Mixing
Game - First person to throw consecutive heads
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Player A and Player B play are playing a game. Player A goes first, and tosses a coin. Player B then tosses his first coin. Player A then tosses his second and so on until the game ends. The first person to toss consecutive heads wins. What is the probability that Player A wins?
For those curious, this question was asked in a finance interview.
probability
asked 21 hours ago
Connolly Devin
1
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Connolly Devin is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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up vote
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down vote
favorite
Player A and Player B play are playing a game. Player A goes first, and tosses a coin. Player B then tosses his first coin. Player A then tosses his second and so on until the game ends. The first person to toss consecutive heads wins. What is the probability that Player A wins?
For those curious, this question was asked in a finance interview.
probability
asked 21 hours ago
Connolly Devin
1
New contributor
Connolly Devin is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
What are your thoughts?
– lulu
21 hours ago
Originally got the answer 4/7 through P(Win) = P(HH) + P(Not HH)P(Not HH)P(HH) and so on, and using sum of an infinite series. Then realised that the outcomes TT,HT and TH are not equivalent, as TH then needs just one more head to win.
– Connolly Devin
21 hours ago
2
This sort of problem is hard to attack with geometric series, for the reason you point out. Best to do it with states. Label a state according to; whose turn it is, how many more Heads $A$ needs and how many more Heads $B$ needs. There aren't very many states. Then look at the transitions between them.
– lulu
20 hours ago
I was looking at the Markov chain method (have some exposure to Stochastic modelling and the likes), but was hoping there may be a trick to it - especially given it was asked as 1 of 10 questions during a short interview.
– Connolly Devin
20 hours ago
Having both conducted (and taken) lots of such interviews, it's often the case that the interviewer is really after a robust and reasoned analysis of the problem. Yes, carrying through the steps might take a while but typically the interviewer will stop you at the outline stage.
– lulu
20 hours ago
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
Player A and Player B play are playing a game. Player A goes first, and tosses a coin. Player B then tosses his first coin. Player A then tosses his second and so on until the game ends. The first person to toss consecutive heads wins. What is the probability that Player A wins?
For those curious, this question was asked in a finance interview.
probability
asked 21 hours ago
Connolly Devin
1
New contributor
Connolly Devin is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Player A and Player B play are playing a game. Player A goes first, and tosses a coin. Player B then tosses his first coin. Player A then tosses his second and so on until the game ends. The first person to toss consecutive heads wins. What is the probability that Player A wins?
For those curious, this question was asked in a finance interview.
probability
probability
asked 21 hours ago
Connolly Devin
1
New contributor
Connolly Devin is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 21 hours ago
Connolly Devin
1
New contributor
Connolly Devin is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 21 hours ago
Connolly Devin
1
New contributor
Connolly Devin is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 21 hours ago
Connolly Devin
1
asked 21 hours ago
Connolly Devin
1
1
New contributor
Connolly Devin is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
Connolly Devin is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Connolly Devin is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
What are your thoughts?
– lulu
21 hours ago
Originally got the answer 4/7 through P(Win) = P(HH) + P(Not HH)P(Not HH)P(HH) and so on, and using sum of an infinite series. Then realised that the outcomes TT,HT and TH are not equivalent, as TH then needs just one more head to win.
– Connolly Devin
21 hours ago
2
This sort of problem is hard to attack with geometric series, for the reason you point out. Best to do it with states. Label a state according to; whose turn it is, how many more Heads $A$ needs and how many more Heads $B$ needs. There aren't very many states. Then look at the transitions between them.
– lulu
20 hours ago
I was looking at the Markov chain method (have some exposure to Stochastic modelling and the likes), but was hoping there may be a trick to it - especially given it was asked as 1 of 10 questions during a short interview.
– Connolly Devin
20 hours ago
Having both conducted (and taken) lots of such interviews, it's often the case that the interviewer is really after a robust and reasoned analysis of the problem. Yes, carrying through the steps might take a while but typically the interviewer will stop you at the outline stage.
– lulu
20 hours ago
add a comment |
What are your thoughts?
– lulu
21 hours ago
Originally got the answer 4/7 through P(Win) = P(HH) + P(Not HH)P(Not HH)P(HH) and so on, and using sum of an infinite series. Then realised that the outcomes TT,HT and TH are not equivalent, as TH then needs just one more head to win.
– Connolly Devin
21 hours ago
2
This sort of problem is hard to attack with geometric series, for the reason you point out. Best to do it with states. Label a state according to; whose turn it is, how many more Heads $A$ needs and how many more Heads $B$ needs. There aren't very many states. Then look at the transitions between them.
– lulu
20 hours ago
I was looking at the Markov chain method (have some exposure to Stochastic modelling and the likes), but was hoping there may be a trick to it - especially given it was asked as 1 of 10 questions during a short interview.
– Connolly Devin
20 hours ago
Having both conducted (and taken) lots of such interviews, it's often the case that the interviewer is really after a robust and reasoned analysis of the problem. Yes, carrying through the steps might take a while but typically the interviewer will stop you at the outline stage.
– lulu
20 hours ago
What are your thoughts?
– lulu
21 hours ago
What are your thoughts?
– lulu
21 hours ago
Originally got the answer 4/7 through P(Win) = P(HH) + P(Not HH)P(Not HH)P(HH) and so on, and using sum of an infinite series. Then realised that the outcomes TT,HT and TH are not equivalent, as TH then needs just one more head to win.
– Connolly Devin
21 hours ago
Originally got the answer 4/7 through P(Win) = P(HH) + P(Not HH)P(Not HH)P(HH) and so on, and using sum of an infinite series. Then realised that the outcomes TT,HT and TH are not equivalent, as TH then needs just one more head to win.
– Connolly Devin
21 hours ago
2
2
This sort of problem is hard to attack with geometric series, for the reason you point out. Best to do it with states. Label a state according to; whose turn it is, how many more Heads $A$ needs and how many more Heads $B$ needs. There aren't very many states. Then look at the transitions between them.
– lulu
20 hours ago
This sort of problem is hard to attack with geometric series, for the reason you point out. Best to do it with states. Label a state according to; whose turn it is, how many more Heads $A$ needs and how many more Heads $B$ needs. There aren't very many states. Then look at the transitions between them.
– lulu
20 hours ago
I was looking at the Markov chain method (have some exposure to Stochastic modelling and the likes), but was hoping there may be a trick to it - especially given it was asked as 1 of 10 questions during a short interview.
– Connolly Devin
20 hours ago
I was looking at the Markov chain method (have some exposure to Stochastic modelling and the likes), but was hoping there may be a trick to it - especially given it was asked as 1 of 10 questions during a short interview.
– Connolly Devin
20 hours ago
Having both conducted (and taken) lots of such interviews, it's often the case that the interviewer is really after a robust and reasoned analysis of the problem. Yes, carrying through the steps might take a while but typically the interviewer will stop you at the outline stage.
– lulu
20 hours ago
Having both conducted (and taken) lots of such interviews, it's often the case that the interviewer is really after a robust and reasoned analysis of the problem. Yes, carrying through the steps might take a while but typically the interviewer will stop you at the outline stage.
– lulu
20 hours ago
add a comment |
1 Answer
1
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oldest
votes
up vote
0
down vote
Denote by ${tt pxy}$ the probability that $A$ finally wins when the game is in the following state:
- person $Pin{A,B}$ is the next to throw his coin,
$A$ has $xin{0,1}$ useful heads on the stack,
$B$ has $yin{0,1}$ useful heads on the stack.
The problem is to determine ${tt a00}$.
There are $8$ nonterminal states, leading to a system of $8$ linear equations. I let Mathematica solve it. The answer is ${tt a00}={14over25}$. Here is the output:
answered 19 hours ago
Christian Blatter
170k7111323
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
Denote by ${tt pxy}$ the probability that $A$ finally wins when the game is in the following state:
- person $Pin{A,B}$ is the next to throw his coin,
$A$ has $xin{0,1}$ useful heads on the stack,
$B$ has $yin{0,1}$ useful heads on the stack.
The problem is to determine ${tt a00}$.
There are $8$ nonterminal states, leading to a system of $8$ linear equations. I let Mathematica solve it. The answer is ${tt a00}={14over25}$. Here is the output:
answered 19 hours ago
Christian Blatter
170k7111323
add a comment |
up vote
0
down vote
Denote by ${tt pxy}$ the probability that $A$ finally wins when the game is in the following state:
- person $Pin{A,B}$ is the next to throw his coin,
$A$ has $xin{0,1}$ useful heads on the stack,
$B$ has $yin{0,1}$ useful heads on the stack.
The problem is to determine ${tt a00}$.
There are $8$ nonterminal states, leading to a system of $8$ linear equations. I let Mathematica solve it. The answer is ${tt a00}={14over25}$. Here is the output:
answered 19 hours ago
Christian Blatter
170k7111323
add a comment |
up vote
0
down vote
up vote
0
down vote
Denote by ${tt pxy}$ the probability that $A$ finally wins when the game is in the following state:
- person $Pin{A,B}$ is the next to throw his coin,
$A$ has $xin{0,1}$ useful heads on the stack,
$B$ has $yin{0,1}$ useful heads on the stack.
The problem is to determine ${tt a00}$.
There are $8$ nonterminal states, leading to a system of $8$ linear equations. I let Mathematica solve it. The answer is ${tt a00}={14over25}$. Here is the output:
answered 19 hours ago
Christian Blatter
170k7111323
Denote by ${tt pxy}$ the probability that $A$ finally wins when the game is in the following state:
- person $Pin{A,B}$ is the next to throw his coin,
$A$ has $xin{0,1}$ useful heads on the stack,
$B$ has $yin{0,1}$ useful heads on the stack.
The problem is to determine ${tt a00}$.
There are $8$ nonterminal states, leading to a system of $8$ linear equations. I let Mathematica solve it. The answer is ${tt a00}={14over25}$. Here is the output:
answered 19 hours ago
Christian Blatter
170k7111323
edited 2 hours ago
answered 19 hours ago
Christian Blatter
170k7111323
answered 19 hours ago
Christian Blatter
170k7111323
answered 19 hours ago
Christian Blatter
170k7111323
170k7111323
add a comment |
add a comment |
Connolly Devin is a new contributor. Be nice, and check out our Code of Conduct.
Connolly Devin is a new contributor. Be nice, and check out our Code of Conduct.
Connolly Devin is a new contributor. Be nice, and check out our Code of Conduct.
Connolly Devin is a new contributor. Be nice, and check out our Code of Conduct.
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What are your thoughts?
– lulu
21 hours ago
Originally got the answer 4/7 through P(Win) = P(HH) + P(Not HH)P(Not HH)P(HH) and so on, and using sum of an infinite series. Then realised that the outcomes TT,HT and TH are not equivalent, as TH then needs just one more head to win.
– Connolly Devin
21 hours ago
2
This sort of problem is hard to attack with geometric series, for the reason you point out. Best to do it with states. Label a state according to; whose turn it is, how many more Heads $A$ needs and how many more Heads $B$ needs. There aren't very many states. Then look at the transitions between them.
– lulu
20 hours ago
I was looking at the Markov chain method (have some exposure to Stochastic modelling and the likes), but was hoping there may be a trick to it - especially given it was asked as 1 of 10 questions during a short interview.
– Connolly Devin
20 hours ago
Having both conducted (and taken) lots of such interviews, it's often the case that the interviewer is really after a robust and reasoned analysis of the problem. Yes, carrying through the steps might take a while but typically the interviewer will stop you at the outline stage.
– lulu
20 hours ago