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I toss a balanced coin until the number of heads I get equals the number of tails? What's the chance I never stop?

I have tried considering the reverse event and a recursive reasoning but nothing conclusive. A close question has already been asked here but I don't know the Markov formalism that is used.

Jam
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    Can you evaluate the probability you will toss the coin more than a million times? – TheHolyJoker Jan 24 '20 at 14:07
  • It feels as hard as the original problem. –  Jan 24 '20 at 14:09
  • My hunch is that you will always stop eventually but the number of pairs of tosses required will have a long tail. – Oscar Lanzi Jan 24 '20 at 14:13
  • The problem for sequences of flips of finite length, $n$, is slightly more complicated, since we could stop flipping at subsequences before $n$. However, when $n$ is infinite, whenever we have equal $H,T$, we can keep flipping and ask the same question by taking the stopping point as a new starting point, by symmetry: $\underbrace{\big[HTH\ldots TH\big]}{\text{stopped sequence}}\big[\underbrace{TH\ldots}{?}\big]$. Therefore, we can simplify the problem by considering all infinite sequences of flips and asking whether they all have equal $H,T$ at some point. – Jam Jan 24 '20 at 14:55
  • To give a combinatoric association: Rephrased to the terminology of random walks, if we count head as a step in direction (1,1), and tail as a step in direction (1,-1), the number of flip-sequences that has at any given point in time at least as many heads as tails is called a Dyck-path. Using this as a starting point, it's not that difficult to reach the conclusion given in the answers. – Sudix Jan 24 '20 at 15:03

2 Answers2

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Th probability is zero that you will continue forever. The Wikipedia article on the random walk cites the result that with probability $1$ you will visit any specific point on the number line, which includes zero, but does not prove it.

Ross Millikan
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    I did a simulation on MS Excel where you toss the coin 1000 times. In 100 replications of this simulation I reached a stopping point 98 times. Since the probability of continuing decreases only as $1/n^{1/2}$ where $n$ is the number of tosses allowed, I take this numerical resukt as consistentcwith the answer. – Oscar Lanzi Jan 24 '20 at 14:20
  • https://math.stackexchange.com/questions/536/proving-that-1-and-2-d-simple-symmetric-random-walks-return-to-the-origin-with and also https://www.math.leidenuniv.nl/scripties/BachLeenman.pdf – Maximilian Janisch Jan 24 '20 at 14:27
  • @OscarLanzi I agree the simulation is evidence that it becomes very unlikely to continue forever, but I don't think simulation is a very good tool for evaluating infinite sequences and asymptotic probabilities. For any n and sufficient replications, you will always find an example of failing to stop, and that is true for any finite n. Any simulation you choose to run with finite n will come to the opposite conclusion as infinite n - with finite n, you will fail to stop with nonzero probability. – Nuclear Hoagie Jan 24 '20 at 14:40
  • @NuclearWang The case of $n\to\infty$ is a limiting case, so the behaviour of large $n$ should provide a heuristic estimation of the limit, if it exists. If $p(n)$ fluctuates wildly, it would suggest that there is no limit. But this is not the case, as Oscar's computation would show that $p(n)$ smoothly approaches zero. Indeed, it doesn't prove that $p(n)$ doesn't have unpredictable behaviour for large $n$, but it suggests that this is unlikely. – Jam Jan 24 '20 at 14:44
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The probability of returning to the origin equal to the probability of returning back to the origin in the one-dimensional random walk problem, which is well-known to be $1$. Hence, the probability of not returning is $0$.

Ekesh Kumar
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