Poker dice Probability calculator

[Yodhur]

Hello,

I am not sure of the way to calculate a precise probability in a poker dice game.
The poker dice game use 5 dice (1/6 probabilty of one figure to come out in one throw).
more info: https://en.wikipedia.org/wiki/Poker_dice

I want to calculate the probability of a particular set of 2 different pair being thrown out.

I want to calculate the odds of one pair being thrown out (for example 66), then the second is the pair directly behind (in the example 55) and the last dice is the figure again rigth behind (4). so 66554 in my example. (I want to calculate odd of the first pair being N, the second N-1 and the last dice N-2.

I know there is 7776 different combination with 5 dices (6^5).
I know there is simply 4 possible way to arrange the two pair I want to get: 66554 - 55443 - 44332 -33221. (I fugure out just by trying).

The solution cannot be simply, 4/7776?

I calculate that the first pair as (5 2) way to be arrange in 5 throw, then the second want has (3 2) in the 3 remaining dice and the fifth dice is determine (2 1).
Then I calculate 4 * (5 2) * (3 2) /7776=1.54%

Am I correct, I am not sure at all.

I want also to calculate the odds of a 2 pair rising (one dice N, first pair N+1, second pair N+2 for example 65544).
Is this the same calculation? I think so.

Would someone be so kind to ligth my way?
I am a bit lost.

Regards and thank you in advance,
Yodhur

 

[Romsek]

since you have two pairs in your target roll you have 4 ways of rolling each of them.

\(\displaystyle p=\dfrac{16}{7776} = \dfrac{1}{486}\)

 

[JeffM]

How many ways can I get two aces, two kings, and one queen.

How many ways can I get exactly two aces?

[MATH]\dbinom{5}{2} = 10.[/MATH]
Let's check.

AAXXX, AXAXX, AXXAX, AXXXA, XAAXX, XAXAX, XAXXA, XXAAX, XXAXA, and XXXAA.

How many ways can I get exactly two kings given that I got exactly two aces?

[MATH]\dbinom{3}{2} = 3.[/MATH]
Let's check. Suppose I got XAAXX. Then I could have KAAKX, KAAXK, or XAAKK.

So we have 3 * 10 ways to roll exactly two aces and two kings. But that leaves only one way to roll a queen if we already have two aces and two kings.

So there are 30 ways to roll two aces, two kings, and one queen.

Similar logic applies to two kings, two queens, and one jack, or two queens, two jacks, and one 10, or two jacks, two 10's , and one 9.

So we have (5 * 30) = 5 * 5 * 6 ways to get the situation you want. How many total possibilities are there? Obviously 6^5.

Therefore the probability that you want is

[MATH]\dfrac{5 * 5 * 6}{6^5} = \dfrac{25}{1296} \approx 0.4\%.[/MATH]
Now this is worrisome because I get a different answer from romsek, and he is almost invariably correct.

 

[JeffM]

In the previous post, that should be

[MATH]4 * 30 = 20 * 6[/MATH] ways to get the situation desired.

So the probability is

[MATH]\dfrac{20 * 6}{6^5} = \dfrac{20}{1296} = \dfrac{10}{648} \approx 1.5\%.[/MATH]

 

[HallsofIvy]

My standard rant: "Dices" is a verb that means "cuts something into small cubes". "Dice", as a noun, is itself the plural of the singular "die". You cannot have "5 dices", rather "5 dice" instead of "one die".