1

A coin is biased so that the probability of obtaining a head is $p$, where $0 < p < 1$. If the sequence $HHH$ occurs first then Player $A$ wins, and if the sequence $HTH$ occurs first then player $B$ wins. The coin is tossed until one player wins. For what value of $p$ is the game fair? i.e. Probability of Player $A$ and Player $B$ winning are equal.

I've attempted to research on Penney's game with biased coins, but there doesn't appear to be any information on it. Personally, I'm not well versed with probability, just know the basics, so any help would be useful. I've seen ideas of using Markov Chains but I've never used them myself.

TShiong
  • 1,257
LegitFakeID
  • 15
  • 4
  • Have you ever heard of penney's game? Have you seen any calculations associated with penney's game? What sort of tools do you have available to you / would you like to use? Markov Chains? Recursion? – JMoravitz Sep 22 '20 at 18:29
  • 1
    Related: https://math.stackexchange.com/questions/18405/a-question-related-to-two-competing-patterns-in-coin-tossing/18503#18503 and https://math.stackexchange.com/questions/66670/probability-about-a-coin-games?noredirect=1&lq=1 – JMoravitz Sep 22 '20 at 18:32
  • I was researching into penney's game, but I wasn't able to find any information of Penney's game using a biased coin. I was looking to solve this using any method possible really, although I am not experienced with Markov chains, but willing to learn! – LegitFakeID Sep 22 '20 at 18:32
  • Please provide context for the question, showing in some details the own tries and the own progress makes potential answerers know what is clear, and how deeply they should go and type in detail. Having a good question deserves presenting it in a good shape. So please provide context. https://math.meta.stackexchange.com/questions/9959/how-to-ask-a-good-question – dan_fulea Sep 22 '20 at 18:44
  • Apologies, it is my first post on mathstackexchange, I will repost after making further attempts @dan_fulea – LegitFakeID Sep 22 '20 at 18:46
  • There is no problem, welcome here, i was trying only to avoid some many downvotes that would probably come, if there is no edit in sight. You have a good question! The question may become valuable for the community. Some smart answers may come, and even if i would have any, i may see a better one. So there is a point to attract attention, so that good solutions are shown, this is a benefit for the community. So as it is, the question is in great danger of being closed. Just provide a quick source, some references... – dan_fulea Sep 22 '20 at 18:50
  • I think the links @JMoravitz has provided discuss very similar problems at length with multiple solutions. You should be able to use it to show your attempt. – Math Lover Sep 22 '20 at 20:19

1 Answers1

3

Flip the coin, obtaining some number of $\color{blue} T$'s until you see the first $\color{blue} H$. Once this happens, it becomes possible for Player A or B to win in the next two flips. Then flip the coin twice, obtaining a sequence of two letters. If we see $\color{red}{HH}$, Player A wins. If we see $\color{red}{TH}$, Player B wins. If we see $\color{red}{TT}$, then neither player can win in the next flip, so we start this process over. If we see $\color{red}{HT}$, then flip once more. Player B wins if we see $\color{green}{H}$; otherwise, we start over. For example, $$ \color{blue}{TTTTTTH} \color{red}{TH} $$ results in Player B winning, but $$ \color{blue}{TTTTTTTH} \color{red}{HT} \color{green}{T}$$ results in us starting over. The probability that Player A wins in one of these rounds is the probability of seeing $\color{red}{HH}$, which is $p^2$. The probability that Player B wins is the probability of seeing $\color{red}{TH}$ or $\color{red}{HT} \color{green}{H}$, which is $(1-p)p + p(1-p)p$. Therefore, if we repeat this until some player wins, the probability Player A wins is $$ \frac{p^2}{p^2 + (1-p)p + p(1-p)p} .$$

nahp
  • 621
  • That gives you the probability for Player A winning, what value of p would make the game fair? ie Player A and B have the same chance of winning – LegitFakeID Sep 22 '20 at 20:52
  • @LegitFakeID For player A and B to have the same probability of winning, the probability of A winning must be $\frac{1}{2}$. Simply set that probability equal to $\frac12$ and solve for $p$. – Varun Vejalla Sep 22 '20 at 21:08