Mixing
Probability of each player scores at least once in a simple random game
Clash Royale CLAN TAG#URR8PPP
up vote
0
down vote
favorite
Let’s suppose a very simple game that lasts 3 seconds only.
A player generates a random number between 1 and 3 in the first second.
If the generated number is equal to 3, the player scores and s(he) generates a new random number for the next second.
Otherwise, the generated number is incremented for the next second.
For example, the player generated 2 in the first second. In the next second, the generated value will be incremented to 3 and the player scores and a new value is generated for the last second.
In the best case, the player generated 3 every second.
The worst case is when the player generates 1 at the first second (the player scores just once in the last second)
Let’s suppose two players are playing this game.
There is a special rule when there are multiple players.
If both have value 3 at the same time, no one scores.
For instance, when both players generate 1 in the first second, they will have the value 3 in the last second at the same time.
What is the probability that both players score at least once during these 3 seconds of game?
Thanks in advance.
probability
asked Jul 27 at 6:49
Rodrigo Teles
31
add a comment |Â
up vote
0
down vote
favorite
Let’s suppose a very simple game that lasts 3 seconds only.
A player generates a random number between 1 and 3 in the first second.
If the generated number is equal to 3, the player scores and s(he) generates a new random number for the next second.
Otherwise, the generated number is incremented for the next second.
For example, the player generated 2 in the first second. In the next second, the generated value will be incremented to 3 and the player scores and a new value is generated for the last second.
In the best case, the player generated 3 every second.
The worst case is when the player generates 1 at the first second (the player scores just once in the last second)
Let’s suppose two players are playing this game.
There is a special rule when there are multiple players.
If both have value 3 at the same time, no one scores.
For instance, when both players generate 1 in the first second, they will have the value 3 in the last second at the same time.
What is the probability that both players score at least once during these 3 seconds of game?
Thanks in advance.
probability
asked Jul 27 at 6:49
Rodrigo Teles
31
By "number" you mean *natural number"? What about $2.7664455519$? If so, the chance that the player chooses $3$ is $0$.
– David G. Stork
Jul 27 at 7:09
What happens if e.g. both players generate a 3 in first round? They do not score then, but I guess the 3 will not be incremented (correct me if I am wrong). Do both generate a new number then?
– drhab
Jul 27 at 7:09
If they generate 3 in the first round they don't score. They generate a new number for the next round.
– Rodrigo Teles
Jul 27 at 7:32
How about first two rounds both generate 2? Will this score and leave 1 OR no score?
– Xiaonan
Jul 27 at 10:08
Hi, in this case the value would be 3 in the second round, no score, but they would have one last opportunity in generating a 3 (one of them) in the last round.
– Rodrigo Teles
Jul 27 at 11:55
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
Let’s suppose a very simple game that lasts 3 seconds only.
A player generates a random number between 1 and 3 in the first second.
If the generated number is equal to 3, the player scores and s(he) generates a new random number for the next second.
Otherwise, the generated number is incremented for the next second.
For example, the player generated 2 in the first second. In the next second, the generated value will be incremented to 3 and the player scores and a new value is generated for the last second.
In the best case, the player generated 3 every second.
The worst case is when the player generates 1 at the first second (the player scores just once in the last second)
Let’s suppose two players are playing this game.
There is a special rule when there are multiple players.
If both have value 3 at the same time, no one scores.
For instance, when both players generate 1 in the first second, they will have the value 3 in the last second at the same time.
What is the probability that both players score at least once during these 3 seconds of game?
Thanks in advance.
probability
asked Jul 27 at 6:49
Rodrigo Teles
31
Let’s suppose a very simple game that lasts 3 seconds only.
A player generates a random number between 1 and 3 in the first second.
If the generated number is equal to 3, the player scores and s(he) generates a new random number for the next second.
Otherwise, the generated number is incremented for the next second.
For example, the player generated 2 in the first second. In the next second, the generated value will be incremented to 3 and the player scores and a new value is generated for the last second.
In the best case, the player generated 3 every second.
The worst case is when the player generates 1 at the first second (the player scores just once in the last second)
Let’s suppose two players are playing this game.
There is a special rule when there are multiple players.
If both have value 3 at the same time, no one scores.
For instance, when both players generate 1 in the first second, they will have the value 3 in the last second at the same time.
What is the probability that both players score at least once during these 3 seconds of game?
Thanks in advance.
probability
asked Jul 27 at 6:49
Rodrigo Teles
31
asked Jul 27 at 6:49
Rodrigo Teles
31
asked Jul 27 at 6:49
Rodrigo Teles
31
asked Jul 27 at 6:49
Rodrigo Teles
31
31
By "number" you mean *natural number"? What about $2.7664455519$? If so, the chance that the player chooses $3$ is $0$.
– David G. Stork
Jul 27 at 7:09
What happens if e.g. both players generate a 3 in first round? They do not score then, but I guess the 3 will not be incremented (correct me if I am wrong). Do both generate a new number then?
– drhab
Jul 27 at 7:09
If they generate 3 in the first round they don't score. They generate a new number for the next round.
– Rodrigo Teles
Jul 27 at 7:32
How about first two rounds both generate 2? Will this score and leave 1 OR no score?
– Xiaonan
Jul 27 at 10:08
Hi, in this case the value would be 3 in the second round, no score, but they would have one last opportunity in generating a 3 (one of them) in the last round.
– Rodrigo Teles
Jul 27 at 11:55
add a comment |Â
By "number" you mean *natural number"? What about $2.7664455519$? If so, the chance that the player chooses $3$ is $0$.
– David G. Stork
Jul 27 at 7:09
What happens if e.g. both players generate a 3 in first round? They do not score then, but I guess the 3 will not be incremented (correct me if I am wrong). Do both generate a new number then?
– drhab
Jul 27 at 7:09
If they generate 3 in the first round they don't score. They generate a new number for the next round.
– Rodrigo Teles
Jul 27 at 7:32
How about first two rounds both generate 2? Will this score and leave 1 OR no score?
– Xiaonan
Jul 27 at 10:08
Hi, in this case the value would be 3 in the second round, no score, but they would have one last opportunity in generating a 3 (one of them) in the last round.
– Rodrigo Teles
Jul 27 at 11:55
By "number" you mean *natural number"? What about $2.7664455519$? If so, the chance that the player chooses $3$ is $0$.
– David G. Stork
Jul 27 at 7:09
By "number" you mean *natural number"? What about $2.7664455519$? If so, the chance that the player chooses $3$ is $0$.
– David G. Stork
Jul 27 at 7:09
What happens if e.g. both players generate a 3 in first round? They do not score then, but I guess the 3 will not be incremented (correct me if I am wrong). Do both generate a new number then?
– drhab
Jul 27 at 7:09
What happens if e.g. both players generate a 3 in first round? They do not score then, but I guess the 3 will not be incremented (correct me if I am wrong). Do both generate a new number then?
– drhab
Jul 27 at 7:09
If they generate 3 in the first round they don't score. They generate a new number for the next round.
– Rodrigo Teles
Jul 27 at 7:32
If they generate 3 in the first round they don't score. They generate a new number for the next round.
– Rodrigo Teles
Jul 27 at 7:32
How about first two rounds both generate 2? Will this score and leave 1 OR no score?
– Xiaonan
Jul 27 at 10:08
How about first two rounds both generate 2? Will this score and leave 1 OR no score?
– Xiaonan
Jul 27 at 10:08
Hi, in this case the value would be 3 in the second round, no score, but they would have one last opportunity in generating a 3 (one of them) in the last round.
– Rodrigo Teles
Jul 27 at 11:55
Hi, in this case the value would be 3 in the second round, no score, but they would have one last opportunity in generating a 3 (one of them) in the last round.
– Rodrigo Teles
Jul 27 at 11:55
add a comment |Â
1 Answer
1
active
oldest
votes
up vote
0
down vote
accepted
If we let $A$, $B$ and $O$ denote player $A$ scoring, player $B$ scoring and no one scoring, respectively, then the admissible patterns for the three seconds are $AAB$, $ABA$, $BAA$, $ABO$, $AOB$, $OAB$, and the corresponding patterns with $A$ and $B$ swapped. We can calculate the probability for each of them to occur. Here are the numbers that need to be generated (with $overline3$ denoting “not a $3$†and an empty cell indicating that the player doesn't generate a number in that round):
beginarrayc
&A_1&A_2&A_3&B_1&B_2&B_3&textprobability\hline
AAB&3&3&overline3&1&&&frac281\
ABA&3&2&&2&&overline3&frac281\
BAA&2&&3&3&1&&frac181\
ABO&3&1&&2&&overline3&frac281\
AOB&3&1&&1&&½7\
OAB&2&&overline3&1&&&frac227
endarray
We need to add up all these probabilities and multiply by $2$ for the possibility of swapping $A$ and $B$, so the total probability that both players score at least once is
$$
2left(frac281+frac281+frac181+frac281+frac381+frac681right)=frac3281=frac2^53^4approx39.5%;.
$$
answered Jul 27 at 20:00
joriki
164k10179328
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
accepted
If we let $A$, $B$ and $O$ denote player $A$ scoring, player $B$ scoring and no one scoring, respectively, then the admissible patterns for the three seconds are $AAB$, $ABA$, $BAA$, $ABO$, $AOB$, $OAB$, and the corresponding patterns with $A$ and $B$ swapped. We can calculate the probability for each of them to occur. Here are the numbers that need to be generated (with $overline3$ denoting “not a $3$†and an empty cell indicating that the player doesn't generate a number in that round):
beginarrayc
&A_1&A_2&A_3&B_1&B_2&B_3&textprobability\hline
AAB&3&3&overline3&1&&&frac281\
ABA&3&2&&2&&overline3&frac281\
BAA&2&&3&3&1&&frac181\
ABO&3&1&&2&&overline3&frac281\
AOB&3&1&&1&&½7\
OAB&2&&overline3&1&&&frac227
endarray
We need to add up all these probabilities and multiply by $2$ for the possibility of swapping $A$ and $B$, so the total probability that both players score at least once is
$$
2left(frac281+frac281+frac181+frac281+frac381+frac681right)=frac3281=frac2^53^4approx39.5%;.
$$
answered Jul 27 at 20:00
joriki
164k10179328
add a comment |Â
up vote
0
down vote
accepted
If we let $A$, $B$ and $O$ denote player $A$ scoring, player $B$ scoring and no one scoring, respectively, then the admissible patterns for the three seconds are $AAB$, $ABA$, $BAA$, $ABO$, $AOB$, $OAB$, and the corresponding patterns with $A$ and $B$ swapped. We can calculate the probability for each of them to occur. Here are the numbers that need to be generated (with $overline3$ denoting “not a $3$†and an empty cell indicating that the player doesn't generate a number in that round):
beginarrayc
&A_1&A_2&A_3&B_1&B_2&B_3&textprobability\hline
AAB&3&3&overline3&1&&&frac281\
ABA&3&2&&2&&overline3&frac281\
BAA&2&&3&3&1&&frac181\
ABO&3&1&&2&&overline3&frac281\
AOB&3&1&&1&&½7\
OAB&2&&overline3&1&&&frac227
endarray
We need to add up all these probabilities and multiply by $2$ for the possibility of swapping $A$ and $B$, so the total probability that both players score at least once is
$$
2left(frac281+frac281+frac181+frac281+frac381+frac681right)=frac3281=frac2^53^4approx39.5%;.
$$
answered Jul 27 at 20:00
joriki
164k10179328
add a comment |Â
up vote
0
down vote
accepted
up vote
0
down vote
accepted
If we let $A$, $B$ and $O$ denote player $A$ scoring, player $B$ scoring and no one scoring, respectively, then the admissible patterns for the three seconds are $AAB$, $ABA$, $BAA$, $ABO$, $AOB$, $OAB$, and the corresponding patterns with $A$ and $B$ swapped. We can calculate the probability for each of them to occur. Here are the numbers that need to be generated (with $overline3$ denoting “not a $3$†and an empty cell indicating that the player doesn't generate a number in that round):
beginarrayc
&A_1&A_2&A_3&B_1&B_2&B_3&textprobability\hline
AAB&3&3&overline3&1&&&frac281\
ABA&3&2&&2&&overline3&frac281\
BAA&2&&3&3&1&&frac181\
ABO&3&1&&2&&overline3&frac281\
AOB&3&1&&1&&½7\
OAB&2&&overline3&1&&&frac227
endarray
We need to add up all these probabilities and multiply by $2$ for the possibility of swapping $A$ and $B$, so the total probability that both players score at least once is
$$
2left(frac281+frac281+frac181+frac281+frac381+frac681right)=frac3281=frac2^53^4approx39.5%;.
$$
answered Jul 27 at 20:00
joriki
164k10179328
If we let $A$, $B$ and $O$ denote player $A$ scoring, player $B$ scoring and no one scoring, respectively, then the admissible patterns for the three seconds are $AAB$, $ABA$, $BAA$, $ABO$, $AOB$, $OAB$, and the corresponding patterns with $A$ and $B$ swapped. We can calculate the probability for each of them to occur. Here are the numbers that need to be generated (with $overline3$ denoting “not a $3$†and an empty cell indicating that the player doesn't generate a number in that round):
beginarrayc
&A_1&A_2&A_3&B_1&B_2&B_3&textprobability\hline
AAB&3&3&overline3&1&&&frac281\
ABA&3&2&&2&&overline3&frac281\
BAA&2&&3&3&1&&frac181\
ABO&3&1&&2&&overline3&frac281\
AOB&3&1&&1&&½7\
OAB&2&&overline3&1&&&frac227
endarray
We need to add up all these probabilities and multiply by $2$ for the possibility of swapping $A$ and $B$, so the total probability that both players score at least once is
$$
2left(frac281+frac281+frac181+frac281+frac381+frac681right)=frac3281=frac2^53^4approx39.5%;.
$$
answered Jul 27 at 20:00
joriki
164k10179328
answered Jul 27 at 20:00
joriki
164k10179328
answered Jul 27 at 20:00
joriki
164k10179328
answered Jul 27 at 20:00
joriki
164k10179328
164k10179328
add a comment |Â
add a comment |Â
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%2f2864118%2fprobability-of-each-player-scores-at-least-once-in-a-simple-random-game%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
By "number" you mean *natural number"? What about $2.7664455519$? If so, the chance that the player chooses $3$ is $0$.
– David G. Stork
Jul 27 at 7:09
What happens if e.g. both players generate a 3 in first round? They do not score then, but I guess the 3 will not be incremented (correct me if I am wrong). Do both generate a new number then?
– drhab
Jul 27 at 7:09
If they generate 3 in the first round they don't score. They generate a new number for the next round.
– Rodrigo Teles
Jul 27 at 7:32
How about first two rounds both generate 2? Will this score and leave 1 OR no score?
– Xiaonan
Jul 27 at 10:08
Hi, in this case the value would be 3 in the second round, no score, but they would have one last opportunity in generating a 3 (one of them) in the last round.
– Rodrigo Teles
Jul 27 at 11:55