confirm my answer plz?

[Fly]

You throw a dice 5 times, what is the probablity of making a full house? (ex 12112)

I'm pretty sure I'm right but would like some confirmation.

<font color="white">
There are 6*5 full houses and (6+5-1) Choose (6-1) hands (combos w/ rep). So P = 30 / ( 10 choose 5).
</font>

[BruceZ]

[ QUOTE ]
You throw a dice 5 times, what is the probablity of making a full house? (ex 12112)

I'm pretty sure I'm right but would like some confirmation.

<font color="white">
There are 6*5 full houses and (6+5-1) Choose (6-1) hands (combos w/ rep). So P = 30 / ( 10 choose 5).
</font>

[/ QUOTE ]

No. There are 6*5*C(5,2) full houses if we count each ordering of the dice, since the pair can come in C(5,2) positions. The total number of ways to throw the dice is 6^5, so the probability of a full house is

6*5*C(5,2) / 6^5 =~ 3.86%.

I understand that you are ignoring order, but your expression evaluates to about 11.9%. If you explain your C(10,5) term, I will explain why it isn't right.

[jay_shark]

Label each die a1,a2,a3,a4,a5 . Choose any three , this can happen 5c3 ways . Once you've made your three picks , there are 6 different ranks of three of a kinds . Then for the other two , it can be occupied by any of the 5 remaining pairs .

ie 11122,11133,11144,11155,11166 . But the 1's can be arranged in 5c3 ways . The total number of ways of making a full house over every scenario is 5c3*5*6. The denominator is just 6^5 . Therefore the probability of making a full house is:
5c3*5*6/6^5=0.0385 or 3.85%

[Fly]

[ QUOTE ]
[ QUOTE ]
You throw a dice 5 times, what is the probablity of making a full house? (ex 12112)

I'm pretty sure I'm right but would like some confirmation.

<font color="white">
There are 6*5 full houses and (6+5-1) Choose (6-1) hands (combos w/ rep). So P = 30 / ( 10 choose 5).
</font>

[/ QUOTE ]

No. There are 6*5*C(5,2) full houses if we count each ordering of the dice, since the pair can come in C(5,2) positions. The total number of ways to throw the dice is 6^5, so the probability of a full house is

6*5*C(5,2) / 6^5 =~ 3.86%.

I understand that you are ignoring order, but your expression evaluates to about 11.9%. If you explain your C(10,5) term, I will explain why it isn't right.

[/ QUOTE ]

Well, I was trying to count the total number of hands s.t that order didn't matter (ex 12345 = 54321). So C(10,5) is the number of combinations of size 5 from a set of size 6 w/ repetition. And then I did the same for the top for example treating 11122 and 22111 as the same.

[Fly]

What I mean is C(10,5) should be the total number of unique poker hands, it seems to me that 6^5 overcounts since it counts for example 12345 and 54321 as different hands.

For example if were talking about a regular deck of cards, the number of poker hands is C(52,5) not 52^5.

[jay_shark]

Fly , Here is an example with two dice .

You may have {11,12,13,14,15,16}
{21,22,23,24,25,26}
.
.
.
{61,62,63,64,65,66} = 36 total pairs

So if you put a sticker on each die and label it a1,a2 , then a combination like 11, tells you that die 1 is a 1 and die two is a 1. Similarly 12 tells us that die 1 is a 1 and die 2 is a 2 . This means that if you're interested in knowing the number of ways of flopping pairs when you roll two dice it must be 6/36 since there are 6 pairs {11,22,33,44,55,66} and 36 total combinations as given above .

This means that xy is different than yx for y not equal to x because the first number tells us about die 1 and the second number tells us about die 2 .

In poker if we deal out two cards , then AcKc is the same hand as kcAc . So the number of two card combos must be 52*51/2 since each pair is duplicated twice . Moreover , when we deal cards out in poker , we are not allowed to replace the cards for each card dealt out . That is , if we deal one card to one player , then there are only 51 cards remaining and NOT 52. The only time it would be 52^5 like you suggested is if we care about the order of the cards in which we receive them and that for each card dealt we open up a new deck of cards .

[Fly]

Jay, I understand your solution and accept it. I'm just trying to figure out why my solution is incorrect. Do you atleast agree that the number of "hands" that can be generated by throwing a die 5 times is C(10,5) since a hand is only determined by the number of times each "card" (1,2,3,4,5) appears and not the order? If so, is there a way to generate a solution involving that number?

[jason1990]

You are right that, when you ignore order, there are 30 full houses and C(10,5) possible hands. But these hands are not equally likely. So you cannot conclude that the probability is 30/C(10,5). On the other hand, if you do not ignore order, then there are 6^5 possible hands, and they ARE equally likely. That is why one usually does not ignore order in problems like this, even though you might not care about the order in the final analysis.

A simpler analogy is this. Flip two coins. If we ignore order, then there are three possible outcomes: (1) both heads, (2) both tails, or (3) one head and one tail. But that does not mean that outcome (1) will occur 1/3 of the time.