I have a really hard math question. Well at least to me?

Question

Here is what i am trying to do. Lets say i have 14 blocks. One each block it is labled from 1 - 14. now i have a machine that is going to pick 4 of them by random but there will be a first second third and forth. Now here is the question. I am tring to find out the odds of picking all 4 in the correct order. PLEASE help me out

Answer 1

Easy. Assuming I understand this correctly--the chance of picking any one block is equal to the others, and each correct block must be picked out in the correct order...

The chance the first will be picked out correctly is 1/14. The chance of picking the second one out correctly (assuming the first was picked correctly) is 1/13. The next two are 1/12 and 1/11. Therefore the odds of picking out the entire sequence correctly is 1/14 * 1/13 * 1/12 * 1/11, or 1/14 * 13 * 12 * 11, which is 1/24024.

Answer 2

Other answerers seem to have interpreted your question differently than I, so let me explain how I interpreted it first.

You ask what the odds are of "picking all 4 in the correct order." I assume that you mean that you want to know the probability that the four blocks selected are selected in increasing order (whatever the numbers are). For example, if the machine picks 3, 7, 8, and 12 (in that order), that would be a successful outcome; if the machine picks 2, 4, 6, 14 (in that order), that would be a successful outcome; if the machine picks 7, 3, 8, 12, then that would be an unsuccessful outcome.

The problem has a certain symmetry which you can exploit in coming to a solution.

The machine will pick four blocks randomly. Suppose for a moment that it picks four particular numbers; say 1, 2, 3, 4. Now, there are 4! = 24 different *orders* in which it could pick those particular blocks. Each of these orders are equally likely. So if it picks these four numbers, then the probability it will pick them in order is 1/24.

But there is nothing special about those particular numbers; the same reasoning holds no matter which four blocks are selected. No matter which four are selected, the probability of them being in increasing order is 1/24.

So the probability that the machine picks blocks numbered in increasing order is 1/24.

Answer 3

This is a permutation question.

Simply put, There is only 1 way of putting certain 4 blocks out of 14 blocks available in a certain order. Knowing that the odd of picking all 4 picked blocks in correct order will be 1 out of all possible way of picking 4 blocks in certain order from 14 available blocks.

How many ways can you pick 4 blocks out of 14 available blocks?

For the first one, you have a choice of 14.
For the second, you have a choice of 13 (14 - 1 "the one that's already picked as the first one")
For the third, you have a choice of 12 (13 - 1 "the one that's already picked as the second one")
For the fourth, you have a choice of 11 (12 - 1 "the one that's already picked as the third one")

So the answer is 1/(14*13*12*11) = 1/24024 = 0.0000416250 meaning your odds of picking all 4 in the correct order is NOT VERY GOOD!!!!

Answer 4

If your machine has to go in order, then the machine must pick a number 11 or lower, or else otherwise you would run out of numbers before you could get four picked out. So that means the machine has a 11/14 chance for the first pull. Then for the next one, you have one less block, but there is only one block it can pick, so the odd is 1/13, and the third would be 1/12, and the fourth would be 1/11.

So if you multiply those four fractions together you get
11/24024 chance, or 1 in 2184

Answer 5

First, we need to find the number of different arrangements of 4 bricks chosen from 14 without replacement. Since order matters we will calculate the permutation 14 choose 4.

14! / (14 - 4)!
= 14! / 10!
= (14)(13)(12)(11)(10!) / 10!
= (14)(13)(12)(11)
= 24,024

We could also use the fundamental principle of counting (no replacement of bricks after each choice):

(14 ways to choose first brick)(13 w t c second b)(12 w t c third b)(11 w t c fourth b)
= (14)(13)(12)(11)
= 24,024

The probability of choosing 1 set of four bricks, in a particular order, from 14 bricks (without replacement) is 1 / 24, 024.

Answer 6

TheMathemagician's answer is spot on.

Whatever the 4 blocks chosen, there is a 1/24 chance of them being chosen in order.

If you meant that the machine should choose 1,2,3,4 in that order, then the first two guys would be correct.

Answer 7

Since you specify order,
P = (1/14)(1/13)(1/12)(1/11) = 1/24024
Odds against are 24,023 to 1

Answer 8

I think it is probability of an event followed by another without replacement.
sol:

(1/14)X(1/13)X(1/12)X(1/11) =

1 out of 24024 chances...
or .004% oohh.. too bad. =(