This is from Ross.
I know how to solve everything but (d).
The book answer is $\frac{123}{35}$.
There are $4$ different types of coupons, the first $2$ of which compose one group and the second $2$ another group. Each new coupon obtained is type $i$ with probability $p_i$, where $p_1 = p_2 = 1/8$ and $p_3 = p_4 = 3/8$. Find the expected number of coupons that one must obtain to have at least one of
(a) all 4 types; (b) all the types of the first group; (c) all the types of the second group;
(d) all the types of either group.
My attempt
Let $X =$ number of coupons needed to collect all the types of either group.
Let's number the coupons $1$ through $4$ where $1$ and $2$ are part of the first group and $3$ and $4$ are part of the second. Let's say that the notation $4132$ represents the order of newest types seen. For example if you observe $44111332$ then the order of newest types seen was $4132$.
To get the $E[X]$ I'll condition on every arrangements where the game stops once you have all the types of either group.
$$E[X] = E[X \mid 12]P(12) +
E[X \mid 132]P(132) +
E[X \mid 134]P(134) +
E[X \mid 142]P(142) +
E[X \mid 143]P(143) +
E[X \mid 21]P(21) +
E[X \mid 231]P(231) +
E[X \mid 234]P(234) +
E[X \mid 241]P(241) +
E[X \mid 243]P(243) +
E[X \mid 312]P(312) +
E[X \mid 314]P(314) +
E[X \mid 321]P(321) +
E[X \mid 324]P(324) +
E[X \mid 34]P(34) +
E[X \mid 412]P(412) +
E[X \mid 413]P(413) +
E[X \mid 421]P(421) +
E[X \mid 423]P(423) +
E[X \mid 43]P(43)$$
For example to get the first term, $E[X \mid 12] = 1 + \frac{2}{1}$ where the logic is given that you know the coupon types will appear as $12$, the number of coupons needed to get $1$ must be one and then to get $2$ it's the expected value of a geometric RV with $p=\frac{1/8}{1/8+1/8} = \frac{1}{2}$.
To get $P(12)$ we use the multiplication rule. $P(12) = P(1) P(1 \mid 2) = \frac{1}{8} \frac{1}{7} = \frac{1}{56}$. So $E[X \mid 12]P(12) = (1 + \frac{2}{1}) \frac{1}{56}$.
To get the second term, $E[X \mid 132] = 1 + \frac{4}{3} + \frac{4}{1}$ where you know the number of coupons needed to collect $1$ and $2$ is one and to get the middle it's a geometric RV with $p = \frac{3/8}{3/8+1/8} = 3/4$.
To get $P(132) = P(1) P(3 \mid 1) P(2 \mid 13) = \frac{1}{8} \frac{3}{7} \frac{1}{4}$.
Next put it in a python script and get the answer.
Unfortunately I get $3.4 \neq \frac{123}{35} = 3.51428$.
My questions
Can you point out where I'm going wrong? Also, what is the more elegant approach? Thanks.
Best Answer
I would say there are five states, $a,b,c,d,e$, where $a$ has no coupons, $b$ has one coupon from group $1$ and none from group $2$, $c$ has one coupon from group $2$ and none from group $1$, $d$ has one coupon from each group, and $e$ has two coupons from one group and is final.
If you are in state $d$ you finish with probability $\frac 12$ so the expected number of draws from $d$ is $2$.
If you are in state $c$ you finish with probability $\frac 38$, go to $d$ with probability $\frac 14$ and stay in $c$ with probability $\frac 38$. The expected number of draws from $c$ is $$E(c)=1\cdot \frac38 + (1+E(d))\cdot \frac14 +(1+E(c))\cdot \frac 38\\ \frac 58E(c)=\frac 32\\E(c)=\frac {12}5$$ If you are in state $b$ you finish with probability $\frac 18$, go to $d$ with probability $\frac 34$ and stay in $b$ with probability $\frac 18$. The expected number of draws from $b$ is $$E(b)=1\cdot \frac 18+(1+E(d))\cdot \frac 34+(1+E(b))\frac 18\\ \frac 78E(b)=\frac 52\\E(b)=\frac {20}7$$ Finally from $a$ you go to $b$ with probability $\frac 14$ and to $c$ with probability $\frac 34$ so $$E(a)=(1+E(b))\frac 14+(1+E(c))\frac 34\\=\frac {27}{28}+\frac {51}{20}\\=\frac {123}{35}\approx 3.514$$