What Are the Probabilities of Matching Gloves in a Random Distribution?

[evinda]

Hey again! (Wasntme)

We have $6$ pairs of gants,all of them have a different colour.
We share the $6$ right gants randomly to $6$ people,then we share the $6$ left gants randomly to the same people.
Find the possibility:

• Nobody has a pair of the same colour
• Everybody has a pair of the same colour
• Only one gets a pair of the same colour
• at least $2$ get a pair of the same colour

That's what I have tried (Blush):Let $D_n=|\{ \sigma \in S_n: \sigma(i) \neq i \forall i \in [n]\}|$

• $6! \cdot D_6$
• $6!$
• $6! \cdot 5!$
• $6!-D_2$

Could you tell me if it is right?

 

[I like Serena]

Hey again! (Wasntme)

We have $6$ pairs of gants,all of them have a different colour.
We share the $6$ right gants randomly to $6$ people,then we share the $6$ left gants randomly to the same people.
Find the possibility:

• Nobody has a pair of the same colour
• Everybody has a pair of the same colour
• Only one gets a pair of the same colour
• at least $2$ get a pair of the same colour

Hey! (Mmm)

Hmm. What do you mean by "possibility"?
Do you mean the number of possibilities?
Or do you mean the probability? (Wondering)

And what is a "gant"? (Thinking)

That's what I have tried (Blush):Let $D_n=|\{ \sigma \in S_n: \sigma(i) \neq i \forall i \in [n]\}|$

Could you tell me if it is right?

• $6! \cdot D_6$
• $6!$

Right and right! (Sun)

• $6! \cdot 5!$

Hmm... suppose the first person gets a pair.
How many possibilities to divide the remaining gants? (Thinking)

• $6!-D_2$

Hmm... $D_2$ is the number of possibilities that 2 gants are divided differently... which is 1 possibility.
I don't think that will work! (Doh)

 

[soroban]

Hello, evinda!

We have 6 pairs of gants, each pair a different color.
We share the 6 right gants randomly to 6 people,
then we share the 6 left gants randomly to the same people.

Find the probability that:
(a) Nobody has a pair of the same color.
(b) Everyone has pair of the same color.
(c) Exactly one gets a pair of the same color.
(d) At least two get a pair of the same color.

A derangement is an distribution
in which no one has a matching pair.

Let d(n) = number of derangements of n objects.

There are: 6! = 720 possible outcomes.

(a) P(\text{No matches}) \:=\:\frac{d(6)}{6!} \:=\:\frac{265}{720} \:=\:\frac{53}{144}(b) There is one way in which there are 6 matches.
P(\text{6 matches}) \:=\:\frac{1}{720}(c) There are 6 choices for the person with the match.
The other 5 gants must be deranged.
P(\text{1 match}) \:=\;\frac{6\cdot d(5)}{6!} \:=\:\frac{6\cdot 44}{720} \:=\:\frac{11}{30}(d) The opposite of "at least two" is "0 or 1".
We already know: P(0) =\tfrac{265}{720},\;P(1) = \tfrac{264}{720}
Hence: P(\text{0 or 1}) \:=\:\tfrac{265}{720}+\tfrac{264}{720} \:=\:\tfrac{529}{720}
Therefore: .P(\text{2 or more}) \:=\:1 - \frac{529}{720} \:=\:\frac{191}{720}

 

[Nono713]

Also I think "gant" means "glove" here. Probably translation error :p