Expected profit of my simple board game
up vote
3
down vote
favorite
How to play:
Use 1 host and at least 1 player
Each player has to toss fair six-sided dice to go to goal.
If the player is at the 35th cell and tosses 2 or more, he can go to goal aa same as he tosses 1.
If the player reaches the goal in 9 tossing or less, the host has to pay to that player 1$ per 1 tossing less than 10.
For example, if the player reaches the goal in 7 tossing, host have to pay 3$ to that player.
If the player reaches the goal in 11 tossing or more, that player has to pay to host 1$ per 1 tossing more than 10.
For example, if the player reaches the goal in 12 tossing, that player has to pay 2$ to host.
If the player reaches the goal in 10 tossing, no one has to pay.
Each game will end only if the player reaches the goal.
Player can't pay 1$ and start new game if he can't reaches the goal in 11th tossing.
What is expected profit of host per player for each game?
As much as I know for this game, The expected value in rolling a six-sided die is 3.5.
The expected value of distance in 10 tossing is 35-cell but the goal is at 36-cell distance so expected profit of host is positive. If the goal is at 35th-cell, expected profit of host is 0. But I have no idea to calculate.
combinatorics contest-math puzzle combinatorial-game-theory
add a comment |
up vote
3
down vote
favorite
How to play:
Use 1 host and at least 1 player
Each player has to toss fair six-sided dice to go to goal.
If the player is at the 35th cell and tosses 2 or more, he can go to goal aa same as he tosses 1.
If the player reaches the goal in 9 tossing or less, the host has to pay to that player 1$ per 1 tossing less than 10.
For example, if the player reaches the goal in 7 tossing, host have to pay 3$ to that player.
If the player reaches the goal in 11 tossing or more, that player has to pay to host 1$ per 1 tossing more than 10.
For example, if the player reaches the goal in 12 tossing, that player has to pay 2$ to host.
If the player reaches the goal in 10 tossing, no one has to pay.
Each game will end only if the player reaches the goal.
Player can't pay 1$ and start new game if he can't reaches the goal in 11th tossing.
What is expected profit of host per player for each game?
As much as I know for this game, The expected value in rolling a six-sided die is 3.5.
The expected value of distance in 10 tossing is 35-cell but the goal is at 36-cell distance so expected profit of host is positive. If the goal is at 35th-cell, expected profit of host is 0. But I have no idea to calculate.
combinatorics contest-math puzzle combinatorial-game-theory
add a comment |
up vote
3
down vote
favorite
up vote
3
down vote
favorite
How to play:
Use 1 host and at least 1 player
Each player has to toss fair six-sided dice to go to goal.
If the player is at the 35th cell and tosses 2 or more, he can go to goal aa same as he tosses 1.
If the player reaches the goal in 9 tossing or less, the host has to pay to that player 1$ per 1 tossing less than 10.
For example, if the player reaches the goal in 7 tossing, host have to pay 3$ to that player.
If the player reaches the goal in 11 tossing or more, that player has to pay to host 1$ per 1 tossing more than 10.
For example, if the player reaches the goal in 12 tossing, that player has to pay 2$ to host.
If the player reaches the goal in 10 tossing, no one has to pay.
Each game will end only if the player reaches the goal.
Player can't pay 1$ and start new game if he can't reaches the goal in 11th tossing.
What is expected profit of host per player for each game?
As much as I know for this game, The expected value in rolling a six-sided die is 3.5.
The expected value of distance in 10 tossing is 35-cell but the goal is at 36-cell distance so expected profit of host is positive. If the goal is at 35th-cell, expected profit of host is 0. But I have no idea to calculate.
combinatorics contest-math puzzle combinatorial-game-theory
How to play:
Use 1 host and at least 1 player
Each player has to toss fair six-sided dice to go to goal.
If the player is at the 35th cell and tosses 2 or more, he can go to goal aa same as he tosses 1.
If the player reaches the goal in 9 tossing or less, the host has to pay to that player 1$ per 1 tossing less than 10.
For example, if the player reaches the goal in 7 tossing, host have to pay 3$ to that player.
If the player reaches the goal in 11 tossing or more, that player has to pay to host 1$ per 1 tossing more than 10.
For example, if the player reaches the goal in 12 tossing, that player has to pay 2$ to host.
If the player reaches the goal in 10 tossing, no one has to pay.
Each game will end only if the player reaches the goal.
Player can't pay 1$ and start new game if he can't reaches the goal in 11th tossing.
What is expected profit of host per player for each game?
As much as I know for this game, The expected value in rolling a six-sided die is 3.5.
The expected value of distance in 10 tossing is 35-cell but the goal is at 36-cell distance so expected profit of host is positive. If the goal is at 35th-cell, expected profit of host is 0. But I have no idea to calculate.
combinatorics contest-math puzzle combinatorial-game-theory
combinatorics contest-math puzzle combinatorial-game-theory
edited Nov 22 at 10:27
Tralala
739124
739124
asked Nov 22 at 6:17
uesdto signin
184
184
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
up vote
2
down vote
accepted
Here is the exact computation, taking all possible games into account. Idea: Consider the polynomial
$$p_j(x):=(x+x^2+x^3+x^4+x^5+x^6)^j .$$ The coefficient $[x^k]p_j(x)$ gives the number of $j$-tosses histories that bring the player exactly to square $k$. Since I'm not interested in squares $kgeq36$ I truncate $p_j(x)$ after the $x^{35}$ term. In this way I obtain the "truncated series" ${tt s[j]}$. The sum $sum_{k=0}^{35} [x^k]p_j(x)$ counts the number of games that are not over after $j$ tosses. Dividing this sum by $6^j$ gives the probability $p(j)$ that the game is not yet over after $j$ tosses, and $q(j):=p(j-1)-p(j)$ is the probability that the game ends with the $j^{rm th}$ toss. The expected gain for the host then is $sum_{j=1}^{36} (j-10)q(j)$.
If the goal is at square $35$ instead of $36$ the corresponding value is $0.476195$, and for $34$ it is $0.190481$ in favor of the host. In any case I suggest you write your own program and tune the parameters as desired.
I don't understand the program much. But is the expected profit per player for each game really 0.761905? I think it should be around 0-0.2. If it is really 0.761905, if the goal is at 35th-cell or even 34th-cell, the expected profit will still be positive.
– uesdto signin
Nov 22 at 9:41
$0.761905$ is the expected profit of host per game, as demanded in the question.
– Christian Blatter
Nov 22 at 10:08
Could you calculate the expected profit for case the goal is at 35th-cell and 34th-cell, please ?
– uesdto signin
Nov 22 at 10:47
Thank you very much. It is weird that the expected value in rolling a six-sided die is 3.5. But if the goal is at square 34, the expected profit for host is still positive. I think it should be negative. If I write program as same as you write in mathematica, will I get answer as same as you get (Is there anything else)?
– uesdto signin
Nov 22 at 11:48
I hope so. I simulated $1,000,000$ games with goal $=36$ and obtained $0.762758$.
– Christian Blatter
Nov 22 at 12:22
|
show 1 more comment
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
accepted
Here is the exact computation, taking all possible games into account. Idea: Consider the polynomial
$$p_j(x):=(x+x^2+x^3+x^4+x^5+x^6)^j .$$ The coefficient $[x^k]p_j(x)$ gives the number of $j$-tosses histories that bring the player exactly to square $k$. Since I'm not interested in squares $kgeq36$ I truncate $p_j(x)$ after the $x^{35}$ term. In this way I obtain the "truncated series" ${tt s[j]}$. The sum $sum_{k=0}^{35} [x^k]p_j(x)$ counts the number of games that are not over after $j$ tosses. Dividing this sum by $6^j$ gives the probability $p(j)$ that the game is not yet over after $j$ tosses, and $q(j):=p(j-1)-p(j)$ is the probability that the game ends with the $j^{rm th}$ toss. The expected gain for the host then is $sum_{j=1}^{36} (j-10)q(j)$.
If the goal is at square $35$ instead of $36$ the corresponding value is $0.476195$, and for $34$ it is $0.190481$ in favor of the host. In any case I suggest you write your own program and tune the parameters as desired.
I don't understand the program much. But is the expected profit per player for each game really 0.761905? I think it should be around 0-0.2. If it is really 0.761905, if the goal is at 35th-cell or even 34th-cell, the expected profit will still be positive.
– uesdto signin
Nov 22 at 9:41
$0.761905$ is the expected profit of host per game, as demanded in the question.
– Christian Blatter
Nov 22 at 10:08
Could you calculate the expected profit for case the goal is at 35th-cell and 34th-cell, please ?
– uesdto signin
Nov 22 at 10:47
Thank you very much. It is weird that the expected value in rolling a six-sided die is 3.5. But if the goal is at square 34, the expected profit for host is still positive. I think it should be negative. If I write program as same as you write in mathematica, will I get answer as same as you get (Is there anything else)?
– uesdto signin
Nov 22 at 11:48
I hope so. I simulated $1,000,000$ games with goal $=36$ and obtained $0.762758$.
– Christian Blatter
Nov 22 at 12:22
|
show 1 more comment
up vote
2
down vote
accepted
Here is the exact computation, taking all possible games into account. Idea: Consider the polynomial
$$p_j(x):=(x+x^2+x^3+x^4+x^5+x^6)^j .$$ The coefficient $[x^k]p_j(x)$ gives the number of $j$-tosses histories that bring the player exactly to square $k$. Since I'm not interested in squares $kgeq36$ I truncate $p_j(x)$ after the $x^{35}$ term. In this way I obtain the "truncated series" ${tt s[j]}$. The sum $sum_{k=0}^{35} [x^k]p_j(x)$ counts the number of games that are not over after $j$ tosses. Dividing this sum by $6^j$ gives the probability $p(j)$ that the game is not yet over after $j$ tosses, and $q(j):=p(j-1)-p(j)$ is the probability that the game ends with the $j^{rm th}$ toss. The expected gain for the host then is $sum_{j=1}^{36} (j-10)q(j)$.
If the goal is at square $35$ instead of $36$ the corresponding value is $0.476195$, and for $34$ it is $0.190481$ in favor of the host. In any case I suggest you write your own program and tune the parameters as desired.
I don't understand the program much. But is the expected profit per player for each game really 0.761905? I think it should be around 0-0.2. If it is really 0.761905, if the goal is at 35th-cell or even 34th-cell, the expected profit will still be positive.
– uesdto signin
Nov 22 at 9:41
$0.761905$ is the expected profit of host per game, as demanded in the question.
– Christian Blatter
Nov 22 at 10:08
Could you calculate the expected profit for case the goal is at 35th-cell and 34th-cell, please ?
– uesdto signin
Nov 22 at 10:47
Thank you very much. It is weird that the expected value in rolling a six-sided die is 3.5. But if the goal is at square 34, the expected profit for host is still positive. I think it should be negative. If I write program as same as you write in mathematica, will I get answer as same as you get (Is there anything else)?
– uesdto signin
Nov 22 at 11:48
I hope so. I simulated $1,000,000$ games with goal $=36$ and obtained $0.762758$.
– Christian Blatter
Nov 22 at 12:22
|
show 1 more comment
up vote
2
down vote
accepted
up vote
2
down vote
accepted
Here is the exact computation, taking all possible games into account. Idea: Consider the polynomial
$$p_j(x):=(x+x^2+x^3+x^4+x^5+x^6)^j .$$ The coefficient $[x^k]p_j(x)$ gives the number of $j$-tosses histories that bring the player exactly to square $k$. Since I'm not interested in squares $kgeq36$ I truncate $p_j(x)$ after the $x^{35}$ term. In this way I obtain the "truncated series" ${tt s[j]}$. The sum $sum_{k=0}^{35} [x^k]p_j(x)$ counts the number of games that are not over after $j$ tosses. Dividing this sum by $6^j$ gives the probability $p(j)$ that the game is not yet over after $j$ tosses, and $q(j):=p(j-1)-p(j)$ is the probability that the game ends with the $j^{rm th}$ toss. The expected gain for the host then is $sum_{j=1}^{36} (j-10)q(j)$.
If the goal is at square $35$ instead of $36$ the corresponding value is $0.476195$, and for $34$ it is $0.190481$ in favor of the host. In any case I suggest you write your own program and tune the parameters as desired.
Here is the exact computation, taking all possible games into account. Idea: Consider the polynomial
$$p_j(x):=(x+x^2+x^3+x^4+x^5+x^6)^j .$$ The coefficient $[x^k]p_j(x)$ gives the number of $j$-tosses histories that bring the player exactly to square $k$. Since I'm not interested in squares $kgeq36$ I truncate $p_j(x)$ after the $x^{35}$ term. In this way I obtain the "truncated series" ${tt s[j]}$. The sum $sum_{k=0}^{35} [x^k]p_j(x)$ counts the number of games that are not over after $j$ tosses. Dividing this sum by $6^j$ gives the probability $p(j)$ that the game is not yet over after $j$ tosses, and $q(j):=p(j-1)-p(j)$ is the probability that the game ends with the $j^{rm th}$ toss. The expected gain for the host then is $sum_{j=1}^{36} (j-10)q(j)$.
If the goal is at square $35$ instead of $36$ the corresponding value is $0.476195$, and for $34$ it is $0.190481$ in favor of the host. In any case I suggest you write your own program and tune the parameters as desired.
edited Nov 22 at 11:13
answered Nov 22 at 8:57
Christian Blatter
171k7111325
171k7111325
I don't understand the program much. But is the expected profit per player for each game really 0.761905? I think it should be around 0-0.2. If it is really 0.761905, if the goal is at 35th-cell or even 34th-cell, the expected profit will still be positive.
– uesdto signin
Nov 22 at 9:41
$0.761905$ is the expected profit of host per game, as demanded in the question.
– Christian Blatter
Nov 22 at 10:08
Could you calculate the expected profit for case the goal is at 35th-cell and 34th-cell, please ?
– uesdto signin
Nov 22 at 10:47
Thank you very much. It is weird that the expected value in rolling a six-sided die is 3.5. But if the goal is at square 34, the expected profit for host is still positive. I think it should be negative. If I write program as same as you write in mathematica, will I get answer as same as you get (Is there anything else)?
– uesdto signin
Nov 22 at 11:48
I hope so. I simulated $1,000,000$ games with goal $=36$ and obtained $0.762758$.
– Christian Blatter
Nov 22 at 12:22
|
show 1 more comment
I don't understand the program much. But is the expected profit per player for each game really 0.761905? I think it should be around 0-0.2. If it is really 0.761905, if the goal is at 35th-cell or even 34th-cell, the expected profit will still be positive.
– uesdto signin
Nov 22 at 9:41
$0.761905$ is the expected profit of host per game, as demanded in the question.
– Christian Blatter
Nov 22 at 10:08
Could you calculate the expected profit for case the goal is at 35th-cell and 34th-cell, please ?
– uesdto signin
Nov 22 at 10:47
Thank you very much. It is weird that the expected value in rolling a six-sided die is 3.5. But if the goal is at square 34, the expected profit for host is still positive. I think it should be negative. If I write program as same as you write in mathematica, will I get answer as same as you get (Is there anything else)?
– uesdto signin
Nov 22 at 11:48
I hope so. I simulated $1,000,000$ games with goal $=36$ and obtained $0.762758$.
– Christian Blatter
Nov 22 at 12:22
I don't understand the program much. But is the expected profit per player for each game really 0.761905? I think it should be around 0-0.2. If it is really 0.761905, if the goal is at 35th-cell or even 34th-cell, the expected profit will still be positive.
– uesdto signin
Nov 22 at 9:41
I don't understand the program much. But is the expected profit per player for each game really 0.761905? I think it should be around 0-0.2. If it is really 0.761905, if the goal is at 35th-cell or even 34th-cell, the expected profit will still be positive.
– uesdto signin
Nov 22 at 9:41
$0.761905$ is the expected profit of host per game, as demanded in the question.
– Christian Blatter
Nov 22 at 10:08
$0.761905$ is the expected profit of host per game, as demanded in the question.
– Christian Blatter
Nov 22 at 10:08
Could you calculate the expected profit for case the goal is at 35th-cell and 34th-cell, please ?
– uesdto signin
Nov 22 at 10:47
Could you calculate the expected profit for case the goal is at 35th-cell and 34th-cell, please ?
– uesdto signin
Nov 22 at 10:47
Thank you very much. It is weird that the expected value in rolling a six-sided die is 3.5. But if the goal is at square 34, the expected profit for host is still positive. I think it should be negative. If I write program as same as you write in mathematica, will I get answer as same as you get (Is there anything else)?
– uesdto signin
Nov 22 at 11:48
Thank you very much. It is weird that the expected value in rolling a six-sided die is 3.5. But if the goal is at square 34, the expected profit for host is still positive. I think it should be negative. If I write program as same as you write in mathematica, will I get answer as same as you get (Is there anything else)?
– uesdto signin
Nov 22 at 11:48
I hope so. I simulated $1,000,000$ games with goal $=36$ and obtained $0.762758$.
– Christian Blatter
Nov 22 at 12:22
I hope so. I simulated $1,000,000$ games with goal $=36$ and obtained $0.762758$.
– Christian Blatter
Nov 22 at 12:22
|
show 1 more comment
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