Number of expected winners in a game involving three kinds of marbles
Clash Royale CLAN TAG#URR8PPP
up vote
1
down vote
favorite
In a bag there are some red marbles, some blue marbles, and some green marbles.
There is at least one marble of each color in the bag.
Richard, Bob and George ignore the exact numbers of marbles of each color in the bag, and they decide to play a game.
Richard bets to get, in $N$ trials with replacement, at least one red marble, Bob to get at least one blue marble, and George to get at least one green marble.
However, we additionally know that, in $N$ trials, the probability for Richard to win is exactly equal to the probability for Bob to lose.
Denoting with $Xin1,2,3$ the number of boys which can win (*), what is the expected value of the random variable "In $N$ trials, $X$ boys win?"
(*) $Xneq 0$, because we know that there is at least one ball of each kind in the bag.
I tried to use the linearity of the expectation value, illustrated e.g. in this post, but I did not arrive anywhere. Moreover, I suspect that the additional information does not play any role, but I don't know how to prove it.
I apologize in case the question might be not perfectly posed. However, thanks a lot for your suggestions!
probability combinatorics random-variables gambling
asked Aug 15 at 10:33
Andrea Prunotto
1,215623
add a comment |Â
up vote
1
down vote
favorite
In a bag there are some red marbles, some blue marbles, and some green marbles.
There is at least one marble of each color in the bag.
Richard, Bob and George ignore the exact numbers of marbles of each color in the bag, and they decide to play a game.
Richard bets to get, in $N$ trials with replacement, at least one red marble, Bob to get at least one blue marble, and George to get at least one green marble.
However, we additionally know that, in $N$ trials, the probability for Richard to win is exactly equal to the probability for Bob to lose.
Denoting with $Xin1,2,3$ the number of boys which can win (*), what is the expected value of the random variable "In $N$ trials, $X$ boys win?"
(*) $Xneq 0$, because we know that there is at least one ball of each kind in the bag.
I tried to use the linearity of the expectation value, illustrated e.g. in this post, but I did not arrive anywhere. Moreover, I suspect that the additional information does not play any role, but I don't know how to prove it.
I apologize in case the question might be not perfectly posed. However, thanks a lot for your suggestions!
probability combinatorics random-variables gambling
asked Aug 15 at 10:33
Andrea Prunotto
1,215623
The conclusion that $N$ must be $2$ made headlines in the New York Times in the 1990s.
– Empy2
Aug 15 at 11:10
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
In a bag there are some red marbles, some blue marbles, and some green marbles.
There is at least one marble of each color in the bag.
Richard, Bob and George ignore the exact numbers of marbles of each color in the bag, and they decide to play a game.
Richard bets to get, in $N$ trials with replacement, at least one red marble, Bob to get at least one blue marble, and George to get at least one green marble.
However, we additionally know that, in $N$ trials, the probability for Richard to win is exactly equal to the probability for Bob to lose.
Denoting with $Xin1,2,3$ the number of boys which can win (*), what is the expected value of the random variable "In $N$ trials, $X$ boys win?"
(*) $Xneq 0$, because we know that there is at least one ball of each kind in the bag.
I tried to use the linearity of the expectation value, illustrated e.g. in this post, but I did not arrive anywhere. Moreover, I suspect that the additional information does not play any role, but I don't know how to prove it.
I apologize in case the question might be not perfectly posed. However, thanks a lot for your suggestions!
probability combinatorics random-variables gambling
asked Aug 15 at 10:33
Andrea Prunotto
1,215623
In a bag there are some red marbles, some blue marbles, and some green marbles.
There is at least one marble of each color in the bag.
Richard, Bob and George ignore the exact numbers of marbles of each color in the bag, and they decide to play a game.
Richard bets to get, in $N$ trials with replacement, at least one red marble, Bob to get at least one blue marble, and George to get at least one green marble.
However, we additionally know that, in $N$ trials, the probability for Richard to win is exactly equal to the probability for Bob to lose.
Denoting with $Xin1,2,3$ the number of boys which can win (*), what is the expected value of the random variable "In $N$ trials, $X$ boys win?"
(*) $Xneq 0$, because we know that there is at least one ball of each kind in the bag.
I tried to use the linearity of the expectation value, illustrated e.g. in this post, but I did not arrive anywhere. Moreover, I suspect that the additional information does not play any role, but I don't know how to prove it.
I apologize in case the question might be not perfectly posed. However, thanks a lot for your suggestions!
probability combinatorics random-variables gambling
asked Aug 15 at 10:33
Andrea Prunotto
1,215623
edited Aug 15 at 10:46
asked Aug 15 at 10:33
Andrea Prunotto
1,215623
asked Aug 15 at 10:33
Andrea Prunotto
1,215623
asked Aug 15 at 10:33
Andrea Prunotto
1,215623
1,215623
The conclusion that $N$ must be $2$ made headlines in the New York Times in the 1990s.
– Empy2
Aug 15 at 11:10
add a comment |Â
The conclusion that $N$ must be $2$ made headlines in the New York Times in the 1990s.
– Empy2
Aug 15 at 11:10
The conclusion that $N$ must be $2$ made headlines in the New York Times in the 1990s.
– Empy2
Aug 15 at 11:10
The conclusion that $N$ must be $2$ made headlines in the New York Times in the 1990s.
– Empy2
Aug 15 at 11:10
add a comment |Â
1 Answer
1
active
oldest
votes
up vote
1
down vote
accepted
The expected value, as you mentioned, is the sum of individual winning probabilities by linearity. Let there are $R$ red, $B$ blue, and $G$ green marbles in the bag. The probability of getting at least $1$ red marbles in $N$ i.i.d. trials is
$$ 1 - left(frac B + G R + B + Gright)^N$$
and similar for the other two colors. The condition given implies that
$$ 1 - left(frac B + G R + B + Gright)^N = left(frac R + G R + B + Gright)^N$$
Summing the three winning probabilities gives you the expectation:
$$ 3 - frac (B + G)^N + (R + G)^N + (R + B)^N (R + B + G)^N = 2 - frac (R + B)^N (R + B + G)^N$$
where we used the condition to simplify the expectation.
answered Aug 15 at 11:07
BGM
3,615148
Thanks a lot BGM!
– Andrea Prunotto
Aug 15 at 11:13
But, then, the conclusion is that the expected number of winners is less than 2, i.e. one boy? Sorry if the question is silly!
– Andrea Prunotto
Aug 15 at 11:16
1
On average only one of Richard and Bob will win, so X's average is at most 2. The average over many games doesn't have to be a whole number.
– Empy2
Aug 15 at 11:21
I see now. But... what about George? He's also included in the $2-epsilon$, $0<epsilon<1$, right?
– Andrea Prunotto
Aug 15 at 11:24
From the first equation, $(B+G)^N+(R+G)^N=(R+B+G)^N$. Why can you conclude $N=2$?
– Empy2
Aug 15 at 11:26
 |Â
show 5 more comments
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
The expected value, as you mentioned, is the sum of individual winning probabilities by linearity. Let there are $R$ red, $B$ blue, and $G$ green marbles in the bag. The probability of getting at least $1$ red marbles in $N$ i.i.d. trials is
$$ 1 - left(frac B + G R + B + Gright)^N$$
and similar for the other two colors. The condition given implies that
$$ 1 - left(frac B + G R + B + Gright)^N = left(frac R + G R + B + Gright)^N$$
Summing the three winning probabilities gives you the expectation:
$$ 3 - frac (B + G)^N + (R + G)^N + (R + B)^N (R + B + G)^N = 2 - frac (R + B)^N (R + B + G)^N$$
where we used the condition to simplify the expectation.
answered Aug 15 at 11:07
BGM
3,615148
Thanks a lot BGM!
– Andrea Prunotto
Aug 15 at 11:13
But, then, the conclusion is that the expected number of winners is less than 2, i.e. one boy? Sorry if the question is silly!
– Andrea Prunotto
Aug 15 at 11:16
1
On average only one of Richard and Bob will win, so X's average is at most 2. The average over many games doesn't have to be a whole number.
– Empy2
Aug 15 at 11:21
I see now. But... what about George? He's also included in the $2-epsilon$, $0<epsilon<1$, right?
– Andrea Prunotto
Aug 15 at 11:24
From the first equation, $(B+G)^N+(R+G)^N=(R+B+G)^N$. Why can you conclude $N=2$?
– Empy2
Aug 15 at 11:26
 |Â
show 5 more comments
up vote
1
down vote
accepted
The expected value, as you mentioned, is the sum of individual winning probabilities by linearity. Let there are $R$ red, $B$ blue, and $G$ green marbles in the bag. The probability of getting at least $1$ red marbles in $N$ i.i.d. trials is
$$ 1 - left(frac B + G R + B + Gright)^N$$
and similar for the other two colors. The condition given implies that
$$ 1 - left(frac B + G R + B + Gright)^N = left(frac R + G R + B + Gright)^N$$
Summing the three winning probabilities gives you the expectation:
$$ 3 - frac (B + G)^N + (R + G)^N + (R + B)^N (R + B + G)^N = 2 - frac (R + B)^N (R + B + G)^N$$
where we used the condition to simplify the expectation.
answered Aug 15 at 11:07
BGM
3,615148
Thanks a lot BGM!
– Andrea Prunotto
Aug 15 at 11:13
But, then, the conclusion is that the expected number of winners is less than 2, i.e. one boy? Sorry if the question is silly!
– Andrea Prunotto
Aug 15 at 11:16
1
On average only one of Richard and Bob will win, so X's average is at most 2. The average over many games doesn't have to be a whole number.
– Empy2
Aug 15 at 11:21
I see now. But... what about George? He's also included in the $2-epsilon$, $0<epsilon<1$, right?
– Andrea Prunotto
Aug 15 at 11:24
From the first equation, $(B+G)^N+(R+G)^N=(R+B+G)^N$. Why can you conclude $N=2$?
– Empy2
Aug 15 at 11:26
 |Â
show 5 more comments
up vote
1
down vote
accepted
up vote
1
down vote
accepted
The expected value, as you mentioned, is the sum of individual winning probabilities by linearity. Let there are $R$ red, $B$ blue, and $G$ green marbles in the bag. The probability of getting at least $1$ red marbles in $N$ i.i.d. trials is
$$ 1 - left(frac B + G R + B + Gright)^N$$
and similar for the other two colors. The condition given implies that
$$ 1 - left(frac B + G R + B + Gright)^N = left(frac R + G R + B + Gright)^N$$
Summing the three winning probabilities gives you the expectation:
$$ 3 - frac (B + G)^N + (R + G)^N + (R + B)^N (R + B + G)^N = 2 - frac (R + B)^N (R + B + G)^N$$
where we used the condition to simplify the expectation.
answered Aug 15 at 11:07
BGM
3,615148
The expected value, as you mentioned, is the sum of individual winning probabilities by linearity. Let there are $R$ red, $B$ blue, and $G$ green marbles in the bag. The probability of getting at least $1$ red marbles in $N$ i.i.d. trials is
$$ 1 - left(frac B + G R + B + Gright)^N$$
and similar for the other two colors. The condition given implies that
$$ 1 - left(frac B + G R + B + Gright)^N = left(frac R + G R + B + Gright)^N$$
Summing the three winning probabilities gives you the expectation:
$$ 3 - frac (B + G)^N + (R + G)^N + (R + B)^N (R + B + G)^N = 2 - frac (R + B)^N (R + B + G)^N$$
where we used the condition to simplify the expectation.
answered Aug 15 at 11:07
BGM
3,615148
answered Aug 15 at 11:07
BGM
3,615148
answered Aug 15 at 11:07
BGM
3,615148
answered Aug 15 at 11:07
BGM
3,615148
3,615148
Thanks a lot BGM!
– Andrea Prunotto
Aug 15 at 11:13
But, then, the conclusion is that the expected number of winners is less than 2, i.e. one boy? Sorry if the question is silly!
– Andrea Prunotto
Aug 15 at 11:16
1
On average only one of Richard and Bob will win, so X's average is at most 2. The average over many games doesn't have to be a whole number.
– Empy2
Aug 15 at 11:21
I see now. But... what about George? He's also included in the $2-epsilon$, $0<epsilon<1$, right?
– Andrea Prunotto
Aug 15 at 11:24
From the first equation, $(B+G)^N+(R+G)^N=(R+B+G)^N$. Why can you conclude $N=2$?
– Empy2
Aug 15 at 11:26
 |Â
show 5 more comments
Thanks a lot BGM!
– Andrea Prunotto
Aug 15 at 11:13
But, then, the conclusion is that the expected number of winners is less than 2, i.e. one boy? Sorry if the question is silly!
– Andrea Prunotto
Aug 15 at 11:16
1
On average only one of Richard and Bob will win, so X's average is at most 2. The average over many games doesn't have to be a whole number.
– Empy2
Aug 15 at 11:21
I see now. But... what about George? He's also included in the $2-epsilon$, $0<epsilon<1$, right?
– Andrea Prunotto
Aug 15 at 11:24
From the first equation, $(B+G)^N+(R+G)^N=(R+B+G)^N$. Why can you conclude $N=2$?
– Empy2
Aug 15 at 11:26
Thanks a lot BGM!
– Andrea Prunotto
Aug 15 at 11:13
Thanks a lot BGM!
– Andrea Prunotto
Aug 15 at 11:13
But, then, the conclusion is that the expected number of winners is less than 2, i.e. one boy? Sorry if the question is silly!
– Andrea Prunotto
Aug 15 at 11:16
But, then, the conclusion is that the expected number of winners is less than 2, i.e. one boy? Sorry if the question is silly!
– Andrea Prunotto
Aug 15 at 11:16
1
1
On average only one of Richard and Bob will win, so X's average is at most 2. The average over many games doesn't have to be a whole number.
– Empy2
Aug 15 at 11:21
On average only one of Richard and Bob will win, so X's average is at most 2. The average over many games doesn't have to be a whole number.
– Empy2
Aug 15 at 11:21
I see now. But... what about George? He's also included in the $2-epsilon$, $0<epsilon<1$, right?
– Andrea Prunotto
Aug 15 at 11:24
I see now. But... what about George? He's also included in the $2-epsilon$, $0<epsilon<1$, right?
– Andrea Prunotto
Aug 15 at 11:24
From the first equation, $(B+G)^N+(R+G)^N=(R+B+G)^N$. Why can you conclude $N=2$?
– Empy2
Aug 15 at 11:26
From the first equation, $(B+G)^N+(R+G)^N=(R+B+G)^N$. Why can you conclude $N=2$?
– Empy2
Aug 15 at 11:26
 |Â
show 5 more comments
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
StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2883451%2fnumber-of-expected-winners-in-a-game-involving-three-kinds-of-marbles%23new-answer', 'question_page');
);
Post as a guest
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
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
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
The conclusion that $N$ must be $2$ made headlines in the New York Times in the 1990s.
– Empy2
Aug 15 at 11:10