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Suppose I have 8 indistinguishable white flowers and 2 indistinguishable red flowers. Out of all the distinguishable arrangements, what's probability of selecting an arrangement with at least 6 consecutive white flowers.

We have $\frac{10!}{8!2!} = 45$ arrangements. If I treat a block of 6 white flowers as one object, then we have 6w,w,w,r,r. This gives $\frac{5!}{2!2!} = 30$, but we shouldn't over count when the other white flowers are adjacent to the block of 6w. This means we must remove all the 7 white arrangements, so 7w,w,r,r gives $\frac{4!}{2!} = 12$, but we over count the 8 blocks of white flowers. Hence, $\frac{3!}{2!} = 3$, so $12 - 3 = 9$, and $30 - 9 = 21$. So, the probabilty is: $\frac{21}{45} = \frac{7}{15}$.

I have two questions, did I get it right? And, is there a better way to do this if I did?

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  • $\begingroup$ That numbed seems very high. Imagine the $8$ white ones in a row, with gaps. Each red can go in any gap, independently, with equal probability. There are nine gaps and, even for one red, more than half break the desired streak. $\endgroup$
    – lulu
    Commented Jun 21 at 16:35

2 Answers 2

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I would count arrangements with at least six consecutive white flowers by conditioning on the longest white block:

Case 1: Longest White block is $8$ (denoted $W^8$). There are three ways this could happen: $RRW^8$, $RW^8R$, and $W^8RR$.

Case 2: Longest White block is $7$. There are six ways this could happen: $RWRW^7$, $WRRW^7$, $WRW^7R$ and reversals of these.

Case 3: Longest White block is $6$. There are nine ways this could happen: three ways where everything else is to the left of $W^6$; three ways where everything else is to the right of $W^6$; and three ways where there is stuff on both sides of $W^6$).

This all gives a total of $18$ arrangements of interest. So the corresponding probability of such an arrangement is $\frac{18}{45}=\frac{2}{5}$.

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  • $\begingroup$ So, my mistake was that I was still over counting. For 8W it's $3$ which makes 7W have $\frac{4!}{2!} - 2(3) = 6$, and for 6W we have $\frac{5!}{2!2!} - 2(6) - 3(3) = 9$. I.e., it each step I have to increase how many copies of the previous ones I remove. $\endgroup$
    – JavaGoblyn
    Commented Jun 21 at 23:19
  • $\begingroup$ BTW, $\frac{18}{45} = \frac{2}{5}$. $\endgroup$
    – JavaGoblyn
    Commented Jun 21 at 23:29
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  • Between two red flowers, there are three spaces (including ends) which can be viewed as bins where clumps of white flowers can be inserted, $\;\square R \square R \square$

  • First insert the two "extra" $(>6)$ white flowers using stars and bars in $\binom{2+3-1}{3-1} = 6$ ways

  • The "mandatory" clump of $6$ white flowers can go to any of the $3$ bins, giving favorable ways of $6\cdot3 = 18$ against total ways $\binom{10}2 =45$

  • Thus $Pr = \frac{18}{45} = \frac25$

$\mathtt{To\; generalise}$

Let $r$ = number of red flowers
$w=$ "mandatory" clump of whites ($=6$ here)
$x, <w$= "extra" whites available ($=2$ here), then

$$Pr = \dfrac{(r+1)\times\dbinom{x+r}{r}}{\dbinom{x+w+r}{r}}$$

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