# Tough Probability Question. [closed]

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?

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

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Did, Namaste, Leucippus, Shailesh, Claude Leibovici
If this question can be reworded to fit the rules in the help center, please edit the question.

• Have you heard of Markov chains? – Matthew Leingang Mar 31 '17 at 2:03
• 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? – Jay Sarson Mar 31 '17 at 2:26
• Isaac's answer is probably the way to go – Matthew Leingang Mar 31 '17 at 2:46

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 $\approx65.22$%.