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We have a dude who throws a dice at every time he reaches an intersection, and he is starting at the bottom left (if you look closely you can see 'start'). If he throws a 6, he'll move in an eastward direction and if he throws less than 6 he'll move in a northward direction.

You might see this coming, what are the chances that he reaches $A$, $B$ and $C$ after 8 throws? We'll only discuss $A$.

My textbook says the answer is $(\dfrac{1}{6})^2 \times (\dfrac{5}{6})^4 \times \binom{8}{2}$

What I don't understand is the $\binom{8}{2}$ part. We arrive at $A$ when we have $(\dfrac{1}{6})^2 \times (\dfrac{5}{6})^4$, so when add 2 moves, isn't there a (big) possibility that we arrive somewhere else?

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That calculation is actually a mixture of the calculations for $A$ and $B$. In order to get from the starting point to $B$, you must take exactly $2$ steps to the east and $6$ to the north. It doesn’t matter in what order you take those $8$ steps: they will always get you from $\text{start}$ to $B$.

There are $\binom82$ different $2$-element subsets of any $8$-element set, so there are $\binom82$ ways to choose which $2$ of your $8$ steps will go to the east; by default the other $6$ steps will be to the north. However, the probability of any given sequence of $2$ eastward steps and $6$ northward steps is $\left(\frac16\right)^2\left(\frac56\right)^6$, not $\left(\frac16\right)^2\left(\frac56\right)^4$, and the total probability of getting to $B$ is therefore

$$\left(\frac16\right)^2\left(\frac56\right)^6\binom82\;.$$

To get to $A$, on the other hand, you must take one step to the east and four to the north. There are $\binom51=5$ ways to place the eastward step in the string of $5$ steps: it can be first, second, third, fourth, or last. The probability of any specific string of $5$ steps, one to the east and four to the north, is $\left(\frac16\right)^1\left(\frac56\right)^4$, so the total probability of reaching $A$ is

$$\left(\frac16\right)^1\left(\frac56\right)^4\binom51\;.$$

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So, the book is wrong with their answer? –  ZafarS Dec 25 '12 at 12:11
    
Oh, nevermind, stupid question. –  ZafarS Dec 25 '12 at 12:13
    
@ZafarS: It sure is. Spectacularly wrong, in fact. –  Brian M. Scott Dec 25 '12 at 12:13
    
Well I'm sorry but I forgot to add something (the answer in the book is still incorrect though); it must be after 8 throws. This is straight forward for points $B$ and $C$, but how you handle $A$? You have 3 steps left, what can you do with those? –  ZafarS Dec 25 '12 at 12:27
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@ZafarS: The probability of reaching $A$ after exactly $8$ throws is $0$, since you can’t do it. The probability of passing through $A$ in the course of $8$ throws is the same as the probability of reaching $A$ in the first $5$ throws, as given in my answer; the remaining $3$ throws can then be anything at all. –  Brian M. Scott Dec 25 '12 at 12:31

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