Probability of profit.
$begingroup$
Consider this game:
Person A has 4 coins of 1 pound. Person B has 2 coins of 2 pounds. They are once throwing their own coin. Head-the owner keeps the coin. Tails- the owner gives this coin to the player (A/B). (But this coin doesn't take part(later) in the game). At the end of the game, they count the money. Do A and B have equal chances of getting a profit?
At first, I tried to write out all the possibilities. But this is not effective.
probability
edited Jan 8 at 13:15
amWhy
1
asked Jan 8 at 13:06
pawelKpawelK
527
$endgroup$
add a comment |
$begingroup$
Consider this game:
Person A has 4 coins of 1 pound. Person B has 2 coins of 2 pounds. They are once throwing their own coin. Head-the owner keeps the coin. Tails- the owner gives this coin to the player (A/B). (But this coin doesn't take part(later) in the game). At the end of the game, they count the money. Do A and B have equal chances of getting a profit?
At first, I tried to write out all the possibilities. But this is not effective.
probability
edited Jan 8 at 13:15
amWhy
1
asked Jan 8 at 13:06
pawelKpawelK
527
$endgroup$
add a comment |
$begingroup$
Consider this game:
Person A has 4 coins of 1 pound. Person B has 2 coins of 2 pounds. They are once throwing their own coin. Head-the owner keeps the coin. Tails- the owner gives this coin to the player (A/B). (But this coin doesn't take part(later) in the game). At the end of the game, they count the money. Do A and B have equal chances of getting a profit?
At first, I tried to write out all the possibilities. But this is not effective.
probability
edited Jan 8 at 13:15
amWhy
1
asked Jan 8 at 13:06
pawelKpawelK
527
$endgroup$
Consider this game:
Person A has 4 coins of 1 pound. Person B has 2 coins of 2 pounds. They are once throwing their own coin. Head-the owner keeps the coin. Tails- the owner gives this coin to the player (A/B). (But this coin doesn't take part(later) in the game). At the end of the game, they count the money. Do A and B have equal chances of getting a profit?
At first, I tried to write out all the possibilities. But this is not effective.
probability
probability
edited Jan 8 at 13:15
amWhy
1
asked Jan 8 at 13:06
pawelKpawelK
527
edited Jan 8 at 13:15
amWhy
1
asked Jan 8 at 13:06
pawelKpawelK
527
edited Jan 8 at 13:15
amWhy
1
edited Jan 8 at 13:15
amWhy
1
edited Jan 8 at 13:15
amWhy
1
1
asked Jan 8 at 13:06
pawelKpawelK
527
asked Jan 8 at 13:06
pawelKpawelK
527
asked Jan 8 at 13:06
pawelKpawelK
527
527
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
All we care about is the number of heads and tails each player throws.
For $A$ there are $5$ cases, as $A$ might throw from $0$ to $4$ tails. The probabilities are $$(.0625,.25,.375,.25,.0625)$$
For $B$ there are $3$ cases, as $B$ might throw from $0$ to $2$ tails. The probabilities are $$(.25,.5,.25)$$
Let's describe an outcome of the game as a pair $(a,b)$ where $A$ throws $a$ Tails and $B$ throws $b$ Tails. There are $5times 3=15$ possible outcomes.
$A$ turns a profit in the scenarios $$(0,1),,(0,2),, (1,1),,(1,2),, (2,2),,(3,2)$$
The probabilities of those cases sum to $boxed {.390625}$
$B$ turns a profit in the scenarios $$(1,0),,(2,0),,(3,0),,(4,0),,(3,1),,(4,1)$$ The probabilities of those cases also sum to $boxed {.390625}$
Just as a consistency test, ties occur in the scenarios $$(0,0),,(2,1),,(4,2)$$ The probabilities of those cases sum to $boxed {.21875}$ and the three total probabilities do indeed sum to $1$.
answered Jan 8 at 13:28
lulululu
42.2k25080
$endgroup$
add a comment |
$begingroup$
Person A throws 4 coins, and they has a $50%$ chance of keeping each one, so they expect, on average, to end the game with $4cdot0.5=2$ coins, worth 2 pounds. Similarly, Person B throws 2 coins, also with a $50%$ chance of keeping each one, and so they expect to end the game with $2cdot0.5=1$ coin, worth 2 pounds. So they both expect to end the game with 2 pounds worth of coins, meaning they each have equal chances to gain a profit from the game.
answered Jan 8 at 13:17
Calvin GodfreyCalvin Godfrey
633311
$endgroup$
$begingroup$
It's not obvious at least not to me, why saying the game has expectation $0$ for each means that they have the same chance of getting a profit. That is not true generally (though it is in this case)...imagine a game in which I have a $frac 34$ chance of getting $100$ from you and you have a $frac 14$ chance of getting $300$ from me. That has expectation $0$ for both of us but I have a much greater chance of getting a profit.
$endgroup$
– lulu
Jan 8 at 13:34
add a 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',
autoActivateHeartbeat: false,
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%2f3066142%2fprobability-of-profit%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
All we care about is the number of heads and tails each player throws.
For $A$ there are $5$ cases, as $A$ might throw from $0$ to $4$ tails. The probabilities are $$(.0625,.25,.375,.25,.0625)$$
For $B$ there are $3$ cases, as $B$ might throw from $0$ to $2$ tails. The probabilities are $$(.25,.5,.25)$$
Let's describe an outcome of the game as a pair $(a,b)$ where $A$ throws $a$ Tails and $B$ throws $b$ Tails. There are $5times 3=15$ possible outcomes.
$A$ turns a profit in the scenarios $$(0,1),,(0,2),, (1,1),,(1,2),, (2,2),,(3,2)$$
The probabilities of those cases sum to $boxed {.390625}$
$B$ turns a profit in the scenarios $$(1,0),,(2,0),,(3,0),,(4,0),,(3,1),,(4,1)$$ The probabilities of those cases also sum to $boxed {.390625}$
Just as a consistency test, ties occur in the scenarios $$(0,0),,(2,1),,(4,2)$$ The probabilities of those cases sum to $boxed {.21875}$ and the three total probabilities do indeed sum to $1$.
answered Jan 8 at 13:28
lulululu
42.2k25080
$endgroup$
add a comment |
$begingroup$
All we care about is the number of heads and tails each player throws.
For $A$ there are $5$ cases, as $A$ might throw from $0$ to $4$ tails. The probabilities are $$(.0625,.25,.375,.25,.0625)$$
For $B$ there are $3$ cases, as $B$ might throw from $0$ to $2$ tails. The probabilities are $$(.25,.5,.25)$$
Let's describe an outcome of the game as a pair $(a,b)$ where $A$ throws $a$ Tails and $B$ throws $b$ Tails. There are $5times 3=15$ possible outcomes.
$A$ turns a profit in the scenarios $$(0,1),,(0,2),, (1,1),,(1,2),, (2,2),,(3,2)$$
The probabilities of those cases sum to $boxed {.390625}$
$B$ turns a profit in the scenarios $$(1,0),,(2,0),,(3,0),,(4,0),,(3,1),,(4,1)$$ The probabilities of those cases also sum to $boxed {.390625}$
Just as a consistency test, ties occur in the scenarios $$(0,0),,(2,1),,(4,2)$$ The probabilities of those cases sum to $boxed {.21875}$ and the three total probabilities do indeed sum to $1$.
answered Jan 8 at 13:28
lulululu
42.2k25080
$endgroup$
add a comment |
$begingroup$
All we care about is the number of heads and tails each player throws.
For $A$ there are $5$ cases, as $A$ might throw from $0$ to $4$ tails. The probabilities are $$(.0625,.25,.375,.25,.0625)$$
For $B$ there are $3$ cases, as $B$ might throw from $0$ to $2$ tails. The probabilities are $$(.25,.5,.25)$$
Let's describe an outcome of the game as a pair $(a,b)$ where $A$ throws $a$ Tails and $B$ throws $b$ Tails. There are $5times 3=15$ possible outcomes.
$A$ turns a profit in the scenarios $$(0,1),,(0,2),, (1,1),,(1,2),, (2,2),,(3,2)$$
The probabilities of those cases sum to $boxed {.390625}$
$B$ turns a profit in the scenarios $$(1,0),,(2,0),,(3,0),,(4,0),,(3,1),,(4,1)$$ The probabilities of those cases also sum to $boxed {.390625}$
Just as a consistency test, ties occur in the scenarios $$(0,0),,(2,1),,(4,2)$$ The probabilities of those cases sum to $boxed {.21875}$ and the three total probabilities do indeed sum to $1$.
answered Jan 8 at 13:28
lulululu
42.2k25080
$endgroup$
All we care about is the number of heads and tails each player throws.
For $A$ there are $5$ cases, as $A$ might throw from $0$ to $4$ tails. The probabilities are $$(.0625,.25,.375,.25,.0625)$$
For $B$ there are $3$ cases, as $B$ might throw from $0$ to $2$ tails. The probabilities are $$(.25,.5,.25)$$
Let's describe an outcome of the game as a pair $(a,b)$ where $A$ throws $a$ Tails and $B$ throws $b$ Tails. There are $5times 3=15$ possible outcomes.
$A$ turns a profit in the scenarios $$(0,1),,(0,2),, (1,1),,(1,2),, (2,2),,(3,2)$$
The probabilities of those cases sum to $boxed {.390625}$
$B$ turns a profit in the scenarios $$(1,0),,(2,0),,(3,0),,(4,0),,(3,1),,(4,1)$$ The probabilities of those cases also sum to $boxed {.390625}$
Just as a consistency test, ties occur in the scenarios $$(0,0),,(2,1),,(4,2)$$ The probabilities of those cases sum to $boxed {.21875}$ and the three total probabilities do indeed sum to $1$.
answered Jan 8 at 13:28
lulululu
42.2k25080
edited Jan 8 at 13:36
answered Jan 8 at 13:28
lulululu
42.2k25080
answered Jan 8 at 13:28
lulululu
42.2k25080
answered Jan 8 at 13:28
lulululu
42.2k25080
42.2k25080
add a comment |
add a comment |
$begingroup$
Person A throws 4 coins, and they has a $50%$ chance of keeping each one, so they expect, on average, to end the game with $4cdot0.5=2$ coins, worth 2 pounds. Similarly, Person B throws 2 coins, also with a $50%$ chance of keeping each one, and so they expect to end the game with $2cdot0.5=1$ coin, worth 2 pounds. So they both expect to end the game with 2 pounds worth of coins, meaning they each have equal chances to gain a profit from the game.
answered Jan 8 at 13:17
Calvin GodfreyCalvin Godfrey
633311
$endgroup$
$begingroup$
It's not obvious at least not to me, why saying the game has expectation $0$ for each means that they have the same chance of getting a profit. That is not true generally (though it is in this case)...imagine a game in which I have a $frac 34$ chance of getting $100$ from you and you have a $frac 14$ chance of getting $300$ from me. That has expectation $0$ for both of us but I have a much greater chance of getting a profit.
$endgroup$
– lulu
Jan 8 at 13:34
add a comment |
$begingroup$
Person A throws 4 coins, and they has a $50%$ chance of keeping each one, so they expect, on average, to end the game with $4cdot0.5=2$ coins, worth 2 pounds. Similarly, Person B throws 2 coins, also with a $50%$ chance of keeping each one, and so they expect to end the game with $2cdot0.5=1$ coin, worth 2 pounds. So they both expect to end the game with 2 pounds worth of coins, meaning they each have equal chances to gain a profit from the game.
answered Jan 8 at 13:17
Calvin GodfreyCalvin Godfrey
633311
$endgroup$
$begingroup$
It's not obvious at least not to me, why saying the game has expectation $0$ for each means that they have the same chance of getting a profit. That is not true generally (though it is in this case)...imagine a game in which I have a $frac 34$ chance of getting $100$ from you and you have a $frac 14$ chance of getting $300$ from me. That has expectation $0$ for both of us but I have a much greater chance of getting a profit.
$endgroup$
– lulu
Jan 8 at 13:34
add a comment |
$begingroup$
Person A throws 4 coins, and they has a $50%$ chance of keeping each one, so they expect, on average, to end the game with $4cdot0.5=2$ coins, worth 2 pounds. Similarly, Person B throws 2 coins, also with a $50%$ chance of keeping each one, and so they expect to end the game with $2cdot0.5=1$ coin, worth 2 pounds. So they both expect to end the game with 2 pounds worth of coins, meaning they each have equal chances to gain a profit from the game.
answered Jan 8 at 13:17
Calvin GodfreyCalvin Godfrey
633311
$endgroup$
Person A throws 4 coins, and they has a $50%$ chance of keeping each one, so they expect, on average, to end the game with $4cdot0.5=2$ coins, worth 2 pounds. Similarly, Person B throws 2 coins, also with a $50%$ chance of keeping each one, and so they expect to end the game with $2cdot0.5=1$ coin, worth 2 pounds. So they both expect to end the game with 2 pounds worth of coins, meaning they each have equal chances to gain a profit from the game.
answered Jan 8 at 13:17
Calvin GodfreyCalvin Godfrey
633311
answered Jan 8 at 13:17
Calvin GodfreyCalvin Godfrey
633311
answered Jan 8 at 13:17
Calvin GodfreyCalvin Godfrey
633311
answered Jan 8 at 13:17
Calvin GodfreyCalvin Godfrey
633311
633311
$begingroup$
It's not obvious at least not to me, why saying the game has expectation $0$ for each means that they have the same chance of getting a profit. That is not true generally (though it is in this case)...imagine a game in which I have a $frac 34$ chance of getting $100$ from you and you have a $frac 14$ chance of getting $300$ from me. That has expectation $0$ for both of us but I have a much greater chance of getting a profit.
$endgroup$
– lulu
Jan 8 at 13:34
add a comment |
$begingroup$
It's not obvious at least not to me, why saying the game has expectation $0$ for each means that they have the same chance of getting a profit. That is not true generally (though it is in this case)...imagine a game in which I have a $frac 34$ chance of getting $100$ from you and you have a $frac 14$ chance of getting $300$ from me. That has expectation $0$ for both of us but I have a much greater chance of getting a profit.
$endgroup$
– lulu
Jan 8 at 13:34
$begingroup$
It's not obvious at least not to me, why saying the game has expectation $0$ for each means that they have the same chance of getting a profit. That is not true generally (though it is in this case)...imagine a game in which I have a $frac 34$ chance of getting $100$ from you and you have a $frac 14$ chance of getting $300$ from me. That has expectation $0$ for both of us but I have a much greater chance of getting a profit.
$endgroup$
– lulu
Jan 8 at 13:34
$begingroup$
It's not obvious at least not to me, why saying the game has expectation $0$ for each means that they have the same chance of getting a profit. That is not true generally (though it is in this case)...imagine a game in which I have a $frac 34$ chance of getting $100$ from you and you have a $frac 14$ chance of getting $300$ from me. That has expectation $0$ for both of us but I have a much greater chance of getting a profit.
$endgroup$
– lulu
Jan 8 at 13:34
add a 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.
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%2f3066142%2fprobability-of-profit%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
