# Thread: Probabilities in games of chance - help me solve this mystery

1. ## Probabilities in games of chance - help me solve this mystery

I am having difficulty solving this problem...

What is the probability of getting more 6's than 5's on four
tosses of a fair six-sided die?

I'm not sure how to set up the different cases...
P(6-6-6-5), P(6-6-5-x) and ???? How do I factor in other case such
as rolling one 6 and no 5's? The answer is 0.325 but I don't know how
to arrive at this answer. Thanks, Nancy

2. Originally Posted by nancydrew
I am having difficulty solving this problem...

What is the probability of getting more 6's than 5's on four
tosses of a fair six-sided die?

I'm not sure how to set up the different cases...
P(6-6-6-5), P(6-6-5-x) and ???? How do I factor in other case such
as rolling one 6 and no 5's? The answer is 0.325 but I don't know how
to arrive at this answer. Thanks, Nancy
You could have 1 six and no fives in 4(4*4*4) way: 4 places for the six 4 other numbers in three places.

You could have 2 sixes and no fives or 1 five in 6[4*4+2*4]: 6 places to put 2 sixes etc.

You could have 3 sixes and no fives or 1 five: 4[4+1]

You could have 4 sixes: 1 way.

Add those up and divide by $\displaystyle 6^4$ and see what you get.

3. ## Aha!

Plato is worthy of his/her name! Merci!

4. Originally Posted by Plato
You could have 1 six and no fives in 4(4*4*4) way: 4 places for the six 4 other numbers in three places.

You could have 2 sixes and no fives or 1 five in 6[4*4+2*4]: 6 places to put 2 sixes etc.

You could have 3 sixes and no fives or 1 five: 4[4+1]

You could have 4 sixes: 1 way.

Add those up and divide by $\displaystyle 6^4$ and see what you get.
Here is another way to proceed. Let a be the probability that there are more 6s than 5s, b be the probability that there are exactly as many sixes as fives (2 each), and c be the probability that there are more 5s then 6s.

Then $\displaystyle b = \frac{4!}{2! 2!} (1/6)^4$
and by symmetry, a = c.

Now use $\displaystyle a + b + c = 1$ to solve for a.

5. ## Re: Probabilities in games of chance - help me solve this mystery

That's great! Never understood how somebody can solve such tasks. All of this mathematics is such difficult for me .