# Very quick probability verification

1. Sep 24, 2006

### quasar987

I got these very easy prob. questions but none of my answers match those of the book except for the 1st and last one. Here's the problem,

Dice poker is played by throwing simultaneously 5 dices. Show that:

a) P(5 different cards)=0.0926
b) P(1 pair)=0.4630
c) P(2 pairs)=0.2315
d) P(3 of a kind)=0.1543
e) P(full house)=0.0386
f) P(4 of a kind)=0.0193
g) P(5 of a kind)=0.0008

Solution:
The fundamental set $\Omega$ is "all possible thrown of the 5 dices", which yields #$\Omega=6^5$. We make the hypothesis that all fundamental events (the elements of $\Omega$) are equiprobable. With this assumption, we can calculate the required probability by taking the ratio of the given event to the cardinality of $\Omega$.

a) P(5 different cards)=6*5*4*3*2/6^5 = 0.0926

b) There are 6 ways of chosing the pair, then 5*4*3 ways to choose the other 3 cards so they are not equal to one another or to the pair. Hence,
P(1 pair)=6*5*4*3/6^5=0.0463.

c) P(2 pairs)= 6*5*4/6^5=0.0154

d) P(3 of a kind)=6*5*4/6^5=0.0154

e) P(full house)=6*5/6^5=0.00386

f) P(4 of a kind)=6*5/6^5=0.00386

g) P(5 of a kind)=6/6^5=0.0008

2. Sep 24, 2006

### shmoe

You're not taking into account the number of ways you can select the dice from the 5 to make your hand. e.g. for a 4 of a kind, you have 6*5 ways to select which number comes up 4 times and which comes up once, and you have (5 choose 1)=5 ways to select which die gets the unique number.

You are treating your sample space as though the die are labeled and unique, the 6^5 outcomes counts 1,1,1,1,2 and 1,1,1,2,1 as different events and you have to build this in to the number of ways to get the hands since the actual order for the hand irrelevant.

3. Sep 24, 2006

### quasar987

You're absolutely right!

4. Sep 25, 2006

### HallsofIvy

Staff Emeritus
By the way, there is no such thing as "dices". "Dice" is already the plural of "die"

5. Sep 25, 2006

### quasar987

1 die :grumpy: sounds weird :p