Probability between two dice games Two games, both use un-biased 6 sided dice. 
game A, Sam throws one die 4 times. He wins if he rolls at least a 6
game B,  he has 24 turns, and each time he rolls two dice simultaneously. He wins if he rolls at least one "double six"
Which game is Sam most likely to win?
I have a good idea of how to work this out and i have done majority of this i just wanted to double check that i am correct.
What i have done:
game A
sample space = $24$ $(1,2,3,4,5,6 * 4)$
E = all outcomes of rolling a 5
|E| = 4 $((6),(6),(6),(6))$
probability (6) = $1/6$
Game B
Sample space = |S|=$36*4 = 864$
E = all outcomes with (6,6)
|E| = $26$
probability (6,6) = $24/864$ = $1/36$
Is my working out correct?
 A: Your sample spaces are not correct.  For the first one, you have six possible rolls each time, so the size of the sample space is $6^4$, but that is not needed.  The chance of getting at least one six in four rolls is one minus the chance of getting no sixes.  What is the chance of not getting a six on one roll?  Now you need to succeed at that four times in a row.  Similarly for the second problem, what is the chance of not getting double sixes on one roll (of two dice)?  To not get a double six in 24 rolls, you need to succeed at this 24 times in a row.
A: The sample space is to some degree at our disposal. However, it is convenient, if we can manage it, to have a sample space in which all outcomes are equally likely.
For Game A, a convenient space is the set of all ordered quadruples $(a,b,c,d)$, where each of the symbols ranges over the numbers $1$ to $6$. The numbers $a,b,c,d$ record, in order, the result of the first toss, the second, and so on.
The sample space has $6^4$ elements. We count the favourables, the sequences which have at least one $6$. It is easier to count the sequences that have no $6$. There are $5^4$ of them.
So the probability of at least one $6$ is $\frac{6^4-5^4}{6^4}$. This is approximately $0.517746913$. That is a good approximation of the probability of winning Game A.
Now we find the probability of at least one double $6$ in $24$ tosses. Here the sample space is the set of all sequences of length $24$, where the elements of the sequence  are all ordered pairs $(x,y)$, where $x$ records the number on the gree die, and $y$ records the number on the blue die. There are $36^{24}$ such sequences.
There are $35$ equally likely ways to throw two dice and not get a double $6$. So there are $35^{24}$ "bad" sequences of length $24$ consisting of bad double-throws only. Thus the probability that in $24$ double throws there will be at least one double $6$ is $\frac{36^{24}-35^{24}}{26^{24}}$. Thus the probability of winning Game B  is about $0.491403876$.
Game A is somewhat more favourable for Sam than Game B.   
