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Probability that one trial is a successful trial for events related to an Urn Problem
Clash Royale CLAN TAG#URR8PPP
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Consider an urn $C$ which contains three (distinguishable) kinds of elements, say $alpha>0$ elements of kind $A$, $beta>0$ elements of kind $B$, and $gamma>0$ elements of kind $G$. We perform $n>0$ trials (with replacement) of one element at a time from the urn $C$.
We define two events:
- $L^AB_n$: "to get at least one element of kind $A$ and at least one element of kind $G$, in $n$ trials"
- $E_n^G$: "to get $n$ elements of kind $G$, in $n$ trials"
I would like to study the probability that one trial, out of $n$, is a successful trial for these events, separately.
My reasoning, based on the definition of probability as the ratio between favorable and possible cases, is: there are $n$ possible trials (e.g. $n$ possible cases), but not all of them represent a favorable case.
In particular:
- There are $n-1$ trials which represent a favorable case, out of $n$, for the success of the event $L^AB_n$, although not all of these possible cases have the same chance to occur. In fact
$$
P(L^AB_k)=1-left(fracalpha+gammaalpha+beta+gammaright)^k-left(fracbeta+gammaalpha+beta+gammaright)^k+left(fracgammaalpha+beta+gammaright)^k.
$$ - There is only one trial which represents a favorable case, out of $n$, for the the success of the event $E^G_n$, and
$$
P(E^G_n,k)=left(fracgammaalpha+beta+gammaright)^ndelta_n,k,
$$
where $P(E^G_n,k)$ is the probability to get a success for $E^G_n$ at the $k$-th trial, and $delta_n,k$ is the Kroneker symbol.
I point out that I am interested to know the probability that one of the $n$ trials is a success, and not on the probability of the events to occur in the course of the $n$ trials (these probabilities are already calculated above). In other words, I am looking for a probability in the form
$$
fractext$m$ favorable trialstext$n$ possible trials.
$$
I would say that:
- The probability that one trial is a success for the event $L^AB_n$ is
$$
fracsum_k=2^n[P(L^AB_k)cdot 1]n,
$$
rather than simply $fracn-1n$, because we must weight each of the $n-1$ favorable trials at the numerator with the probability that the event $L^AB_n$ occurs in correspondence of each trial.
- The probability that one trial is a success for the event $E_n^G$ is
$$
fracsum_k=1^n[P(E^G_n,k)cdot 1]n=fracleft(fracgammaalpha+beta+gammaright)^nn,
$$
rather than simply $frac1n$, since the probability that the only possible favorable case (i.e. the last trial) is a success is $left(fracgammaalpha+beta+gammaright)^n$.
Is this reasoning correct?
I am almost sure that it is not correct! Therefore I need your help to understand the conceptual mistakes that I am doing here. I however apologize for naivety and silly errors.
NOTE: I posted some other question with a similar topic: for instance, Time to first success in case of increasing probability at each trial, Time to first success of an event-intersection, and Time to first success for a simple event.
probability combinatorics gambling
asked Jul 18 at 18:45
Andrea Prunotto
578215
add a comment |Â
up vote
0
down vote
favorite
Consider an urn $C$ which contains three (distinguishable) kinds of elements, say $alpha>0$ elements of kind $A$, $beta>0$ elements of kind $B$, and $gamma>0$ elements of kind $G$. We perform $n>0$ trials (with replacement) of one element at a time from the urn $C$.
We define two events:
- $L^AB_n$: "to get at least one element of kind $A$ and at least one element of kind $G$, in $n$ trials"
- $E_n^G$: "to get $n$ elements of kind $G$, in $n$ trials"
I would like to study the probability that one trial, out of $n$, is a successful trial for these events, separately.
My reasoning, based on the definition of probability as the ratio between favorable and possible cases, is: there are $n$ possible trials (e.g. $n$ possible cases), but not all of them represent a favorable case.
In particular:
- There are $n-1$ trials which represent a favorable case, out of $n$, for the success of the event $L^AB_n$, although not all of these possible cases have the same chance to occur. In fact
$$
P(L^AB_k)=1-left(fracalpha+gammaalpha+beta+gammaright)^k-left(fracbeta+gammaalpha+beta+gammaright)^k+left(fracgammaalpha+beta+gammaright)^k.
$$ - There is only one trial which represents a favorable case, out of $n$, for the the success of the event $E^G_n$, and
$$
P(E^G_n,k)=left(fracgammaalpha+beta+gammaright)^ndelta_n,k,
$$
where $P(E^G_n,k)$ is the probability to get a success for $E^G_n$ at the $k$-th trial, and $delta_n,k$ is the Kroneker symbol.
I point out that I am interested to know the probability that one of the $n$ trials is a success, and not on the probability of the events to occur in the course of the $n$ trials (these probabilities are already calculated above). In other words, I am looking for a probability in the form
$$
fractext$m$ favorable trialstext$n$ possible trials.
$$
I would say that:
- The probability that one trial is a success for the event $L^AB_n$ is
$$
fracsum_k=2^n[P(L^AB_k)cdot 1]n,
$$
rather than simply $fracn-1n$, because we must weight each of the $n-1$ favorable trials at the numerator with the probability that the event $L^AB_n$ occurs in correspondence of each trial.
- The probability that one trial is a success for the event $E_n^G$ is
$$
fracsum_k=1^n[P(E^G_n,k)cdot 1]n=fracleft(fracgammaalpha+beta+gammaright)^nn,
$$
rather than simply $frac1n$, since the probability that the only possible favorable case (i.e. the last trial) is a success is $left(fracgammaalpha+beta+gammaright)^n$.
Is this reasoning correct?
I am almost sure that it is not correct! Therefore I need your help to understand the conceptual mistakes that I am doing here. I however apologize for naivety and silly errors.
NOTE: I posted some other question with a similar topic: for instance, Time to first success in case of increasing probability at each trial, Time to first success of an event-intersection, and Time to first success for a simple event.
probability combinatorics gambling
asked Jul 18 at 18:45
Andrea Prunotto
578215
What does it mean "the probability that one trial, out of n, is a successful trial for these events", i.e. what is $L_k^AB$?
– d.k.o.
Jul 19 at 7:04
@d.k.o. This means that, in correspondence of the trial $k$, you get at least one element of kind $A$ and at least one element of kind $B$. Imagine, for instance, that you got an element of kind $A$ in a previous trial: if, at the trial $k$, you get one element of kind $B$, then this trial represents a "successful one", since you got a success for the event $L_n^AB$.
– Andrea Prunotto
Jul 19 at 7:25
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
Consider an urn $C$ which contains three (distinguishable) kinds of elements, say $alpha>0$ elements of kind $A$, $beta>0$ elements of kind $B$, and $gamma>0$ elements of kind $G$. We perform $n>0$ trials (with replacement) of one element at a time from the urn $C$.
We define two events:
- $L^AB_n$: "to get at least one element of kind $A$ and at least one element of kind $G$, in $n$ trials"
- $E_n^G$: "to get $n$ elements of kind $G$, in $n$ trials"
I would like to study the probability that one trial, out of $n$, is a successful trial for these events, separately.
My reasoning, based on the definition of probability as the ratio between favorable and possible cases, is: there are $n$ possible trials (e.g. $n$ possible cases), but not all of them represent a favorable case.
In particular:
- There are $n-1$ trials which represent a favorable case, out of $n$, for the success of the event $L^AB_n$, although not all of these possible cases have the same chance to occur. In fact
$$
P(L^AB_k)=1-left(fracalpha+gammaalpha+beta+gammaright)^k-left(fracbeta+gammaalpha+beta+gammaright)^k+left(fracgammaalpha+beta+gammaright)^k.
$$ - There is only one trial which represents a favorable case, out of $n$, for the the success of the event $E^G_n$, and
$$
P(E^G_n,k)=left(fracgammaalpha+beta+gammaright)^ndelta_n,k,
$$
where $P(E^G_n,k)$ is the probability to get a success for $E^G_n$ at the $k$-th trial, and $delta_n,k$ is the Kroneker symbol.
I point out that I am interested to know the probability that one of the $n$ trials is a success, and not on the probability of the events to occur in the course of the $n$ trials (these probabilities are already calculated above). In other words, I am looking for a probability in the form
$$
fractext$m$ favorable trialstext$n$ possible trials.
$$
I would say that:
- The probability that one trial is a success for the event $L^AB_n$ is
$$
fracsum_k=2^n[P(L^AB_k)cdot 1]n,
$$
rather than simply $fracn-1n$, because we must weight each of the $n-1$ favorable trials at the numerator with the probability that the event $L^AB_n$ occurs in correspondence of each trial.
- The probability that one trial is a success for the event $E_n^G$ is
$$
fracsum_k=1^n[P(E^G_n,k)cdot 1]n=fracleft(fracgammaalpha+beta+gammaright)^nn,
$$
rather than simply $frac1n$, since the probability that the only possible favorable case (i.e. the last trial) is a success is $left(fracgammaalpha+beta+gammaright)^n$.
Is this reasoning correct?
I am almost sure that it is not correct! Therefore I need your help to understand the conceptual mistakes that I am doing here. I however apologize for naivety and silly errors.
NOTE: I posted some other question with a similar topic: for instance, Time to first success in case of increasing probability at each trial, Time to first success of an event-intersection, and Time to first success for a simple event.
probability combinatorics gambling
asked Jul 18 at 18:45
Andrea Prunotto
578215
Consider an urn $C$ which contains three (distinguishable) kinds of elements, say $alpha>0$ elements of kind $A$, $beta>0$ elements of kind $B$, and $gamma>0$ elements of kind $G$. We perform $n>0$ trials (with replacement) of one element at a time from the urn $C$.
We define two events:
- $L^AB_n$: "to get at least one element of kind $A$ and at least one element of kind $G$, in $n$ trials"
- $E_n^G$: "to get $n$ elements of kind $G$, in $n$ trials"
I would like to study the probability that one trial, out of $n$, is a successful trial for these events, separately.
My reasoning, based on the definition of probability as the ratio between favorable and possible cases, is: there are $n$ possible trials (e.g. $n$ possible cases), but not all of them represent a favorable case.
In particular:
- There are $n-1$ trials which represent a favorable case, out of $n$, for the success of the event $L^AB_n$, although not all of these possible cases have the same chance to occur. In fact
$$
P(L^AB_k)=1-left(fracalpha+gammaalpha+beta+gammaright)^k-left(fracbeta+gammaalpha+beta+gammaright)^k+left(fracgammaalpha+beta+gammaright)^k.
$$ - There is only one trial which represents a favorable case, out of $n$, for the the success of the event $E^G_n$, and
$$
P(E^G_n,k)=left(fracgammaalpha+beta+gammaright)^ndelta_n,k,
$$
where $P(E^G_n,k)$ is the probability to get a success for $E^G_n$ at the $k$-th trial, and $delta_n,k$ is the Kroneker symbol.
I point out that I am interested to know the probability that one of the $n$ trials is a success, and not on the probability of the events to occur in the course of the $n$ trials (these probabilities are already calculated above). In other words, I am looking for a probability in the form
$$
fractext$m$ favorable trialstext$n$ possible trials.
$$
I would say that:
- The probability that one trial is a success for the event $L^AB_n$ is
$$
fracsum_k=2^n[P(L^AB_k)cdot 1]n,
$$
rather than simply $fracn-1n$, because we must weight each of the $n-1$ favorable trials at the numerator with the probability that the event $L^AB_n$ occurs in correspondence of each trial.
- The probability that one trial is a success for the event $E_n^G$ is
$$
fracsum_k=1^n[P(E^G_n,k)cdot 1]n=fracleft(fracgammaalpha+beta+gammaright)^nn,
$$
rather than simply $frac1n$, since the probability that the only possible favorable case (i.e. the last trial) is a success is $left(fracgammaalpha+beta+gammaright)^n$.
Is this reasoning correct?
I am almost sure that it is not correct! Therefore I need your help to understand the conceptual mistakes that I am doing here. I however apologize for naivety and silly errors.
NOTE: I posted some other question with a similar topic: for instance, Time to first success in case of increasing probability at each trial, Time to first success of an event-intersection, and Time to first success for a simple event.
probability combinatorics gambling
asked Jul 18 at 18:45
Andrea Prunotto
578215
edited Jul 19 at 6:41
asked Jul 18 at 18:45
Andrea Prunotto
578215
asked Jul 18 at 18:45
Andrea Prunotto
578215
asked Jul 18 at 18:45
Andrea Prunotto
578215
578215
What does it mean "the probability that one trial, out of n, is a successful trial for these events", i.e. what is $L_k^AB$?
– d.k.o.
Jul 19 at 7:04
@d.k.o. This means that, in correspondence of the trial $k$, you get at least one element of kind $A$ and at least one element of kind $B$. Imagine, for instance, that you got an element of kind $A$ in a previous trial: if, at the trial $k$, you get one element of kind $B$, then this trial represents a "successful one", since you got a success for the event $L_n^AB$.
– Andrea Prunotto
Jul 19 at 7:25
add a comment |Â
What does it mean "the probability that one trial, out of n, is a successful trial for these events", i.e. what is $L_k^AB$?
– d.k.o.
Jul 19 at 7:04
@d.k.o. This means that, in correspondence of the trial $k$, you get at least one element of kind $A$ and at least one element of kind $B$. Imagine, for instance, that you got an element of kind $A$ in a previous trial: if, at the trial $k$, you get one element of kind $B$, then this trial represents a "successful one", since you got a success for the event $L_n^AB$.
– Andrea Prunotto
Jul 19 at 7:25
What does it mean "the probability that one trial, out of n, is a successful trial for these events", i.e. what is $L_k^AB$?
– d.k.o.
Jul 19 at 7:04
What does it mean "the probability that one trial, out of n, is a successful trial for these events", i.e. what is $L_k^AB$?
– d.k.o.
Jul 19 at 7:04
@d.k.o. This means that, in correspondence of the trial $k$, you get at least one element of kind $A$ and at least one element of kind $B$. Imagine, for instance, that you got an element of kind $A$ in a previous trial: if, at the trial $k$, you get one element of kind $B$, then this trial represents a "successful one", since you got a success for the event $L_n^AB$.
– Andrea Prunotto
Jul 19 at 7:25
@d.k.o. This means that, in correspondence of the trial $k$, you get at least one element of kind $A$ and at least one element of kind $B$. Imagine, for instance, that you got an element of kind $A$ in a previous trial: if, at the trial $k$, you get one element of kind $B$, then this trial represents a "successful one", since you got a success for the event $L_n^AB$.
– Andrea Prunotto
Jul 19 at 7:25
add a comment |Â
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What does it mean "the probability that one trial, out of n, is a successful trial for these events", i.e. what is $L_k^AB$?
– d.k.o.
Jul 19 at 7:04
@d.k.o. This means that, in correspondence of the trial $k$, you get at least one element of kind $A$ and at least one element of kind $B$. Imagine, for instance, that you got an element of kind $A$ in a previous trial: if, at the trial $k$, you get one element of kind $B$, then this trial represents a "successful one", since you got a success for the event $L_n^AB$.
– Andrea Prunotto
Jul 19 at 7:25