probability of encountering 5 heads
$begingroup$
find probability that during a process of tossing coins repeatedly one will encounter a run of 5 consecutive heads before encountering 2 tails.
i started by assuming,
H= probability that this event occurs when first toss was Heads
T= probability that this event occurs when first toss was Tails.
What to do next?
probability
asked Dec 14 '18 at 11:50
tanyatanya
105
$endgroup$
add a comment |
$begingroup$
find probability that during a process of tossing coins repeatedly one will encounter a run of 5 consecutive heads before encountering 2 tails.
i started by assuming,
H= probability that this event occurs when first toss was Heads
T= probability that this event occurs when first toss was Tails.
What to do next?
probability
asked Dec 14 '18 at 11:50
tanyatanya
105
$endgroup$
add a comment |
$begingroup$
find probability that during a process of tossing coins repeatedly one will encounter a run of 5 consecutive heads before encountering 2 tails.
i started by assuming,
H= probability that this event occurs when first toss was Heads
T= probability that this event occurs when first toss was Tails.
What to do next?
probability
asked Dec 14 '18 at 11:50
tanyatanya
105
$endgroup$
find probability that during a process of tossing coins repeatedly one will encounter a run of 5 consecutive heads before encountering 2 tails.
i started by assuming,
H= probability that this event occurs when first toss was Heads
T= probability that this event occurs when first toss was Tails.
What to do next?
probability
probability
asked Dec 14 '18 at 11:50
tanyatanya
105
asked Dec 14 '18 at 11:50
tanyatanya
105
asked Dec 14 '18 at 11:50
tanyatanya
105
asked Dec 14 '18 at 11:50
tanyatanya
105
asked Dec 14 '18 at 11:50
tanyatanya
105
105
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
At any point in time as you keep tossing the coin, you can be in 5 different states. Either your last throw was a T, or your last throws were a streak of 1, 2, 3 or 4 H. (You're in a sixth state just as you start the game, but we'll skip that one for now.)
In each state, you have some probability of winning, losing, or transitioning into one of the other 4 states. For each state, there is also a probability of eventually winning from that point on. I'll call those evental winning probabilities $P(T)$ and $P(1H), P(2H), P(3H), P(4H)$.
From here we can set up equations (using the law of total probability). For instance, if you're in the 4H state, then there is a 50% chance you'll win on the next throw, and a 50% chance you'll enter state T. Thus
$$
P(4H) = frac12 + frac12P(T)
$$
Similarily, if you're in state T, you have a 50% chance of losing, and a 50% chance of entering state 1H. Thus
$$
P(T) = frac12P(1H)
$$
You can similarily set up equations for $P(1H), P(2H)$ and $P(3H)$. You now have five equations and five unknowns, so this may be solved.
Once you've done that, you can look at the state you're in as you start the game, and see that the probability of winning from there is
$$
frac12P(T) + frac12P(1H)
$$
answered Dec 14 '18 at 12:00
ArthurArthur
112k7107191
$endgroup$
$begingroup$
can you solve for just one more state say P(2H) ? please.
$endgroup$
– tanya
Dec 14 '18 at 13:09
$begingroup$
okay wait. is this correct- P(2H) = 1/2 P(T) + 1/2 P(3H)
$endgroup$
– tanya
Dec 14 '18 at 13:11
$begingroup$
@tanya Exactly! probability $frac12$ of getting tails and transitioning to state $T$, $frac12$ of getting heads and transitioning to state $3H$.
$endgroup$
– Arthur
Dec 14 '18 at 13:31
$begingroup$
thank you so much.
$endgroup$
– tanya
Dec 14 '18 at 13:41
add a comment |
$begingroup$
To be discerned are the following states:
- 0) No tosses are made yet.
- 1) $H$
- 2) $HH$
- 3) $HHH$
- 4) $HHHH$
- 5) $T$
Here e.g. $HH$ stands for the state that the last two tosses were
heads and before that there was a tail or there were no tosses at
all.
For $i=0,1,2,3,4,5$ let $p_{i}$ denote the probability on the
event that is described in your question under condition that we start
in state $i$.
Then to be found is $p_{0}$ and we have the following equalities:
$p_{i}=frac{1}{2}p_{i+1}+frac{1}{2}p_{5}$ for $i=0,1,2,3$
- $p_{4}=frac{1}{2}+frac{1}{2}p_{5}$
- $p_{5}=frac{1}{2}p_{1}$
These equalities enable you to find $p_0$ and the result will be:$$p_0=frac3{34}$$
answered Dec 14 '18 at 12:16
drhabdrhab
98.7k544129
$endgroup$
add a comment |
Your Answer
StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");
StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "69"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});
function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});
}
});
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3039267%2fprobability-of-encountering-5-heads%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
At any point in time as you keep tossing the coin, you can be in 5 different states. Either your last throw was a T, or your last throws were a streak of 1, 2, 3 or 4 H. (You're in a sixth state just as you start the game, but we'll skip that one for now.)
In each state, you have some probability of winning, losing, or transitioning into one of the other 4 states. For each state, there is also a probability of eventually winning from that point on. I'll call those evental winning probabilities $P(T)$ and $P(1H), P(2H), P(3H), P(4H)$.
From here we can set up equations (using the law of total probability). For instance, if you're in the 4H state, then there is a 50% chance you'll win on the next throw, and a 50% chance you'll enter state T. Thus
$$
P(4H) = frac12 + frac12P(T)
$$
Similarily, if you're in state T, you have a 50% chance of losing, and a 50% chance of entering state 1H. Thus
$$
P(T) = frac12P(1H)
$$
You can similarily set up equations for $P(1H), P(2H)$ and $P(3H)$. You now have five equations and five unknowns, so this may be solved.
Once you've done that, you can look at the state you're in as you start the game, and see that the probability of winning from there is
$$
frac12P(T) + frac12P(1H)
$$
answered Dec 14 '18 at 12:00
ArthurArthur
112k7107191
$endgroup$
$begingroup$
can you solve for just one more state say P(2H) ? please.
$endgroup$
– tanya
Dec 14 '18 at 13:09
$begingroup$
okay wait. is this correct- P(2H) = 1/2 P(T) + 1/2 P(3H)
$endgroup$
– tanya
Dec 14 '18 at 13:11
$begingroup$
@tanya Exactly! probability $frac12$ of getting tails and transitioning to state $T$, $frac12$ of getting heads and transitioning to state $3H$.
$endgroup$
– Arthur
Dec 14 '18 at 13:31
$begingroup$
thank you so much.
$endgroup$
– tanya
Dec 14 '18 at 13:41
add a comment |
$begingroup$
At any point in time as you keep tossing the coin, you can be in 5 different states. Either your last throw was a T, or your last throws were a streak of 1, 2, 3 or 4 H. (You're in a sixth state just as you start the game, but we'll skip that one for now.)
In each state, you have some probability of winning, losing, or transitioning into one of the other 4 states. For each state, there is also a probability of eventually winning from that point on. I'll call those evental winning probabilities $P(T)$ and $P(1H), P(2H), P(3H), P(4H)$.
From here we can set up equations (using the law of total probability). For instance, if you're in the 4H state, then there is a 50% chance you'll win on the next throw, and a 50% chance you'll enter state T. Thus
$$
P(4H) = frac12 + frac12P(T)
$$
Similarily, if you're in state T, you have a 50% chance of losing, and a 50% chance of entering state 1H. Thus
$$
P(T) = frac12P(1H)
$$
You can similarily set up equations for $P(1H), P(2H)$ and $P(3H)$. You now have five equations and five unknowns, so this may be solved.
Once you've done that, you can look at the state you're in as you start the game, and see that the probability of winning from there is
$$
frac12P(T) + frac12P(1H)
$$
answered Dec 14 '18 at 12:00
ArthurArthur
112k7107191
$endgroup$
$begingroup$
can you solve for just one more state say P(2H) ? please.
$endgroup$
– tanya
Dec 14 '18 at 13:09
$begingroup$
okay wait. is this correct- P(2H) = 1/2 P(T) + 1/2 P(3H)
$endgroup$
– tanya
Dec 14 '18 at 13:11
$begingroup$
@tanya Exactly! probability $frac12$ of getting tails and transitioning to state $T$, $frac12$ of getting heads and transitioning to state $3H$.
$endgroup$
– Arthur
Dec 14 '18 at 13:31
$begingroup$
thank you so much.
$endgroup$
– tanya
Dec 14 '18 at 13:41
add a comment |
$begingroup$
At any point in time as you keep tossing the coin, you can be in 5 different states. Either your last throw was a T, or your last throws were a streak of 1, 2, 3 or 4 H. (You're in a sixth state just as you start the game, but we'll skip that one for now.)
In each state, you have some probability of winning, losing, or transitioning into one of the other 4 states. For each state, there is also a probability of eventually winning from that point on. I'll call those evental winning probabilities $P(T)$ and $P(1H), P(2H), P(3H), P(4H)$.
From here we can set up equations (using the law of total probability). For instance, if you're in the 4H state, then there is a 50% chance you'll win on the next throw, and a 50% chance you'll enter state T. Thus
$$
P(4H) = frac12 + frac12P(T)
$$
Similarily, if you're in state T, you have a 50% chance of losing, and a 50% chance of entering state 1H. Thus
$$
P(T) = frac12P(1H)
$$
You can similarily set up equations for $P(1H), P(2H)$ and $P(3H)$. You now have five equations and five unknowns, so this may be solved.
Once you've done that, you can look at the state you're in as you start the game, and see that the probability of winning from there is
$$
frac12P(T) + frac12P(1H)
$$
answered Dec 14 '18 at 12:00
ArthurArthur
112k7107191
$endgroup$
At any point in time as you keep tossing the coin, you can be in 5 different states. Either your last throw was a T, or your last throws were a streak of 1, 2, 3 or 4 H. (You're in a sixth state just as you start the game, but we'll skip that one for now.)
In each state, you have some probability of winning, losing, or transitioning into one of the other 4 states. For each state, there is also a probability of eventually winning from that point on. I'll call those evental winning probabilities $P(T)$ and $P(1H), P(2H), P(3H), P(4H)$.
From here we can set up equations (using the law of total probability). For instance, if you're in the 4H state, then there is a 50% chance you'll win on the next throw, and a 50% chance you'll enter state T. Thus
$$
P(4H) = frac12 + frac12P(T)
$$
Similarily, if you're in state T, you have a 50% chance of losing, and a 50% chance of entering state 1H. Thus
$$
P(T) = frac12P(1H)
$$
You can similarily set up equations for $P(1H), P(2H)$ and $P(3H)$. You now have five equations and five unknowns, so this may be solved.
Once you've done that, you can look at the state you're in as you start the game, and see that the probability of winning from there is
$$
frac12P(T) + frac12P(1H)
$$
answered Dec 14 '18 at 12:00
ArthurArthur
112k7107191
answered Dec 14 '18 at 12:00
ArthurArthur
112k7107191
answered Dec 14 '18 at 12:00
ArthurArthur
112k7107191
answered Dec 14 '18 at 12:00
ArthurArthur
112k7107191
112k7107191
$begingroup$
can you solve for just one more state say P(2H) ? please.
$endgroup$
– tanya
Dec 14 '18 at 13:09
$begingroup$
okay wait. is this correct- P(2H) = 1/2 P(T) + 1/2 P(3H)
$endgroup$
– tanya
Dec 14 '18 at 13:11
$begingroup$
@tanya Exactly! probability $frac12$ of getting tails and transitioning to state $T$, $frac12$ of getting heads and transitioning to state $3H$.
$endgroup$
– Arthur
Dec 14 '18 at 13:31
$begingroup$
thank you so much.
$endgroup$
– tanya
Dec 14 '18 at 13:41
add a comment |
$begingroup$
can you solve for just one more state say P(2H) ? please.
$endgroup$
– tanya
Dec 14 '18 at 13:09
$begingroup$
okay wait. is this correct- P(2H) = 1/2 P(T) + 1/2 P(3H)
$endgroup$
– tanya
Dec 14 '18 at 13:11
$begingroup$
@tanya Exactly! probability $frac12$ of getting tails and transitioning to state $T$, $frac12$ of getting heads and transitioning to state $3H$.
$endgroup$
– Arthur
Dec 14 '18 at 13:31
$begingroup$
thank you so much.
$endgroup$
– tanya
Dec 14 '18 at 13:41
$begingroup$
can you solve for just one more state say P(2H) ? please.
$endgroup$
– tanya
Dec 14 '18 at 13:09
$begingroup$
can you solve for just one more state say P(2H) ? please.
$endgroup$
– tanya
Dec 14 '18 at 13:09
$begingroup$
okay wait. is this correct- P(2H) = 1/2 P(T) + 1/2 P(3H)
$endgroup$
– tanya
Dec 14 '18 at 13:11
$begingroup$
okay wait. is this correct- P(2H) = 1/2 P(T) + 1/2 P(3H)
$endgroup$
– tanya
Dec 14 '18 at 13:11
$begingroup$
@tanya Exactly! probability $frac12$ of getting tails and transitioning to state $T$, $frac12$ of getting heads and transitioning to state $3H$.
$endgroup$
– Arthur
Dec 14 '18 at 13:31
$begingroup$
@tanya Exactly! probability $frac12$ of getting tails and transitioning to state $T$, $frac12$ of getting heads and transitioning to state $3H$.
$endgroup$
– Arthur
Dec 14 '18 at 13:31
$begingroup$
thank you so much.
$endgroup$
– tanya
Dec 14 '18 at 13:41
$begingroup$
thank you so much.
$endgroup$
– tanya
Dec 14 '18 at 13:41
add a comment |
$begingroup$
To be discerned are the following states:
- 0) No tosses are made yet.
- 1) $H$
- 2) $HH$
- 3) $HHH$
- 4) $HHHH$
- 5) $T$
Here e.g. $HH$ stands for the state that the last two tosses were
heads and before that there was a tail or there were no tosses at
all.
For $i=0,1,2,3,4,5$ let $p_{i}$ denote the probability on the
event that is described in your question under condition that we start
in state $i$.
Then to be found is $p_{0}$ and we have the following equalities:
$p_{i}=frac{1}{2}p_{i+1}+frac{1}{2}p_{5}$ for $i=0,1,2,3$
- $p_{4}=frac{1}{2}+frac{1}{2}p_{5}$
- $p_{5}=frac{1}{2}p_{1}$
These equalities enable you to find $p_0$ and the result will be:$$p_0=frac3{34}$$
answered Dec 14 '18 at 12:16
drhabdrhab
98.7k544129
$endgroup$
add a comment |
$begingroup$
To be discerned are the following states:
- 0) No tosses are made yet.
- 1) $H$
- 2) $HH$
- 3) $HHH$
- 4) $HHHH$
- 5) $T$
Here e.g. $HH$ stands for the state that the last two tosses were
heads and before that there was a tail or there were no tosses at
all.
For $i=0,1,2,3,4,5$ let $p_{i}$ denote the probability on the
event that is described in your question under condition that we start
in state $i$.
Then to be found is $p_{0}$ and we have the following equalities:
$p_{i}=frac{1}{2}p_{i+1}+frac{1}{2}p_{5}$ for $i=0,1,2,3$
- $p_{4}=frac{1}{2}+frac{1}{2}p_{5}$
- $p_{5}=frac{1}{2}p_{1}$
These equalities enable you to find $p_0$ and the result will be:$$p_0=frac3{34}$$
answered Dec 14 '18 at 12:16
drhabdrhab
98.7k544129
$endgroup$
add a comment |
$begingroup$
To be discerned are the following states:
- 0) No tosses are made yet.
- 1) $H$
- 2) $HH$
- 3) $HHH$
- 4) $HHHH$
- 5) $T$
Here e.g. $HH$ stands for the state that the last two tosses were
heads and before that there was a tail or there were no tosses at
all.
For $i=0,1,2,3,4,5$ let $p_{i}$ denote the probability on the
event that is described in your question under condition that we start
in state $i$.
Then to be found is $p_{0}$ and we have the following equalities:
$p_{i}=frac{1}{2}p_{i+1}+frac{1}{2}p_{5}$ for $i=0,1,2,3$
- $p_{4}=frac{1}{2}+frac{1}{2}p_{5}$
- $p_{5}=frac{1}{2}p_{1}$
These equalities enable you to find $p_0$ and the result will be:$$p_0=frac3{34}$$
answered Dec 14 '18 at 12:16
drhabdrhab
98.7k544129
$endgroup$
To be discerned are the following states:
- 0) No tosses are made yet.
- 1) $H$
- 2) $HH$
- 3) $HHH$
- 4) $HHHH$
- 5) $T$
Here e.g. $HH$ stands for the state that the last two tosses were
heads and before that there was a tail or there were no tosses at
all.
For $i=0,1,2,3,4,5$ let $p_{i}$ denote the probability on the
event that is described in your question under condition that we start
in state $i$.
Then to be found is $p_{0}$ and we have the following equalities:
$p_{i}=frac{1}{2}p_{i+1}+frac{1}{2}p_{5}$ for $i=0,1,2,3$
- $p_{4}=frac{1}{2}+frac{1}{2}p_{5}$
- $p_{5}=frac{1}{2}p_{1}$
These equalities enable you to find $p_0$ and the result will be:$$p_0=frac3{34}$$
answered Dec 14 '18 at 12:16
drhabdrhab
98.7k544129
edited Dec 14 '18 at 12:35
answered Dec 14 '18 at 12:16
drhabdrhab
98.7k544129
answered Dec 14 '18 at 12:16
drhabdrhab
98.7k544129
answered Dec 14 '18 at 12:16
drhabdrhab
98.7k544129
98.7k544129
add a comment |
add a comment |
Thanks for contributing an answer to Mathematics Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3039267%2fprobability-of-encountering-5-heads%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
