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
edited Nov 22 at 10:27
Vee Hua Zhi
776124
asked Nov 22 at 6:17
uesdto signin
184
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
edited Nov 22 at 10:27
Vee Hua Zhi
776124
asked Nov 22 at 6:17
uesdto signin
184
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
edited Nov 22 at 10:27
Vee Hua Zhi
776124
asked Nov 22 at 6:17
uesdto signin
184
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
Vee Hua Zhi
776124
asked Nov 22 at 6:17
uesdto signin
184
edited Nov 22 at 10:27
Vee Hua Zhi
776124
asked Nov 22 at 6:17
uesdto signin
184
edited Nov 22 at 10:27
Vee Hua Zhi
776124
edited Nov 22 at 10:27
Vee Hua Zhi
776124
edited Nov 22 at 10:27
Vee Hua Zhi
776124
776124
asked Nov 22 at 6:17
uesdto signin
184
asked Nov 22 at 6:17
uesdto signin
184
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.
answered Nov 22 at 8:57
Christian Blatter
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
Your Answer
StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");
StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "69"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});
function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});
}
});
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
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3008802%2fexpected-profit-of-my-simple-board-game%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
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.
answered Nov 22 at 8:57
Christian Blatter
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
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.
answered Nov 22 at 8:57
Christian Blatter
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
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.
answered Nov 22 at 8:57
Christian Blatter
171k7111325
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.
answered Nov 22 at 8:57
Christian Blatter
171k7111325
edited Nov 22 at 11:13
answered Nov 22 at 8:57
Christian Blatter
171k7111325
answered Nov 22 at 8:57
Christian Blatter
171k7111325
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
Thanks for contributing an answer to Mathematics Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Some of your past answers have not been well-received, and you're in danger of being blocked from answering.
Please pay close attention to the following guidance:
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
To learn more, see our tips on writing great answers.
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
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3008802%2fexpected-profit-of-my-simple-board-game%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
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
Required, but never shown
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
Required, but never shown
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
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
