Probability

[akhpad]

Bill has a small deck of 12 playing cards made up of only 2 suits of 6 cards each. Each of the 6 cards within a suit has a different value from 1 to 6; thus, for each value from 1 to 6, there are two cards in the deck with that value. Bill likes to play a game in which he shuffles the deck, turns over 4 cards, and looks for pairs of cards that have the same value. What is the chance that Bill finds at least one pair of cards that have the same value?

Solution:
(1, 1), (2, 2), (3, 3), (4, 4), (5, 5), (6, 6)

Method 1:
First find out chances of getting all cards different
Probability of selecting 1st card = 12/12
Probability of selecting 2st card = 10/11
Probability of selecting 3st card = 8/10
Probability of selecting 4st card = 6/9

Probability of selecting
Probability of selecting (all 4 cards different) = (12/12) * (10/11) * (8/10) * (6/9) = 16/33
Probability of selecting (at least one pain) = 1 - 16/33 = 17/33

Method 2:
Total combination with one or more pair = 6 * (10C2) - 6C2 = 5*51
Probability of selecting at least one pain = 5*51 / 495 = 17/33

Method 3:
Total combination with all different = 12 * 10 * 8 * 6 / 4! =240
Total no of selections = 12C4 = 495
Probability of selecting at least one pain = (495 - 240) / 495 = 17/33

Hence required probability = 17/33

My Q: Why do we need to divide by 4! in Method 3?

[sanju09]

Bill has a small deck of 12 playing cards made up of only 2 suits of 6 cards each. Each of the 6 cards within a suit has a different value from 1 to 6; thus, for each value from 1 to 6, there are two cards in the deck with that value. Bill likes to play a game in which he shuffles the deck, turns over 4 cards, and looks for pairs of cards that have the same value. What is the chance that Bill finds at least one pair of cards that have the same value?

Solution:
(1, 1), (2, 2), (3, 3), (4, 4), (5, 5), (6, 6)

Method 1:
First find out chances of getting all cards different
Probability of selecting 1st card = 12/12
Probability of selecting 2st card = 10/11
Probability of selecting 3st card = 8/10
Probability of selecting 4st card = 6/9

Probability of selecting
Probability of selecting (all 4 cards different) = (12/12) * (10/11) * (8/10) * (6/9) = 16/33
Probability of selecting (at least one pain) = 1 - 16/33 = 17/33

Method 2:
Total combination with one or more pair = 6 * (10C2) - 6C2 = 5*51
Probability of selecting at least one pain = 5*51 / 495 = 17/33

Method 3:
Total combination with all different = 12 * 10 * 8 * 6 / 4! =240
Total no of selections = 12C4 = 495
Probability of selecting at least one pain = (495 - 240) / 495 = 17/33

Hence required probability = 17/33

My Q: Why do we need to divide by 4! in Method 3?

You have shown a great piece of work, Akhilesh.

Well, in your method 3, 12 * 10 * 8 * 6 shows all possible arrangements of the four chosen cards, whereas we are looking for total combinations of those four cards, and that nCr = nPr/r!, therefore we need to divide 12 * 10 * 8 * 6 by 4! for the correct answer to the question.

[harshavardhanc]

Well, in your method 3, 12 * 10 * 8 * 6 shows all possible arrangements of the four chosen cards, whereas we are looking for total combinations of those four cards, and that nCr = nPr/r!, therefore we need to divide 12 * 10 * 8 * 6 by 4! for the correct answer to the question.

did you mean 12*10*8*6 is some nPr ? How?

[sanju09]

Well, in your method 3, 12 * 10 * 8 * 6 shows all possible arrangements of the four chosen cards, whereas we are looking for total combinations of those four cards, and that nCr = nPr/r!, therefore we need to divide 12 * 10 * 8 * 6 by 4! for the correct answer to the question.

did you mean 12*10*8*6 is some nPr ? How?

sort of

there are four place numbers being filled in 12, 10, 8, and 6 number of ways in any order. Hence, all arrangements were considered here, we need to divide it by 4! in order to get the favorite combinations only.

[akhpad]

Your answer to my question looks good. Thanks

Some more comment from other people please.