For someone to pick the same toy:

Jon picks car1 and car2

Given 3 people




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:

  1. Only Jon gets the same toys
  2. Only Thomas gets the same toys
  3. 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.

  • $\begingroup$ Sorry for editing... I found a mistake and fixed it. $\endgroup$ – Tony Tannous Jan 21 '18 at 9:37
  • $\begingroup$ By "someone getting the same toy" do you mean "two people have a common toy", "two people get the same two toys" or "a person gets two equal toys"? It is not clear. $\endgroup$ – Crostul Jan 21 '18 at 9:41
  • $\begingroup$ @Crostul, sorry, it means that Jon got car1 and car2 for example. $\endgroup$ – Tony Tannous Jan 21 '18 at 9:41
  • So Hans picks a toy, and he can pick any;
  • now Thomas picks a toy, and he should choose a different toy among 5 available, $4/5$;
  • now Jon picks a toy, and he can choose either the same toy as Thomas ($1/4$) or a different toy from the toys chosen by Hans and Thomas ($2/4$); these 2 cases should be considered separately, and here is your mistake.

The probability of the 1st scenario (only Hans picks the same toy) is $$\frac{4}{5}\left(\frac{1}{4}\cdot\frac{1}{3}+\frac{2}{4}\cdot\frac{1}{3}\cdot\frac{1}{2}\right)=\frac{2}{15}$$

PS: The probability that at least one player gets the same toy is $$3\cdot\frac{2}{15}+\frac{1}{15}=\frac{7}{15}$$


Hint: calculate the complement, with some not too hard insights it is far easier to work out.

Answer below:

Thera are $3! $ combinatioms of all different toys, each of which can be either Toy1 or Toy2, so we have to multiply by $2^3$ and we get $48$ combinations.

The total number of possible combinations is $6×5×4$ so the solution is $72/120=13/20$


Successful situations: Either only one guy gets $2$ of the same thing or everyone gets $2$ of the same thing.

One guy gets 2 of the same thing: $3\times3=9$ ways

Everyone gets 2 of the same thing: $3!=6$ ways

Total number of ways: The above two scenarios, and the case where no one gets the pair, which is also $3!=6$ ways. So total number of ways is $6+6+9=21$.

Answer: $\frac{Successful}{Total}=\frac{15}{21}=\frac{5}{7}$.

  • $\begingroup$ think that the first one picks one toy... Then he has a chance 1/5 of the other toys to be the pair of the first he got... Already (1/5)=20%... If he didn't make it... The other two players has 0%? $\endgroup$ – koleygr Jan 21 '18 at 11:46
  • $\begingroup$ I found the correct solution: 5/7 $\endgroup$ – QuIcKmAtHs Jan 21 '18 at 14:18
  • $\begingroup$ I also listed out to check $\endgroup$ – QuIcKmAtHs Jan 21 '18 at 14:18
  • $\begingroup$ The total number of equally likely ways to distribute the toys is $\frac{6!}{2\cdot 2 \cdot2}=90$ $\endgroup$ – kludg Jan 21 '18 at 14:50

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)


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)


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:

$P_C=P_{A->fail}\cdot P_{B->same\ pair\ as\ A}$


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.

  • $\begingroup$ How is this possible that there is more than 50% chance of getting 2 of the same toys? $\endgroup$ – QuIcKmAtHs Jan 21 '18 at 10:44
  • $\begingroup$ For the first "player" the chance is lower ... much lower... But if he will not do it... the rest of the players have more chances to do it (1/3) and this added to the first chances will give a big probability for the result $\endgroup$ – koleygr Jan 21 '18 at 10:48
  • $\begingroup$ @XcoderX: But may be you are right... May be I had to multiply every next probability with the probability of failure for the previous... (But this will keep the result close to 50% anyway) $\endgroup$ – koleygr Jan 21 '18 at 10:54
  • $\begingroup$ The chances of all players to get the same toy are equal. In the end, the problem can be solved using combinatorics only, or 'by counting' so that all players enter symmetrically, but it is a bit tedious. You've got $P_A=1/5$, but this probability includes the case that all players got the same toys. $\endgroup$ – kludg Jan 21 '18 at 13:22
  • $\begingroup$ @kludg we say the same things with a different way... I just made a mistake in the last lines calculations... I had \frac{2}{15} and made it \frac{2}{30} when tried to do the addition... In other case our results are the same.. \frac{14}{30}=\frac{7}{15}... If you look my way more carefully, you will see that my events are independent and this is why I adding the probabilities... even if in first is contained the case of all of them have a pair, I am multiplying with the case that A had not a pair before add it to the probability that A had a pair.... etc (+1 by the way) $\endgroup$ – koleygr Jan 21 '18 at 14:22

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