Estimating Probability for first appearance of HT before HH and HH before HT using Markov chains Announcing the arrival of Valued Associate #679: Cesar Manara Planned maintenance scheduled April 23, 2019 at 00:00UTC (8:00pm US/Eastern)Probability with Markov chainsEstimate the probability using Markov chainsSupply the transition matrix for these (possible) Markov chainsProbability of a fair sequence of tosses ending on two successive tails given the first toss was a head?Probability of a coin tossed using Markov ChainsFinding chance of winning $A$ is in a alternate throw a fair coin.Probability chains similar to Markov chains but being functions of probability vector.Calculating probability using Markov chainMarkov Chain and ExpectationIntuition behind positive recurrent and null recurrent Markov Chains

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Estimating Probability for first appearance of HT before HH and HH before HT using Markov chains



Announcing the arrival of Valued Associate #679: Cesar Manara
Planned maintenance scheduled April 23, 2019 at 00:00UTC (8:00pm US/Eastern)Probability with Markov chainsEstimate the probability using Markov chainsSupply the transition matrix for these (possible) Markov chainsProbability of a fair sequence of tosses ending on two successive tails given the first toss was a head?Probability of a coin tossed using Markov ChainsFinding chance of winning $A$ is in a alternate throw a fair coin.Probability chains similar to Markov chains but being functions of probability vector.Calculating probability using Markov chainMarkov Chain and ExpectationIntuition behind positive recurrent and null recurrent Markov Chains










0












$begingroup$


Consider the case where a coin is tossed repeatedly, I win when HH appears for the first time and you win when HT appears for the first time.



I know that one of us will win when the first Head appears so we have a 50-50 chance of winning.



But I recently studied about Markov chain and wanted to know whether it will be applicable here and if it is, I want to know how you can use it calculate the probability who has more probability to win.

I'm new to Markov chains and please help me solve this question










share|cite|improve this question









$endgroup$











  • $begingroup$
    The approach you sketch is a Markov chain approach. You are ignoring all the $T's$ you throw before getting that first $H$. That's because the initial Markov state goes back to itself every time you throw a $T$. Once you've thrown an $H$ you are in a new state. Now, whatever happens, the game ends on the next throw.
    $endgroup$
    – lulu
    Mar 27 at 16:59










  • $begingroup$
    To add to @lulu’s comment, you’ll end up with a Markov chain with two absorbing states. The probabilities that you’re looking for are the corresponding absorption probabilities.
    $endgroup$
    – amd
    Mar 27 at 18:40
















0












$begingroup$


Consider the case where a coin is tossed repeatedly, I win when HH appears for the first time and you win when HT appears for the first time.



I know that one of us will win when the first Head appears so we have a 50-50 chance of winning.



But I recently studied about Markov chain and wanted to know whether it will be applicable here and if it is, I want to know how you can use it calculate the probability who has more probability to win.

I'm new to Markov chains and please help me solve this question










share|cite|improve this question









$endgroup$











  • $begingroup$
    The approach you sketch is a Markov chain approach. You are ignoring all the $T's$ you throw before getting that first $H$. That's because the initial Markov state goes back to itself every time you throw a $T$. Once you've thrown an $H$ you are in a new state. Now, whatever happens, the game ends on the next throw.
    $endgroup$
    – lulu
    Mar 27 at 16:59










  • $begingroup$
    To add to @lulu’s comment, you’ll end up with a Markov chain with two absorbing states. The probabilities that you’re looking for are the corresponding absorption probabilities.
    $endgroup$
    – amd
    Mar 27 at 18:40














0












0








0





$begingroup$


Consider the case where a coin is tossed repeatedly, I win when HH appears for the first time and you win when HT appears for the first time.



I know that one of us will win when the first Head appears so we have a 50-50 chance of winning.



But I recently studied about Markov chain and wanted to know whether it will be applicable here and if it is, I want to know how you can use it calculate the probability who has more probability to win.

I'm new to Markov chains and please help me solve this question










share|cite|improve this question









$endgroup$




Consider the case where a coin is tossed repeatedly, I win when HH appears for the first time and you win when HT appears for the first time.



I know that one of us will win when the first Head appears so we have a 50-50 chance of winning.



But I recently studied about Markov chain and wanted to know whether it will be applicable here and if it is, I want to know how you can use it calculate the probability who has more probability to win.

I'm new to Markov chains and please help me solve this question







probability markov-chains






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Mar 27 at 16:51









8056780567

64




64











  • $begingroup$
    The approach you sketch is a Markov chain approach. You are ignoring all the $T's$ you throw before getting that first $H$. That's because the initial Markov state goes back to itself every time you throw a $T$. Once you've thrown an $H$ you are in a new state. Now, whatever happens, the game ends on the next throw.
    $endgroup$
    – lulu
    Mar 27 at 16:59










  • $begingroup$
    To add to @lulu’s comment, you’ll end up with a Markov chain with two absorbing states. The probabilities that you’re looking for are the corresponding absorption probabilities.
    $endgroup$
    – amd
    Mar 27 at 18:40

















  • $begingroup$
    The approach you sketch is a Markov chain approach. You are ignoring all the $T's$ you throw before getting that first $H$. That's because the initial Markov state goes back to itself every time you throw a $T$. Once you've thrown an $H$ you are in a new state. Now, whatever happens, the game ends on the next throw.
    $endgroup$
    – lulu
    Mar 27 at 16:59










  • $begingroup$
    To add to @lulu’s comment, you’ll end up with a Markov chain with two absorbing states. The probabilities that you’re looking for are the corresponding absorption probabilities.
    $endgroup$
    – amd
    Mar 27 at 18:40
















$begingroup$
The approach you sketch is a Markov chain approach. You are ignoring all the $T's$ you throw before getting that first $H$. That's because the initial Markov state goes back to itself every time you throw a $T$. Once you've thrown an $H$ you are in a new state. Now, whatever happens, the game ends on the next throw.
$endgroup$
– lulu
Mar 27 at 16:59




$begingroup$
The approach you sketch is a Markov chain approach. You are ignoring all the $T's$ you throw before getting that first $H$. That's because the initial Markov state goes back to itself every time you throw a $T$. Once you've thrown an $H$ you are in a new state. Now, whatever happens, the game ends on the next throw.
$endgroup$
– lulu
Mar 27 at 16:59












$begingroup$
To add to @lulu’s comment, you’ll end up with a Markov chain with two absorbing states. The probabilities that you’re looking for are the corresponding absorption probabilities.
$endgroup$
– amd
Mar 27 at 18:40





$begingroup$
To add to @lulu’s comment, you’ll end up with a Markov chain with two absorbing states. The probabilities that you’re looking for are the corresponding absorption probabilities.
$endgroup$
– amd
Mar 27 at 18:40











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