Find the probablity in the square and ball problem. [closed]
$begingroup$
A ball is thrown randomly from each vertex of a square and reaches the adjacent vertex. At each time the probably that the ball goes clockwise is $P$.
If we start at point A and we keep throwing until we land back at point A, what is the probability that we return to A from the opposite vertex we started at?
Lets assume that the initial throw is not counterclockwise.
In fact, I have got the answer of this question which is as follows.
However I don't understand why does these terms with 2 appear in the solutions. Please clarify this bit.
Note: P1 means that it never takes a counterclockwise movement. P2 means the ball goes only one time in the counterclockwise direction and etc.
$P_{1} = P^{4}$
$P_{2} = P^{5}. (1-P).2$
$ P_{3} = P^{6}. (1-P)^{2}.2^{2}$
$vdots $
$P_{n} = P^{n+3}. (1-P)^{n-1}.2^{n-1}$
probability
asked Dec 29 '18 at 21:06
ElectricmanElectricman
18012
$endgroup$
closed as unclear what you're asking by Eevee Trainer, KReiser, Leucippus, Cesareo, Brian Borchers Dec 30 '18 at 2:54
Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.
add a comment |
$begingroup$
A ball is thrown randomly from each vertex of a square and reaches the adjacent vertex. At each time the probably that the ball goes clockwise is $P$.
If we start at point A and we keep throwing until we land back at point A, what is the probability that we return to A from the opposite vertex we started at?
Lets assume that the initial throw is not counterclockwise.
In fact, I have got the answer of this question which is as follows.
However I don't understand why does these terms with 2 appear in the solutions. Please clarify this bit.
Note: P1 means that it never takes a counterclockwise movement. P2 means the ball goes only one time in the counterclockwise direction and etc.
$P_{1} = P^{4}$
$P_{2} = P^{5}. (1-P).2$
$ P_{3} = P^{6}. (1-P)^{2}.2^{2}$
$vdots $
$P_{n} = P^{n+3}. (1-P)^{n-1}.2^{n-1}$
probability
asked Dec 29 '18 at 21:06
ElectricmanElectricman
18012
$endgroup$
closed as unclear what you're asking by Eevee Trainer, KReiser, Leucippus, Cesareo, Brian Borchers Dec 30 '18 at 2:54
Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.
3
$begingroup$
Please explain the reason of down voting, so I modify my question!
$endgroup$
– Electricman
Dec 29 '18 at 21:13
1
$begingroup$
You have a serious typo. "vertex" not "vortex". (I didn't down vote!)
$endgroup$
– herb steinberg
Dec 29 '18 at 22:47
$begingroup$
@Electricman your question was down-voted, because it is unclear what exactly you are asking. We need you to we need you to specify your points of confusion so that we know how to answer your question.
$endgroup$
– clathratus
Dec 30 '18 at 1:05
add a comment |
$begingroup$
A ball is thrown randomly from each vertex of a square and reaches the adjacent vertex. At each time the probably that the ball goes clockwise is $P$.
If we start at point A and we keep throwing until we land back at point A, what is the probability that we return to A from the opposite vertex we started at?
Lets assume that the initial throw is not counterclockwise.
In fact, I have got the answer of this question which is as follows.
However I don't understand why does these terms with 2 appear in the solutions. Please clarify this bit.
Note: P1 means that it never takes a counterclockwise movement. P2 means the ball goes only one time in the counterclockwise direction and etc.
$P_{1} = P^{4}$
$P_{2} = P^{5}. (1-P).2$
$ P_{3} = P^{6}. (1-P)^{2}.2^{2}$
$vdots $
$P_{n} = P^{n+3}. (1-P)^{n-1}.2^{n-1}$
probability
asked Dec 29 '18 at 21:06
ElectricmanElectricman
18012
$endgroup$
A ball is thrown randomly from each vertex of a square and reaches the adjacent vertex. At each time the probably that the ball goes clockwise is $P$.
If we start at point A and we keep throwing until we land back at point A, what is the probability that we return to A from the opposite vertex we started at?
Lets assume that the initial throw is not counterclockwise.
In fact, I have got the answer of this question which is as follows.
However I don't understand why does these terms with 2 appear in the solutions. Please clarify this bit.
Note: P1 means that it never takes a counterclockwise movement. P2 means the ball goes only one time in the counterclockwise direction and etc.
$P_{1} = P^{4}$
$P_{2} = P^{5}. (1-P).2$
$ P_{3} = P^{6}. (1-P)^{2}.2^{2}$
$vdots $
$P_{n} = P^{n+3}. (1-P)^{n-1}.2^{n-1}$
probability
probability
asked Dec 29 '18 at 21:06
ElectricmanElectricman
18012
asked Dec 29 '18 at 21:06
ElectricmanElectricman
18012
edited Dec 30 '18 at 14:05
Electricman
asked Dec 29 '18 at 21:06
ElectricmanElectricman
18012
asked Dec 29 '18 at 21:06
ElectricmanElectricman
18012
asked Dec 29 '18 at 21:06
ElectricmanElectricman
18012
18012
closed as unclear what you're asking by Eevee Trainer, KReiser, Leucippus, Cesareo, Brian Borchers Dec 30 '18 at 2:54
Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.
closed as unclear what you're asking by Eevee Trainer, KReiser, Leucippus, Cesareo, Brian Borchers Dec 30 '18 at 2:54
Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.
3
$begingroup$
Please explain the reason of down voting, so I modify my question!
$endgroup$
– Electricman
Dec 29 '18 at 21:13
1
$begingroup$
You have a serious typo. "vertex" not "vortex". (I didn't down vote!)
$endgroup$
– herb steinberg
Dec 29 '18 at 22:47
$begingroup$
@Electricman your question was down-voted, because it is unclear what exactly you are asking. We need you to we need you to specify your points of confusion so that we know how to answer your question.
$endgroup$
– clathratus
Dec 30 '18 at 1:05
add a comment |
3
$begingroup$
Please explain the reason of down voting, so I modify my question!
$endgroup$
– Electricman
Dec 29 '18 at 21:13
1
$begingroup$
You have a serious typo. "vertex" not "vortex". (I didn't down vote!)
$endgroup$
– herb steinberg
Dec 29 '18 at 22:47
$begingroup$
@Electricman your question was down-voted, because it is unclear what exactly you are asking. We need you to we need you to specify your points of confusion so that we know how to answer your question.
$endgroup$
– clathratus
Dec 30 '18 at 1:05
3
3
$begingroup$
Please explain the reason of down voting, so I modify my question!
$endgroup$
– Electricman
Dec 29 '18 at 21:13
$begingroup$
Please explain the reason of down voting, so I modify my question!
$endgroup$
– Electricman
Dec 29 '18 at 21:13
1
1
$begingroup$
You have a serious typo. "vertex" not "vortex". (I didn't down vote!)
$endgroup$
– herb steinberg
Dec 29 '18 at 22:47
$begingroup$
You have a serious typo. "vertex" not "vortex". (I didn't down vote!)
$endgroup$
– herb steinberg
Dec 29 '18 at 22:47
$begingroup$
@Electricman your question was down-voted, because it is unclear what exactly you are asking. We need you to we need you to specify your points of confusion so that we know how to answer your question.
$endgroup$
– clathratus
Dec 30 '18 at 1:05
$begingroup$
@Electricman your question was down-voted, because it is unclear what exactly you are asking. We need you to we need you to specify your points of confusion so that we know how to answer your question.
$endgroup$
– clathratus
Dec 30 '18 at 1:05
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
The $2$s are coming in to represent different possible outcomes.
Look at the answer for having only one counter clockwise rotation. We start at $A$ and suppose we go to $C$. We can't go back to $A$ from here or else we are not returning to $A$ from the opposite vertex we reached from the first throw, but rather the same vertex we reached on the first throw. Which means our one counterclockwise throw must happen from $D$ to $B$ or from $B$ to $C$. Because we have two possible outcomes we multiply by $2$. You can also check this with general answers to see why we have included the $2$s. For a quick example, if we have two counterclockwise throws, we only have $2^2 = 4$ ways it can happen
As far as possible down voting comes in, it may be because the problem is phrased slightly problematically. The way you have the question written, it states that :
If we start at point $A$ and we keep throwing until we land back at point $A$, what is the probability that we return to $A$ from the opposite vertex we started at?
It may make more sense if you explain what is meant by this last sentence. I am assuming that we stop throwing once we return to $A$, and the answer you have provided also supports that assumption as well; however it does not take into account the possibility of starting counterclockwise
If we go counterclockwise all four throws we also end up back at $A$ from the opposite vertex we were first thrown at, and this has probability $(1-p)^4$
If we go counterclockwise in $5$ throws with one clockwise throw, we have probability $(1-p)^5 *p * 2$
One last note is that you are asking the probability of this occurring in any amount of throws, so the overall probability that we return to $A$ from the opposite vertex we first are thrown to, is in the form of an infinite series rather than many probabilities partitioned by how many counterclockwise throws we have.
The answer would look something like $$sum_{k=0}^infty p^{4+k} (1-p)^k 2^k +sum_{m=0}^infty (1-p)^{4+m} p^m 2^m $$
answered Dec 29 '18 at 23:00
WaveXWaveX
2,7742822
$endgroup$
1
$begingroup$
Thanks for your answer. I've also edited my question. Would you tell me the 4 possible ways with 2 counterclockwise throw? I think, the two counterclockwise throw can happen just from B to C or just from D to B or one time from D-B and the next time from B-C. So it will be 5 ways and not 4. Where is my mistake? @wavex
$endgroup$
– Electricman
Dec 30 '18 at 14:11
1
$begingroup$
The four ways are: $D$ to $B$ twice in a row (so you get to $D$, go back to $B$, go up to $D$ again, then go back again to $B$ before finally going clockwise the rest of the way), or we could go $B$ to $C$ twice in a row, or we could go $B$ to $C$ then $D$ to $B$, and the fourth, final possibility is going backwards twice in a row from $D$ to $B$ then immediately from $B$ to $C$
$endgroup$
– WaveX
Dec 30 '18 at 16:07
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
The $2$s are coming in to represent different possible outcomes.
Look at the answer for having only one counter clockwise rotation. We start at $A$ and suppose we go to $C$. We can't go back to $A$ from here or else we are not returning to $A$ from the opposite vertex we reached from the first throw, but rather the same vertex we reached on the first throw. Which means our one counterclockwise throw must happen from $D$ to $B$ or from $B$ to $C$. Because we have two possible outcomes we multiply by $2$. You can also check this with general answers to see why we have included the $2$s. For a quick example, if we have two counterclockwise throws, we only have $2^2 = 4$ ways it can happen
As far as possible down voting comes in, it may be because the problem is phrased slightly problematically. The way you have the question written, it states that :
If we start at point $A$ and we keep throwing until we land back at point $A$, what is the probability that we return to $A$ from the opposite vertex we started at?
It may make more sense if you explain what is meant by this last sentence. I am assuming that we stop throwing once we return to $A$, and the answer you have provided also supports that assumption as well; however it does not take into account the possibility of starting counterclockwise
If we go counterclockwise all four throws we also end up back at $A$ from the opposite vertex we were first thrown at, and this has probability $(1-p)^4$
If we go counterclockwise in $5$ throws with one clockwise throw, we have probability $(1-p)^5 *p * 2$
One last note is that you are asking the probability of this occurring in any amount of throws, so the overall probability that we return to $A$ from the opposite vertex we first are thrown to, is in the form of an infinite series rather than many probabilities partitioned by how many counterclockwise throws we have.
The answer would look something like $$sum_{k=0}^infty p^{4+k} (1-p)^k 2^k +sum_{m=0}^infty (1-p)^{4+m} p^m 2^m $$
answered Dec 29 '18 at 23:00
WaveXWaveX
2,7742822
$endgroup$
1
$begingroup$
Thanks for your answer. I've also edited my question. Would you tell me the 4 possible ways with 2 counterclockwise throw? I think, the two counterclockwise throw can happen just from B to C or just from D to B or one time from D-B and the next time from B-C. So it will be 5 ways and not 4. Where is my mistake? @wavex
$endgroup$
– Electricman
Dec 30 '18 at 14:11
1
$begingroup$
The four ways are: $D$ to $B$ twice in a row (so you get to $D$, go back to $B$, go up to $D$ again, then go back again to $B$ before finally going clockwise the rest of the way), or we could go $B$ to $C$ twice in a row, or we could go $B$ to $C$ then $D$ to $B$, and the fourth, final possibility is going backwards twice in a row from $D$ to $B$ then immediately from $B$ to $C$
$endgroup$
– WaveX
Dec 30 '18 at 16:07
add a comment |
$begingroup$
The $2$s are coming in to represent different possible outcomes.
Look at the answer for having only one counter clockwise rotation. We start at $A$ and suppose we go to $C$. We can't go back to $A$ from here or else we are not returning to $A$ from the opposite vertex we reached from the first throw, but rather the same vertex we reached on the first throw. Which means our one counterclockwise throw must happen from $D$ to $B$ or from $B$ to $C$. Because we have two possible outcomes we multiply by $2$. You can also check this with general answers to see why we have included the $2$s. For a quick example, if we have two counterclockwise throws, we only have $2^2 = 4$ ways it can happen
As far as possible down voting comes in, it may be because the problem is phrased slightly problematically. The way you have the question written, it states that :
If we start at point $A$ and we keep throwing until we land back at point $A$, what is the probability that we return to $A$ from the opposite vertex we started at?
It may make more sense if you explain what is meant by this last sentence. I am assuming that we stop throwing once we return to $A$, and the answer you have provided also supports that assumption as well; however it does not take into account the possibility of starting counterclockwise
If we go counterclockwise all four throws we also end up back at $A$ from the opposite vertex we were first thrown at, and this has probability $(1-p)^4$
If we go counterclockwise in $5$ throws with one clockwise throw, we have probability $(1-p)^5 *p * 2$
One last note is that you are asking the probability of this occurring in any amount of throws, so the overall probability that we return to $A$ from the opposite vertex we first are thrown to, is in the form of an infinite series rather than many probabilities partitioned by how many counterclockwise throws we have.
The answer would look something like $$sum_{k=0}^infty p^{4+k} (1-p)^k 2^k +sum_{m=0}^infty (1-p)^{4+m} p^m 2^m $$
answered Dec 29 '18 at 23:00
WaveXWaveX
2,7742822
$endgroup$
1
$begingroup$
Thanks for your answer. I've also edited my question. Would you tell me the 4 possible ways with 2 counterclockwise throw? I think, the two counterclockwise throw can happen just from B to C or just from D to B or one time from D-B and the next time from B-C. So it will be 5 ways and not 4. Where is my mistake? @wavex
$endgroup$
– Electricman
Dec 30 '18 at 14:11
1
$begingroup$
The four ways are: $D$ to $B$ twice in a row (so you get to $D$, go back to $B$, go up to $D$ again, then go back again to $B$ before finally going clockwise the rest of the way), or we could go $B$ to $C$ twice in a row, or we could go $B$ to $C$ then $D$ to $B$, and the fourth, final possibility is going backwards twice in a row from $D$ to $B$ then immediately from $B$ to $C$
$endgroup$
– WaveX
Dec 30 '18 at 16:07
add a comment |
$begingroup$
The $2$s are coming in to represent different possible outcomes.
Look at the answer for having only one counter clockwise rotation. We start at $A$ and suppose we go to $C$. We can't go back to $A$ from here or else we are not returning to $A$ from the opposite vertex we reached from the first throw, but rather the same vertex we reached on the first throw. Which means our one counterclockwise throw must happen from $D$ to $B$ or from $B$ to $C$. Because we have two possible outcomes we multiply by $2$. You can also check this with general answers to see why we have included the $2$s. For a quick example, if we have two counterclockwise throws, we only have $2^2 = 4$ ways it can happen
As far as possible down voting comes in, it may be because the problem is phrased slightly problematically. The way you have the question written, it states that :
If we start at point $A$ and we keep throwing until we land back at point $A$, what is the probability that we return to $A$ from the opposite vertex we started at?
It may make more sense if you explain what is meant by this last sentence. I am assuming that we stop throwing once we return to $A$, and the answer you have provided also supports that assumption as well; however it does not take into account the possibility of starting counterclockwise
If we go counterclockwise all four throws we also end up back at $A$ from the opposite vertex we were first thrown at, and this has probability $(1-p)^4$
If we go counterclockwise in $5$ throws with one clockwise throw, we have probability $(1-p)^5 *p * 2$
One last note is that you are asking the probability of this occurring in any amount of throws, so the overall probability that we return to $A$ from the opposite vertex we first are thrown to, is in the form of an infinite series rather than many probabilities partitioned by how many counterclockwise throws we have.
The answer would look something like $$sum_{k=0}^infty p^{4+k} (1-p)^k 2^k +sum_{m=0}^infty (1-p)^{4+m} p^m 2^m $$
answered Dec 29 '18 at 23:00
WaveXWaveX
2,7742822
$endgroup$
The $2$s are coming in to represent different possible outcomes.
Look at the answer for having only one counter clockwise rotation. We start at $A$ and suppose we go to $C$. We can't go back to $A$ from here or else we are not returning to $A$ from the opposite vertex we reached from the first throw, but rather the same vertex we reached on the first throw. Which means our one counterclockwise throw must happen from $D$ to $B$ or from $B$ to $C$. Because we have two possible outcomes we multiply by $2$. You can also check this with general answers to see why we have included the $2$s. For a quick example, if we have two counterclockwise throws, we only have $2^2 = 4$ ways it can happen
As far as possible down voting comes in, it may be because the problem is phrased slightly problematically. The way you have the question written, it states that :
If we start at point $A$ and we keep throwing until we land back at point $A$, what is the probability that we return to $A$ from the opposite vertex we started at?
It may make more sense if you explain what is meant by this last sentence. I am assuming that we stop throwing once we return to $A$, and the answer you have provided also supports that assumption as well; however it does not take into account the possibility of starting counterclockwise
If we go counterclockwise all four throws we also end up back at $A$ from the opposite vertex we were first thrown at, and this has probability $(1-p)^4$
If we go counterclockwise in $5$ throws with one clockwise throw, we have probability $(1-p)^5 *p * 2$
One last note is that you are asking the probability of this occurring in any amount of throws, so the overall probability that we return to $A$ from the opposite vertex we first are thrown to, is in the form of an infinite series rather than many probabilities partitioned by how many counterclockwise throws we have.
The answer would look something like $$sum_{k=0}^infty p^{4+k} (1-p)^k 2^k +sum_{m=0}^infty (1-p)^{4+m} p^m 2^m $$
answered Dec 29 '18 at 23:00
WaveXWaveX
2,7742822
answered Dec 29 '18 at 23:00
WaveXWaveX
2,7742822
answered Dec 29 '18 at 23:00
WaveXWaveX
2,7742822
answered Dec 29 '18 at 23:00
WaveXWaveX
2,7742822
2,7742822
1
$begingroup$
Thanks for your answer. I've also edited my question. Would you tell me the 4 possible ways with 2 counterclockwise throw? I think, the two counterclockwise throw can happen just from B to C or just from D to B or one time from D-B and the next time from B-C. So it will be 5 ways and not 4. Where is my mistake? @wavex
$endgroup$
– Electricman
Dec 30 '18 at 14:11
1
$begingroup$
The four ways are: $D$ to $B$ twice in a row (so you get to $D$, go back to $B$, go up to $D$ again, then go back again to $B$ before finally going clockwise the rest of the way), or we could go $B$ to $C$ twice in a row, or we could go $B$ to $C$ then $D$ to $B$, and the fourth, final possibility is going backwards twice in a row from $D$ to $B$ then immediately from $B$ to $C$
$endgroup$
– WaveX
Dec 30 '18 at 16:07
add a comment |
1
$begingroup$
Thanks for your answer. I've also edited my question. Would you tell me the 4 possible ways with 2 counterclockwise throw? I think, the two counterclockwise throw can happen just from B to C or just from D to B or one time from D-B and the next time from B-C. So it will be 5 ways and not 4. Where is my mistake? @wavex
$endgroup$
– Electricman
Dec 30 '18 at 14:11
1
$begingroup$
The four ways are: $D$ to $B$ twice in a row (so you get to $D$, go back to $B$, go up to $D$ again, then go back again to $B$ before finally going clockwise the rest of the way), or we could go $B$ to $C$ twice in a row, or we could go $B$ to $C$ then $D$ to $B$, and the fourth, final possibility is going backwards twice in a row from $D$ to $B$ then immediately from $B$ to $C$
$endgroup$
– WaveX
Dec 30 '18 at 16:07
1
1
$begingroup$
Thanks for your answer. I've also edited my question. Would you tell me the 4 possible ways with 2 counterclockwise throw? I think, the two counterclockwise throw can happen just from B to C or just from D to B or one time from D-B and the next time from B-C. So it will be 5 ways and not 4. Where is my mistake? @wavex
$endgroup$
– Electricman
Dec 30 '18 at 14:11
$begingroup$
Thanks for your answer. I've also edited my question. Would you tell me the 4 possible ways with 2 counterclockwise throw? I think, the two counterclockwise throw can happen just from B to C or just from D to B or one time from D-B and the next time from B-C. So it will be 5 ways and not 4. Where is my mistake? @wavex
$endgroup$
– Electricman
Dec 30 '18 at 14:11
1
1
$begingroup$
The four ways are: $D$ to $B$ twice in a row (so you get to $D$, go back to $B$, go up to $D$ again, then go back again to $B$ before finally going clockwise the rest of the way), or we could go $B$ to $C$ twice in a row, or we could go $B$ to $C$ then $D$ to $B$, and the fourth, final possibility is going backwards twice in a row from $D$ to $B$ then immediately from $B$ to $C$
$endgroup$
– WaveX
Dec 30 '18 at 16:07
$begingroup$
The four ways are: $D$ to $B$ twice in a row (so you get to $D$, go back to $B$, go up to $D$ again, then go back again to $B$ before finally going clockwise the rest of the way), or we could go $B$ to $C$ twice in a row, or we could go $B$ to $C$ then $D$ to $B$, and the fourth, final possibility is going backwards twice in a row from $D$ to $B$ then immediately from $B$ to $C$
$endgroup$
– WaveX
Dec 30 '18 at 16:07
add a comment |
3
$begingroup$
Please explain the reason of down voting, so I modify my question!
$endgroup$
– Electricman
Dec 29 '18 at 21:13
1
$begingroup$
You have a serious typo. "vertex" not "vortex". (I didn't down vote!)
$endgroup$
– herb steinberg
Dec 29 '18 at 22:47
$begingroup$
@Electricman your question was down-voted, because it is unclear what exactly you are asking. We need you to we need you to specify your points of confusion so that we know how to answer your question.
$endgroup$
– clathratus
Dec 30 '18 at 1:05