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The question is as follow:

(i) Find the binomial expansion of $(1-x)^{-3}$ up to and including $x^{4}$.

(ii) A player throws a 6-sided fair die at random. If he gets an even number, he loses the game and the game ends. If he gets a "1", "3" or "5" he throws the die again. He wins the game if he gets either "3" or "5" thrice consecutively (eg. 335, 555, 353) and the game ends. Find the exact probability of him winning the game.

I have been thinking about this question for quite a while. Obviously, the author of the question wants us to solve part (ii) with the help of part (i). However, to solve part (ii), it looks more an infinite series to me (the possible combinations of winning the game).

Could anyone contribute to solve this question please?

user42268
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  • To clarify: Your examples all show consecutive $3$s and $5$s, but you don't require them to be consecutive in the text. Does the player win by rolling $3,1,3,5$? – joriki Jun 06 '16 at 09:58
  • Hi Joriki, thanks for the reply! Sorry for not making it clear. In order to win, the player has to roll either 3 or 5 three times consecutively. So 3,1,3,5 is not a win. But 3,1,3,5,5 is a win. – user42268 Jun 06 '16 at 10:01
  • Yes. It is a win. As long as there is consecutive sequence of 3 or 5 three times. – user42268 Jun 06 '16 at 10:03

2 Answers2

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Binomial expansion of $$(1 + x)^a = 1 + ax + \frac{a(a-1)}{2!} x^2 + \frac{a(a-1)(a-2)}{3!} x^3 + \frac{a(a-1)(a-2)(a-3)}{4!} x^4 + {\cal O}(x^5)$$ when $|x| < 1$.

Part 2) I have just written out the pattern of him winning in 3 steps, 4 steps, 5 steps, etc. $$ \Pr(Win) = \frac{1}{3^3} \bigg[ 1 + \Big(\frac{1}{6}\Big) + \Big(\frac{1}{6^2} + \frac{1}{6 \cdot 3}\Big) + \Big(\frac{1}{6^3} + 2 \frac{1}{6^2\cdot3} + \frac{1}{6\cdot 3^2}\Big) + \Big(\frac{1}{6^4} + 3 \frac{1}{6^3\cdot3} + 3\frac{1}{6^2\cdot 3^2}\Big) + \Big(\frac{1}{6^5} + 4 \frac{1}{6^4\cdot3} + 6\frac{1}{6^3\cdot 3^2}\Big) + \cdots \bigg] $$ Do you see what can be done from here?

I wrote them as series which can be summed.

$$\Pr(Win) = \frac{1}{3^3}\bigg[\sum_{n=0}^\infty 6^{-n} + \frac{1}{3}\sum_{k=0}^\infty k 6^{-k} + \frac{1}{3^2} \sum_{h=1}^\infty {h+1 \choose 2}6^{-h} \bigg],$$

Each of which can be summed without too much difficulty.

Merkh
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  • Hi Merkh. Thanks for your help. I am reading the working now. – user42268 Jun 06 '16 at 10:56
  • I can understand why how you wrote down the possible outcomes. For example, if x stands for "3" or "5" and y stands for "1". The third term in the parenthesis ($\frac{1}{6^{3}}$ + $2\frac{1}{6^{2}\times3}$ + $\frac{1}{6\times3^{2}}$) means "yyyxxx", "yxyxxx", "xxyxxx" and "xyyxxx". The sum of series also make sense. My question is: how could we make use of part (i)? Does the result of part (i) help solve part (ii)? Or does part (i) only give us hint to use summation of series to solve part (ii)? That's my main doubt now. – user42268 Jun 06 '16 at 11:00
  • I can understand why how you wrote down the possible outcomes. For example, if x stands for "3" or "5" and y stands for "1". The third term in the parenthesis ($\frac{1}{6^{3}}$ + $2\frac{1}{6^{2}\times3}$ + $\frac{1}{6\times3^{2}}$) means "yyy", "yxy", "xxy" and "xyy". The probability of "xxx" is $\frac{1}{3^{3}}$. The sum of series also make sense. My question is: how could we make use of part (i)? Does the result of part (i) help solve part (ii)? Or does part (i) only give us hint to use summation of series to solve part (ii)? That's my main doubt now. – user42268 Jun 06 '16 at 11:05
  • The binomial expansion up to order $x^4$ is not going to help sum those series. I'll have to think a bit on how to use the first 4 terms of the binomial series can be used in calculating the probability of winning, seeing as there are an infinite number of ways one could win, hence one expects an infinite series in the answer. If I come up with anything of use, I'll post it later. – Merkh Jun 06 '16 at 11:12
  • Thank you! I will work on how to sum the third summation of series. $$\frac{1}{3^2} \sum_{h=1}^\infty {h+1 \choose 2}6^{-h} \bigg],$$ – user42268 Jun 06 '16 at 11:31
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This doesn't really answer your question how to make use of part (i), but since you already accepted an answer that doesn't do that either, perhaps that's not so bad, and perhaps this answer can help someone else answer the original question.

The generating function of three variables $x,y,z$ for sequences of three types of rolls with exactly one roll of type $z$ at the end and with no three consecutive rolls of type $x$ is

$$ z\left(1+x+x^2\right)\sum_{j=0}^\infty\left(y\left(1+x+x^2\right)\right)^j=z\cdot\frac{1+x+x^2}{1-y\left(1+x+x^2\right)}\;. $$

Substituting the probabilities $x=\frac13$, $y=\frac16$, $z=\frac12$ yields the probability

$$ \frac12\cdot\frac{1+\frac13+\frac19}{1-\frac16\left(1+\frac13+\frac19\right)}=\frac{39}{41} $$

for losing the game, and thus $\frac2{41}$ for winning the game.

The expansion of $(1-x)^{-3}$ might conceivably enter into it if $1+x+x^2=\frac{1-x^3}{1-x}$ is used.

joriki
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