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During a Stochastic Process lecture, one course mate asked me a question, which he thought to be related to the Gambler's Ruin Problem. Despite the fact that up till now he hasn't found the right formalism to make this connection, the question itself intrigued me, and I deemed it an important and interesting question. It goes as follows,

Suppose that we have $2n$ balls, of which $n$ are red (non-identical), and the other $n$ are blue (non-identical). We arrange them into a sequence, according to the rule below,

For any integer $1\le i\le 2n$, the first $i$ balls always have more or equal number of red ones compared with blue ones.

The question is, how many such arrangements are possible?

Please any help or hint would be appreciated, thanks!

LYg
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    Don't you mean the number of red balls is at least as the number of blue balls? If otherwise then for $i=2n$ we we have $n$ red and $n$ blue balls, so the condition will be violated. – Stefan4024 Oct 13 '16 at 18:25
  • If so then you can check my answer on this question. The two questions are more or less the same questions, only differently worded. – Stefan4024 Oct 13 '16 at 18:26
  • Yes exactly as you said. I have just edited it. Thank you for your link, I will have a look. – LYg Oct 13 '16 at 18:28
  • Thank you very much Stefan. Exactly as you said, it's the same problem. I haven't heard of Catalan number before, now doing research about it. Can you please give me any material that demonstrates how this number is obtained for the Dyck words problem? @Stefan4024, By the way, if you like to refer the link in the formal answer, I will accept your answer formally and close the session. – LYg Oct 13 '16 at 18:57
  • As the answers aren't exact duplicates I will write an answer in a second – Stefan4024 Oct 13 '16 at 18:59
  • Great, thanks a lot! @Stefan4024 – LYg Oct 13 '16 at 19:02
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    Possible duplicate of Probability for the length of the longest run in $n$ Bernoulli trials – amWhy Oct 14 '16 at 00:30
  • @amWhy Please note that I have added the context of my problem. – LYg Oct 16 '16 at 17:44
  • @amWhy No, I didn't change the question itself, I only added the first paragraph about the context of the question. Yes what Stefan4024 linked to me boils down to the same problem IN ESSENCE as the one I posted, but to be honest it took me quite a while to realize that, because on surface they don't look related. Before I posted my question I tried various key words to search in MathSE in case of ready solution, but couldn't find one, that's why I posted this one for solution. My course mate appreciated it when I sent him the link to my post because it's the direct question from him. – LYg Oct 17 '16 at 18:24
  • @LYgI apologize that I suggested you changed the actual question you posted. With respect to your comment in general, doesn't it feel good that, even if it took awhile, you were able to recognize the shared essence? Sometimes, when skimming the surface, we can miss all that lies below the surface, and how it all connects. I'm glad you and your course-mate found what you're looking for. – amWhy Oct 17 '16 at 18:31
  • @amWhy Oh don't worry, that's fine. I do appreciate your effort to maintain MathSE a good environment to communicate math problems. Yes indeed I was very glad when I recognized the shared essence, and that actually is the beauty of Mathematics, to abstract the key components out of real world questions, and formulate a framework that comprehends all. As you may have noticed if you look at my profile page, this one is my first post at MathSE, with all the rest in PhysicsSE. I have found SE a very good and efficient platform for academic communications. Thank u & ur colleagues for ur service. – LYg Oct 17 '16 at 22:31

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Let denote a red ball with $1$ and a blue ball with $0$. Now writing the sequence of the $2n$ balls will give us a binary sequence. We want to count the number of binary sequence in which any initial sequence has at least the same number of red ball as blue balls. This is exactly the number of Dyck words of length $2n$ and it's given by the Catalan Number:

$$C_n = \frac1{n+1}\binom{2n}{n}$$

The wikipedia page about Catalan Numbers has a good explanation how the Catalan Numbers and Dyck Words are connected.

Stefan4024
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