Probability Combination and Permutations

[Saladsamurai]

Homework Statement

I am getting a little confused as to whether to use the number of permutations or the number of combinations as my sample space when it comes to probability problems, or whether it depends on the situation. Let's look at two of the examples from my text to help:
Example 1:

In the above example, they are using the following formula to determine the sample space:

and thus the sample space consists of distinct combinations.Example 2:

In this example it looks like they are using permutations as the sample space since, for example, elements THH and HHT are not distinct. I will also include the definition of probability, since I think that it might help me clear the confusion:

Any thought on this one?

~Casey

 

[happyg1]

With ONE coin, If you get a T first, you are looking for another T. If that doesn't happen and you still get a T first, then get a H, you are looking for another H for the 2 in a row. Here ORDER MATTERS. That's a permutation. YOu have to count all of them.

In the poker problem, you have 52 cards, you are choosing 5, and of those 5 you have preferences for the J and A.

In the coin problem, it's not 3 choose 2, it's 3 coin tosses with 2 possibilities each. You have to count all of them.

When you are dealt 5 cards, it doesn't matter if you get AJAJJ or JJJAA or JJAAJ etc.

In order to determine the TOTAL number of ways something(coin tosses) can happen, you have to count all of the possibilities. Coin tosses are independent. Dealing cards is not.

Independence and dependence doesn't govern the use of permutations or combinations...Order of the cards on a poker hand is a combination. Tosses of a coin uses permutations to find the total number of outcomes, as does the combination. Then you divide out the repeated outcomes...that's the r! in the denominator. Probability = desired outcome/total possible outcomes.

You have the combination thing posted twice. If there are repeats in your desired outcomes, you divide them out.

I ramble...

 

[vela]

You can recast the coin-flipping problems in terms of combinations. To lose, he has get either one head or two heads, and it doesn't matter which of the three coins comes up heads, i.e. order doesn't matter. The number of ways to get one head is therefore 3c1. He has three slots and has to choose one in which to place the H. Similarly, the number of ways to get two heads is 3c2; he has to choose two slots for the Hs. So he has a total of 3c1+3c2=6 losing outcomes out of a total of 2³ possible outcomes, giving a 3/4 probability of losing.

With this problem, since there are only 8 possible outcomes, it's just easier and clearer to write down all the possibilities and count them by hand, but you can see the connection between the two approaches as follows:

3c0 combinations - TTT
3c1 combinations - HTT, THT, TTH
3c2 combinations - HHT, HTH, THH
3c3 combinations - HHH

 

[Saladsamurai]

Hey guys. Thanks for all of the input. I seem to have confused myself yet again. I thought that the formula:

\left( \begin{array}{c}n\\r\end{array}\right)=<br /> \left( \begin{array}{c}\frac{n!}{r!(n-r)!}\end{array}\right)

gives the number of distinct arrangements of n objects taken r at a time. So I am not sure about this vela

3c0 combinations - TTT
3c1 combinations - HTT, THT, TTH
3c2 combinations - HHT, HTH, THH
3c3 combinations - HHH

For example:

for 3c1 combinations - HTT, THT, TTH

3c1 = (3!)/(1!(3 - 1)!) = 3

I am not sure how this is a legitimate usage of the formula. HTT,THT,TTH are not distinct combinations. Sorry, I am just not sure how you are using the rule appropriately. Or rather, why the answer comes out right.

 

[Office_Shredder]

Hey guys. Thanks for all of the input. I seem to have confused myself yet again. I thought that the formula:

\left( \begin{array}{c}n\\r\end{array}\right)=<br /> \left( \begin{array}{c}\frac{n!}{r!(n-r)!}\end{array}\right)

gives the number of distinct arrangements of n objects taken r at a time. So I am not sure about this vela

This gives the number of ways to pick r objects out of n objects. It says nothing about the number of arrangements.

 

[Saladsamurai]

This gives the number of ways to pick r objects out of n objects. It says nothing about the number of arrangements.

So how is vela using this formula? It seems like saying 3c1 simply gives the number of ways of picking 1 coin from 3. And it says nothing of wether it is heads or tails.

 

[Tedjn]

Yes, almost. It is picking one position out of the 3 in the sequence in which to place the H.

 

[Saladsamurai]

Yes, almost. It is picking one position out of the 3 in the sequence in which to place the H.

Right, but it does not relate the chance of it actually being a heads.

 

[Tedjn]

I haven't read the entire thread, but it seems that vela is just using this to count the number of ways that the player can lose. When all else fails, try to use P(event) = #(ways event occurs)/#(total possibilities).

 

[Saladsamurai]

I haven't read the entire thread, but it seems that vela is just using this to count the number of ways that the player can lose. When all else fails, try to use P(event) = #(ways event occurs)/#(total possibilities).

Gotcha! I misinterpreted his usage.

 

[Office_Shredder]

Right, but it does not relate the chance of it actually being a heads.

Because he's not doing probability yet, he's just counting the number of ways for there to be 1 head (which is the number of ways to pick 1 coin out of the three) and the number of ways to get two heads (which is the number of ways to pick 2 coins out of the three). Only after you're done counting do you divide by the total number of configurations to get the probability

 

[Saladsamurai]

Because he's not doing probability yet, he's just counting the number of ways for there to be 1 head (which is the number of ways to pick 1 coin out of the three) and the number of ways to get two heads (which is the number of ways to pick 2 coins out of the three). Only after you're done counting do you divide by the total number of configurations to get the probability

Hi Office_Shredder I think that Tedjin was correct. If you think about it

<snip> ... the number of ways for there to be 1 head (which is the number of ways to pick 1 coin out of the three) and the number of ways to get two heads (which is the number of ways to pick 2 coins out of the three)

does not make much sense. At least to me. Saying that there is N number of ways to take 1 coin from 3 does not equate to the number of ways for there to be one head. What if all 3 are on tails? Then it does not matter which coin I pull...there are still no H's.

Maybe I am misinterpreting you.

 

[Office_Shredder]

Hi Office_Shredder I think that Tedjin was correct. If you think about it

does not make much sense. At least to me. Saying that there is N number of ways to take 1 coin from 3 does not equate to the number of ways for there to be one head. What if all 3 are on tails? Then it does not matter which coin I pull...there are still no H's.

Maybe I am misinterpreting you.

If you get one head, then one out of the three coins is a head. The number of different ways for this to occur is the number of ways for you to pick one coin (picking the coin to be heads). Then the number of ways for there to be two heads is the number of different ways for you to pick two coins, which will be the two heads.

All I'm doing at this point is counting. There's no probability, there's no chance of having zero heads yet because all I've done is count the number of ways for there to be one head, and the number of ways for there to be two heads.

Now for the probability part: There are 6 ways to get one head or two heads, and there are 8 total possibilities for the three coin flips, each with the same chance. So there is a 6/8 chance that you get one or two heads

 

[Saladsamurai]

OK! I think I am with you now If anyone is interested, I also found http://www.beatthegmat.com/combination-and-permutation-manhattan-t8924.html" in which the use of both combinations and permutations is used for the same problem (4th from last reply posted by Ian Stewart).