Understanding combinations and permutations is critical in calculating gaming odds. To review these concepts I will use examples based on a drawer of socks.

Combinations: The number of ways to choose a subset from an equal or larger group, without regard to order.

Permutations: The number of ways to choose a subset from an equal or larger group, with regard to order.

Drawer 1 consists of 7 pairs of socks, one pair labeled for each day of the week: Monday, Tuesday, etc.. Your mother matches socks of the same day together and rolls them up.

Q1: You are rebellious and pick a pair of socks at random (as opposed to matching the socks to the day you wear them). You do this every day, without replacement. What is the probability you wear the correct socks for the given day all week?

A1: The total number of ways to order 7 socks is 7*6*5*4*3*2*1=5040. This is expressed as 7!. In Excel the function is fact(7). There is only one way to pick the right socks every day. So the probability of picking the right socks every day is 1/5040 = 0.02%.

Q2: What is the probability that you pick the correct socks the first three days?

A2: The number of ways to pick 3 socks out of 7, with respect to order is 7!/(7-3)! = 7!/4! = 5040/24 = 210. In Excel the function is permut(7,3). Another solution is there are 7 possible socks for day 1, 6 for day 2, and 5 for day 3. 7*6*5 = 210. There is only one way to pick the right socks on all three days. So the probability is 1/210 = 0.48%.

Q3: What is the probability the first three days you picks the Monday, Tuesday, and Wednesday socks, but in any order?

A3: The expression for this answer is 7!/((7-3)!*3!) = 7!/(4!*3!) = 5040/(24*6) = 35. The function in Excel is combin(7,3). We know from question 2 there are 7!/(7-4!) = 210 permutations to pick three socks out of 7, with regard to order. We also know from question 1 there are 3!=6 ways to order 3 pairs of socks. If order doesn’t matter we can divide the number of permutations by the number of ways they can be ordered. In this case we divide by 3! So that is why the number of combinations is 7!/(7-3!)/3! = 7!/((7-3)!*3!).

In drawer 2 you have 4 red socks, 3 blue socks, 2 yellow socks, and 1 green sock.

Q4: You pick two socks at random from drawer 2 without replacement what is the probability you pick two of the same color?

A4: The answer is the number of 2-sock combinations or the same color divided by the number of all 2-sock combinations. The number of combinations of 2 red socks is combin(4,2)=4!/(2!*2!) = 6. The number of combinations of 2 blue socks is combin(3,2)=3!/(2!*1!) = 3. The number of 2 yellow sock combinations is 2!/(2!*0!) = 1. The number of 2 green sock combinations is zero, because there is only one green sock. So the total number of combinations of the same color is 6+3+1=10. The total number of ways to pick any two socks out of 10 is combin(10,2)=10!/(8!*2!) = 10*9/2=45. So the probability of choosing two of the same color is 10/45 = 22.22%.

Q5: You pick 4 socks at random from drawer 2 without replacement. What is the probability they are all different colors?

A5: The number of ways to draw 4 socks of different colors is 4 (one for each red sock) * 3 (one for each blue sock) * 2 (one for each yellow sock) * 1 (one for the green sock). So the total number of ways to pick four different colors is 4*3*2*1 = 24. The total number of ways to choose four socks out of 10 is combin(10,4) = (10*9*8*7)/(1*2*3*4) = 210. So the answer is 24/210 = 11.43%.

Q6: You draw one group of 4 socks, one group of 3 socks, one groups of 2 socks, and one group of 1 sock, all without replacement. What is the probability each group consists of socks of all the same color?

A6: The number of ways to draw groups of 4, 3, 2, and 1 is 10!/(4!*3!*2!*1!) = 12,600. There is only one way to have all groups of the same color: {RRRR},{BBB},{YY},{G}. So the probability all groups are of the same color is 1/12600 = 0.000079.

In drawer three you have 4 red socks, 4 blue socks, 4 yellow socks, and 4 green socks.

Q7: You pick 3 socks at random. What is the probability they are all the same color?

A7: There are 4 different colors. Each one has three socks. There is combin(4,3)=4!/(3!*1!) = 4 ways to choose 3 socks out of 4. So there are 4*4=16 ways to choose 3 socks of the same color. There are a total of combin(16,3)=560 ways to choose 3 socks out of 16. So the probability of choosing 3 of the same color is 16/560 = 2.86%.

Q8: You pick three socks at random without replacement. What is the probability of choosing three socks of different colors?

A8: There are combin(4,3)=4 ways of choosing 3 colors out of 4. For each color there are 4 different socks. So there are 4 (for each of the different color combinations) * 4³ (for the different ways to choose one of three socks from each of 3 different specific colors) = 4*64 = 256. We know there are 560 total combinations from question 7. So the probability of choosing 3 different colors is 256/560 = 45.71%

Q9: You pick three socks at random without replacement. What is the probability of exactly two different colors represented?

A9: There are combin(4,2)=6 ways to choose 2 colors out of 4. Of the color with two socks there are combin(4,2)=6 ways to choose 2 of the 4 socks. Of the color with one sock there are 4 ways to choose the one sock. So if we arbitrarily decided which color would have two socks and which would have one then the total number of combinations would be 6*6*4=144. However once we pick two colors (say red and blue) either the red socks or blue socks will have 2. So we must multiply the 144 by 2 because the pair can come from one of two colors. Thus the total number of combinations is 144*2=288. We know from problem 7 there are 560 total combinations. So the probability of having 2 colors represented is 288/560=51.43%.