Chance of pairs, rolling 5 dices simultaneously

• Nov 23rd 2008, 10:22 AM
desperate87
Chance of pairs, rolling 5 dices simultaneously
Hello all,

This is from A first course in probability by Seldon Ross 7th edition.

Chapter 2 Problem 16
Poker dice is played by simultaneously rolling 5 dice. Show that:
a) P { no two alike } = 0.0926
b) P { one pair } = 0.4630
c) P { two pair } = 0.2315
d) P { three alike } = 0.1543
e) P { full house } = 0.0386
f) P { four alike } = 0.0193
g) P { five alike } = 0.0008

Thanks!
• Nov 24th 2008, 11:01 AM
desperate87
Ok I can solve problems a,b and g.

a) P { no two alike } = $\frac{6*5*4*3*2}{6^5}$ = 0.0926

b) P { one pair } = $\frac{6*{5 \choose 2}*5*4*3}{6^5}$ = 0.4630

g) P { five alike } = $\frac{6}{6^5}$ = 0.0008

However the trick I used for b doesn't work for c,d,e and f, and I don't understand
why? This is how I solve b:

The numerical value for the pair can be chosen in 6 ways and can be placed in ${5 \choose 2}$ ways, and then there are 3 'slots' left. The numerical value for the remaining 3 slots can be chosen in 5*4*3 ways. The whole thing is divided by the total number of possibilities which is $6^5$.

But as I said this same line of reasoning doesn't work for c,d,e and f? Am I doing something wrong?
• Nov 24th 2008, 11:28 AM
Plato
Quote:

Originally Posted by desperate87
Poker dice is played by simultaneously rolling 5 dice. Show that:
c) P { two pair } = 0.2315
d) P { three alike } = 0.1543
e) P { full house } = 0.0386
f) P { four alike } = 0.0193

It is just counting. All the dominators $6^5$.

c) Pick the two pairs: ${6 \choose 2}$. Pick the odd die out $4$. Now scramble.
${6 \choose 2}4\frac{5!}{(2!)^2}$

You try the rest!
• Nov 24th 2008, 01:03 PM
desperate87
Thank you Plato
I finally understand it.

I'll post my results when I'm done (Nod)