=6$ ways for the second "player".
Last one 1 way (only two toys left).
Does it matter if the "players" take first one toy and then the second?
No, same results.
Now: 
For the wanted result, 
A) the first has 3 ways between 15 options (to chose between one of the 3 pairs)
$P_A=\frac{3}{15}=\frac{1}{5}$
and the second has :
B1) if first got a pair : 2 ways from the four options... But we don't care (already true statement because of the first and the 2 ways was to chose between every pair left there) 
Edit:
$P_B=P_A\cdot\frac{2}{6}=\frac{1}{5}\cdot\frac{1}{3}=\frac{1}{15}$
(We will not ask for this -first already made true our request-)
B2) first didn't got a pair : 1 way (first destroyed two of the pairs and only one left for the second)
$P_B=(1-P_A)\cdot\frac{1}{6}=\frac{4}{5}\cdot\frac{1}{6}=\frac{4}{30}=\frac{2}{15}$
Last player:
C1) If first or second (or both) did it, we don't care.
C2) If first and second failed The only option is that first and second destroyed the same two pairs... But then the third will pick the last pair anyway:
$P_C=P_{A->fail}\cdot P_{B->same\ pair\ as\ A}$
$P_C=(1-P_A)\cdot\frac{1}{6}=\frac{4}{5}\cdot\frac{1}{6}=\frac{4}{30}=\frac{2}{15}$
Answer: 
first has 3\frac{1}{5} and if he didn't make it second has \frac{2}{15} to keep pair in his hands or \frac{4}{30} to leave pair to the third:
Edit:
$P=\frac{1}{5}+\frac{2}{15}+\frac{4}{30}=\frac{6}{30}+\frac{4}{30}+\frac{4}{30}=\frac{14}{30}=\frac{7}{15}$
About @kludg's comment:
About the same or different probabilities:
A) The probability for each player to be the only one with a pair is the same for all players and equal to ($\frac{2}{15}$) but:


*

*The probability of every player to catch a pair and be the one who will finishing the game is $\frac{1}{5}$ for the first player and then remains $\frac{1}{3}$ after the first player (inside the event left from first failure $=\frac{4}{5}$) where second and third have $\frac{1}{6}$ and \frac{1}{5} each one inside the event they remain ($\frac{4}{5}$ for the B and $\frac{4}{5}\cdot{5}{6}=\frac{2}{3}$ for the player C). Also, in the second step (Player B) the total possibilities inside the 80% of A's failure we have $\frac{1}{3}$ chances to find a (one) pair.


So, by 20% ($\frac{1}{5}$) the game finishing by the first player, then, if not the player B has $\frac{1}{6}$ chances to be the one with the pair... having a total chance $\frac{2}{15}$ and after him (in a case of both previous failure that comes by possibility $\frac{2}{3}$) by $\frac{1}{5}$. [Notice that $\frac{1}{5}\cdot\frac{5}{6}\cdot\frac{4}{5}=\frac{2}{15}$ again]
For somebody need another way of explanation:
--Player A: $P_A=\frac{1}{5}=\frac{3}{15}$ because we can consider he is taking first two toys... and $\frac{1}{5}$ is the paired toy he looks for.
--Player B: Player A have already failed and B is taking one of four toys that left and have to find the pair that is one of 3 that remain. But even the first toy he will take has to be deferent than both of Player A toys. So, he has two ways to take both toys non matching with A ($P_B=2\cdot\frac{2}{4}\cdot\frac{1}{3}=\frac{2}{12}=\frac{1}{6}$ being in an event of $1-P_{A}=\frac{4}{5}$). So, second player has $\frac{4}{30}=\frac{2}{15}$ total chances but also $\frac{1}{6}$ chances after A's failure.
--Player C: If both A,B have failed we are in a case with possibility $P_{fail\{A\&B\}}=\frac{4}{5}\cdot\frac{5}{6}=\frac{2}{3}$. But then he already have no choices... he will take the toys that left for him. The possibility of A and B to took the same (not-paired) pairs. What about these possibilities? failure of A ($\frac{4}{5}$) and the second has to take exactly the same toys that A got: $\frac{2}{4}\cdot\frac{1}{3}=\frac{1}{6}$ again.
Again the total chance for the third player is $\frac{4}{5}\cdot\frac{1}{6}=\frac{2}{15}$ to be the one.
</s>