Combinations of Pairs of Gloves..halp

[yellowjack]

There are 10 pairs of gloves (20 gloves), and 6 are selected at random. Let X be the number of pairs of gloves. Find the EV of X.

I'm having trouble getting all the possible combinations to add up to 20_C_6 = 38,760 . Can someone help me by looking at my numbers?

I distributed the possible combinations by splitting the 20 gloves up into left and right, 2 groups of 10 (i.e. the gloves are labelled L1, L2,..,L10, R1, R2,...,R10)

The 2nd combination for 0 pairs is "choose 5 from the 10 Left gloves, then choose 5 of the remaining Right gloves whose pair has not been chosen". For example, if L1, L2,...,L5 were chosen, one of R6,R7,...,R10 would be chosen with it.

0 pairs
(10_C_6 * 4_C_0) + (10_C_5 * 5_C_1) + (10_C_4 * 6_C_2) + (10_C_3 * 7_C_3) + (10_C_2 * 8_C_4) + (10_C_1 * 9_C_5) + (4_C_0 * 10_C_6) = 13,440

1 pair
(10_C_5 * 5_C_1 * 5_C_0) + (10_C_4 * 4_C_1* 6_C_1) + (10_C_3 * 3_C_1 * 7_C_2) + (10_C_2 * 2_C_1 * 8_C_3) + (10_C_1 * 1_C_1 * 9_C_4) = 20,160

2 pairs
(10_C_4 * 4_C_2 * 6_C_0) + (10_C_3 * 3_C_2 * 7_C_1) + (10_C_2 * 2_C_2 * 8_C_2) = 5,040

3 pairs
10_C_3 * 3_C_3 = 360

20_C_6 = 38,760
The total of combinations from the above is 39,000; 240 too many. Can you spot the flaw in the combinations? Also if you have a different counting scheme that may/may not be more efficient than mine, please post it. Thanks.

[BruceZ]

[ QUOTE ]
There are 10 pairs of gloves (20 gloves), and 6 are selected at random. Let X be the number of pairs of gloves. Find the EV of X.

[/ QUOTE ]

Here's an easy way. The EV is the sum of the probabilities of each pair being selected, which is 10 times the probability that a particular pair is selected, which is
<font color="red">
Edited from:10*C(20,4)/C(20,6) = 1.25.</font>
<font color="blue">Edited to: 10*C(18,4)/C(20,6) =~ 0.7895

[yellowjack]

Hey Bruce,

I don't understand this bit:

"The EV is the sum of the probabilities of each pair being selected, which is 10 times the probability that a particular pair is selected"

I have no clue how you deduced that, can someone elaborate?

Also the EV I got was quite different, although it may be because my work in the original post is incorrect.

[BruceZ]

[ QUOTE ]
Hey Bruce,

I don't understand this bit:

"The EV is the sum of the probabilities of each pair being selected, which is 10 times the probability that a particular pair is selected"

I have no clue how you deduced that, can someone elaborate?

[/ QUOTE ]

The EV of a number of things is always the sum of the probabilities of each thing. Let Xi be a random variable which = 1 if pair i is selected, and 0 if not. The expected value of the sum of Xi from 1 to 6 is the expected value of the number of pairs. This is the sum of Pi*1, where Pi is the probability that the ith pair is selected. All the Pi's are equal in this case since each pair has the same probability of being selected.

Here's an intuitive way to grasp this powerful concept. Suppose employee A works 7 days a week, employee B works 5 days a week, and employee C works 3 days a week, at random. What is the average number of employees on any given day? Well employee 1 is always there, so that's 1. B is there 5/7 of the time, and C is there 3/7 of the time, so the average is 1 + 5/7 + 3/7 = 2 1/7.

[yellowjack]

Thanks /images/graemlins/laugh.gif

[Siegmund]

Here is a simpler way to count.

First, select which matched pairs from the 10 available ones to draw.

Then, select which of the remaining pairs you will draw one glove from.

Then, for each single, select whether it is a left or right glove.

0 pairs: 10C6 * 2^6 = 13,440.

1 pair: 10 * 9C4 * 2^4 = 20,160.

2 pairs: 10C2 * 8C2 * 2^2 = 5,040.

3 pairs: 10C3 = 120.

Your only error was that 3C3=1, not 3, on your "3 pairs" line.

EV: 30600 / 38760 ~ 0.7894 pairs.

The "Expected number of pairs picked = 10 * the chance that any given pair is picked" argument is correct.

The chance of picking one particular pair, however, is 18C4 / 20C6 = 3060 / 38760, rather than the number quoted by the other responders.

[BruceZ]

[ QUOTE ]
The "Expected number of pairs picked = 10 * the chance that any given pair is picked" argument is correct.

The chance of picking one particular pair, however, is 18C4 / 20C6 = 3060 / 38760, rather than the number quoted by the other responders.

[/ QUOTE ]

Damn all those other responders, how could they all have missed that? /images/graemlins/grin.gif I would have figured that out when I fixed the combinatorics in the original post, which I still plan to do later tonight.

Edit: Looks like you already found the combinatoric problem.