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The game rules are simple. You take one fair coin and flip it. You start counting how many times did you get heads and how many times did you get tails. When the number of tails equal the number of heads, the game finishes. What is the probability that the game carries on forever?

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    what have you tried, where are you stuck ? – Surb Sep 29 '22 at 12:01
  • That doesn't seem like a well formulated question. A game can not "go on forever" per se. Do you mean to ask something else such as "What is the probability the game requires more than $1000$ throws? Requires more than $10000$ throws? Requires more than $N$ throws?" – JMoravitz Sep 29 '22 at 12:06
  • It is well defined. You can just take the limit of what you describe. – justabit Sep 29 '22 at 12:11
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    This is a random walk on a line with constant step size. With probability one, it comes back to the origin (infinitely often, if you just keep going). – Gerry Myerson Sep 29 '22 at 12:20
  • Some related questions: https://math.stackexchange.com/questions/2874899/number-of-returns-of-simple-random-walk-until-a-stopping-time and https://math.stackexchange.com/questions/2487268/random-walk-high-probability – Gerry Myerson Sep 29 '22 at 12:30
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    Your "universe" can be placed in bijection with the set of pathes beginning at the origin on the square grid $\mathbb{N}^2$ made of successive East moves $\binom{1}{0}$ and North moves $\binom{0}{1}$ like EENENEEN...which do not intersect the diagonal line. – Jean Marie Sep 29 '22 at 12:34
  • It is almost certain (:)) that the answer is $0$. – Jean Marie Sep 29 '22 at 12:38
  • I did not do much work, but I noticed that the number of coins flipped should be pair and that the probability that half the 2n flipped coins will be modelized by a binomial distribution. P(n when 2n are flipped)=2nCn * (1/n)^2n –  Sep 29 '22 at 12:51
  • so I said the probability of a game ending at 2n flipped coins is the probability of getting n of them to be head. Therefore the probability of a game not ending would be 1-P (where P is the probability of the game finishing). At last i said for a game not to end, it shouldn't end at neither 2n=2;4;6;.. so the probability of a game never ending will be product sum of (1-P) But that is certainly wrong because in this way the probability of having for example in 4 flipped coins of having 2 heads counts the probability of having those combinations: HTTH, HTHT, HHTT, TTHH, THHT,THTH. –  Sep 29 '22 at 12:54
  • as you can see the game would end in 4 cases before we flip the coin 4 times, so that's where I am stuck... –  Sep 29 '22 at 12:58
  • I don't really know how to tag but for jean marie's comment, I have thought about it but I don't really know how I would rigorously write an answer with this proposition –  Sep 29 '22 at 13:03
  • @Aelkas a point of vocabulary : pair (in french) $\to$ even (in english), impair $\to$ odd. – Jean Marie Sep 29 '22 at 13:08
  • I gave you a search term, aelkas, and some links. Did you follow up on any of that? – Gerry Myerson Sep 30 '22 at 02:46
  • Yes, the second might help but it introduces new math theorems that I haven't learned yet –  Sep 30 '22 at 08:15
  • Here's another link you may wish to follow: https://math.stackexchange.com/questions/536/proving-that-1-and-2d-simple-symmetric-random-walks-return-to-the-origin-wi Also, if you want to be sure I see a comment intended for me, you have to put @Gerry in it. – Gerry Myerson Oct 01 '22 at 23:38

1 Answers1

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The probability is zero, as shown by the following simple argument.

For each $n\in\{0,1,2,\dots\}$, let $X_n$ be the quantity (number of heads) - (number of tails) after $n$ flips. WLOG, assume that the first flip is heads, so $X_1=+1$. We will divide the experiment into several "phases" as follows:

  • Phase 1 starts at the beginning, and ends the first time that $X_n$ hits either $0$ or $2$. If it exits via $0$, then the experiment is over. Otherwise, phase $2$ starts.

  • Phase $2$ ends the first time that $X_n$ hits either $0$ or $4$.

  • Phase $3$ ends the first time that $X_n$ hits either $0$ or $8$.

  • Phase $4$ ends the first time that $X_n$ hits either $0$ or $16$.

  • $\quad\vdots$

The point is, by symmetry, each phase has a $50\%$ chance to end with $X_n=0$, by symmetry, independently of previous phases. The only way for the number of heads to never equal tails is for the this $50\%$ event to never occur after infinitely many attempts, the probability of which is obviously zero.

To be complete, you need to show each phase is almost surely finite. This is not hard to do. For example, phase $4$ ends if at any point there are $16$ heads in a row, so if you divide all of the flips into disjoint blocks of $16$, there will certainly be a block of $16$ heads.

Mike Earnest
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