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Box 1 and Box 2 exist. Box 1 contains 4 red balls and 1 blue ball. Box 2 contains 5 blue balls and 3 red balls. Draw from box one first. The next box drawn from is determined by the color of the ball drawn. A red ball will lead you to draw from Box 1, while a blue ball will lead you to draw from box 2. What is the chance a red ball is drawn on the 100th draw?

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closed as off-topic by Did, Namaste, Leucippus, Shailesh, Claude Leibovici Apr 1 '17 at 5:59

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  • 2
    $\begingroup$ Have you heard of Markov chains? $\endgroup$ – Matthew Leingang Mar 31 '17 at 2:03
  • $\begingroup$ I wasnt aware of them, but now that I have educated myself on the matter, how can i form such a recursive formula if thats even what I should be doing? $\endgroup$ – Jay Sarson Mar 31 '17 at 2:26
  • $\begingroup$ Isaac's answer is probably the way to go $\endgroup$ – Matthew Leingang Mar 31 '17 at 2:46
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I assume we draw with replacement, since otherwise there would not be 100 draws.

Anyways, to solve this question, we will need to use something called a matrix. If you don't know what that is, watch a Khan Academy video or something. Pay special attention to how matrices are multiplied, as that is important to constructing something called the transition matrix. $$ \left[ {\begin{array}{cc} a_{1,1} & a_{1,2} \\ a_{2,1} & a_{2,2} \\ \end{array} } \right] $$ Here is a transition matrix, where Entry $a_{i,j}$, is the chance that something in state $i$ will move to state $j$ in the next step. We can iterate this transition matrix by multiplying it by itself, and we can iterate it $n$ times by raising it to the $n$th power.

So, in our case, our starting position is the vector $[1,0]$ since we start with a $100$% chance that we are picking from the first box. And our transition matrix is

$$ \left[ {\begin{array}{cc} 4/5 & 1/5 \\ 3/8 & 5/8 \\ \end{array} } \right] $$

So now we have all the tools we need, and we perform the following computation (preferably on a calculator). $$ [1,0]\cdot \left[ {\begin{array}{cc} 4/5 & 1/5 \\ 3/8 & 5/8 \\ \end{array} } \right]^{100} \approx [.6522, .3478] $$ Thus, our chance of just having picked the red ball last time is simply the chance of us being in the first state. Thus, the probability is $\approx65.22$%.

It may be interesting to note that a matrix to the n power seems to converge to a limiting transition matrix as n goes to infinity. I suggest you investigate this further, as it is probably what this problem would be leading up to.

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