Mixing
If a sports team is down by 1-2 in a best of 7 series, what is their chance of winning (if both teams have a...
$begingroup$
I know that the team has to win 3 of the 4 remaining games. So I thought that from the 4 remaining games, there are a total of $2^4 = 16$ total different outcomes.
Could someone point me towards the right direction to determine the probability of the team winning 3 of the 4 remaining games? Would it be 4C3/16 since the three wins can be chosen amongst the 4 games left?
probability statistics
edited Jan 14 at 21:51
David G. Stork
11k41432
asked Jan 14 at 21:44
Tim WeahTim Weah
865
$endgroup$
|
show 3 more comments
$begingroup$
I know that the team has to win 3 of the 4 remaining games. So I thought that from the 4 remaining games, there are a total of $2^4 = 16$ total different outcomes.
Could someone point me towards the right direction to determine the probability of the team winning 3 of the 4 remaining games? Would it be 4C3/16 since the three wins can be chosen amongst the 4 games left?
probability statistics
edited Jan 14 at 21:51
David G. Stork
11k41432
asked Jan 14 at 21:44
Tim WeahTim Weah
865
$endgroup$
2
$begingroup$
The winning scenarios are $WWW, WWLW, WLWW, LWWW$ so... If it helps, you can think of $WWW$ as short hand for the two scenarios $WWWW$ and $WWWL$.
$endgroup$
– lulu
Jan 14 at 21:48
$begingroup$
You can draw a tree diagram, if you're familiar with that.
$endgroup$
– saulspatz
Jan 14 at 21:50
$begingroup$
Note: your thought, $binom 43times frac 1{16}$, is close but not correct since your team might win the next three (in which case there is no "missing" slot).
$endgroup$
– lulu
Jan 14 at 21:51
$begingroup$
Drawing a tree diagram helped me, but is there a way to approach this problem without a diagram? I'm having difficulty excluding the probabilities that is impossible to occur, such as the team winning all of the remaining four games.
$endgroup$
– Tim Weah
Jan 14 at 21:53
$begingroup$
Either the series ends in exactly $2$ more games, or $3$ more games, or $4$ more games. For each case, calculate the probability each team wins in exactly that number of additional games. Then sum these probabilities for the given team.
$endgroup$
– David G. Stork
Jan 14 at 21:59
|
show 3 more comments
$begingroup$
I know that the team has to win 3 of the 4 remaining games. So I thought that from the 4 remaining games, there are a total of $2^4 = 16$ total different outcomes.
Could someone point me towards the right direction to determine the probability of the team winning 3 of the 4 remaining games? Would it be 4C3/16 since the three wins can be chosen amongst the 4 games left?
probability statistics
edited Jan 14 at 21:51
David G. Stork
11k41432
asked Jan 14 at 21:44
Tim WeahTim Weah
865
$endgroup$
I know that the team has to win 3 of the 4 remaining games. So I thought that from the 4 remaining games, there are a total of $2^4 = 16$ total different outcomes.
Could someone point me towards the right direction to determine the probability of the team winning 3 of the 4 remaining games? Would it be 4C3/16 since the three wins can be chosen amongst the 4 games left?
probability statistics
probability statistics
edited Jan 14 at 21:51
David G. Stork
11k41432
asked Jan 14 at 21:44
Tim WeahTim Weah
865
edited Jan 14 at 21:51
David G. Stork
11k41432
asked Jan 14 at 21:44
Tim WeahTim Weah
865
edited Jan 14 at 21:51
David G. Stork
11k41432
edited Jan 14 at 21:51
David G. Stork
11k41432
edited Jan 14 at 21:51
David G. Stork
11k41432
11k41432
asked Jan 14 at 21:44
Tim WeahTim Weah
865
asked Jan 14 at 21:44
Tim WeahTim Weah
865
asked Jan 14 at 21:44
Tim WeahTim Weah
865
865
2
$begingroup$
The winning scenarios are $WWW, WWLW, WLWW, LWWW$ so... If it helps, you can think of $WWW$ as short hand for the two scenarios $WWWW$ and $WWWL$.
$endgroup$
– lulu
Jan 14 at 21:48
$begingroup$
You can draw a tree diagram, if you're familiar with that.
$endgroup$
– saulspatz
Jan 14 at 21:50
$begingroup$
Note: your thought, $binom 43times frac 1{16}$, is close but not correct since your team might win the next three (in which case there is no "missing" slot).
$endgroup$
– lulu
Jan 14 at 21:51
$begingroup$
Drawing a tree diagram helped me, but is there a way to approach this problem without a diagram? I'm having difficulty excluding the probabilities that is impossible to occur, such as the team winning all of the remaining four games.
$endgroup$
– Tim Weah
Jan 14 at 21:53
$begingroup$
Either the series ends in exactly $2$ more games, or $3$ more games, or $4$ more games. For each case, calculate the probability each team wins in exactly that number of additional games. Then sum these probabilities for the given team.
$endgroup$
– David G. Stork
Jan 14 at 21:59
|
show 3 more comments
2
$begingroup$
The winning scenarios are $WWW, WWLW, WLWW, LWWW$ so... If it helps, you can think of $WWW$ as short hand for the two scenarios $WWWW$ and $WWWL$.
$endgroup$
– lulu
Jan 14 at 21:48
$begingroup$
You can draw a tree diagram, if you're familiar with that.
$endgroup$
– saulspatz
Jan 14 at 21:50
$begingroup$
Note: your thought, $binom 43times frac 1{16}$, is close but not correct since your team might win the next three (in which case there is no "missing" slot).
$endgroup$
– lulu
Jan 14 at 21:51
$begingroup$
Drawing a tree diagram helped me, but is there a way to approach this problem without a diagram? I'm having difficulty excluding the probabilities that is impossible to occur, such as the team winning all of the remaining four games.
$endgroup$
– Tim Weah
Jan 14 at 21:53
$begingroup$
Either the series ends in exactly $2$ more games, or $3$ more games, or $4$ more games. For each case, calculate the probability each team wins in exactly that number of additional games. Then sum these probabilities for the given team.
$endgroup$
– David G. Stork
Jan 14 at 21:59
2
2
$begingroup$
The winning scenarios are $WWW, WWLW, WLWW, LWWW$ so... If it helps, you can think of $WWW$ as short hand for the two scenarios $WWWW$ and $WWWL$.
$endgroup$
– lulu
Jan 14 at 21:48
$begingroup$
The winning scenarios are $WWW, WWLW, WLWW, LWWW$ so... If it helps, you can think of $WWW$ as short hand for the two scenarios $WWWW$ and $WWWL$.
$endgroup$
– lulu
Jan 14 at 21:48
$begingroup$
You can draw a tree diagram, if you're familiar with that.
$endgroup$
– saulspatz
Jan 14 at 21:50
$begingroup$
You can draw a tree diagram, if you're familiar with that.
$endgroup$
– saulspatz
Jan 14 at 21:50
$begingroup$
Note: your thought, $binom 43times frac 1{16}$, is close but not correct since your team might win the next three (in which case there is no "missing" slot).
$endgroup$
– lulu
Jan 14 at 21:51
$begingroup$
Note: your thought, $binom 43times frac 1{16}$, is close but not correct since your team might win the next three (in which case there is no "missing" slot).
$endgroup$
– lulu
Jan 14 at 21:51
$begingroup$
Drawing a tree diagram helped me, but is there a way to approach this problem without a diagram? I'm having difficulty excluding the probabilities that is impossible to occur, such as the team winning all of the remaining four games.
$endgroup$
– Tim Weah
Jan 14 at 21:53
$begingroup$
Drawing a tree diagram helped me, but is there a way to approach this problem without a diagram? I'm having difficulty excluding the probabilities that is impossible to occur, such as the team winning all of the remaining four games.
$endgroup$
– Tim Weah
Jan 14 at 21:53
$begingroup$
Either the series ends in exactly $2$ more games, or $3$ more games, or $4$ more games. For each case, calculate the probability each team wins in exactly that number of additional games. Then sum these probabilities for the given team.
$endgroup$
– David G. Stork
Jan 14 at 21:59
$begingroup$
Either the series ends in exactly $2$ more games, or $3$ more games, or $4$ more games. For each case, calculate the probability each team wins in exactly that number of additional games. Then sum these probabilities for the given team.
$endgroup$
– David G. Stork
Jan 14 at 21:59
|
show 3 more comments
2 Answers
2
active
oldest
votes
$begingroup$
The team that's behind needs to win three of the remaining four games.
One possibility is to explicitly write out all the possible scenarios: WWW, WWLW, WLWW, LWWW. These have probability 1/8, 1/16, 1/16, and 1/16, so the probability of winning is the sum of these, 5/16.
Alternatively, imagine that all four remaining games get played even if they don't need to be. Then the probability that the team that's behind wins at least three of its games is just the probability that a binomial(4, 1/2) random variable is at least 3 - that's
$$ {{4 choose 3} + {4 choose 4} over 2^4} = {4 + 1 over 16} = {5 over 16}$$
answered Jan 14 at 22:01
Michael LugoMichael Lugo
18.1k33576
$endgroup$
$begingroup$
for your second method, aren't you double counting some scenarios, such as WWW and WWWW?
$endgroup$
– Tim Weah
Jan 14 at 22:23
$begingroup$
with a tree diagram, the probability I got is 2/5. The list of scenarios are: WWW, WWLW, WWLL, WLWW, WLWL, WLL, LWWW, LWWL, LWL, LL, where the team wins on 4/10 of those.
$endgroup$
– Tim Weah
Jan 14 at 22:27
$begingroup$
@TimWeah In the second method, WWW doesn't appear, because they are looking at 4-game combinations, and that's only three games. For your calculation of probabilities, are you weighting each scenario by the probability of seeing that scenario?
$endgroup$
– Acccumulation
Jan 14 at 22:54
$begingroup$
@Acccumulation I just used a tree diagram (like the one below), and got 2/5. I'm just trying to figure out a way to show my solution numerically, but am having difficulty.
$endgroup$
– Tim Weah
Jan 14 at 22:57
add a comment |
$begingroup$
Starting with a reply to the OP's request for a tree:
Calculate the probabilities of getting to a "winning" end $42$ and $43$, being sure to add the probabilities along all paths:
$underbrace{(1/2)^3}_{{rm path~to~} 42} + underbrace{3 (1/2)^4}_{{rm paths~to~} 43} = 5/16$
answered Jan 14 at 22:47
David G. StorkDavid G. Stork
11k41432
$endgroup$
$begingroup$
Thats exactly what I have, but (since a tree diagram is not sufficient enough for this course), I'm trying to figure out a way to show this numerically. (not sure if thats the correct term)
$endgroup$
– Tim Weah
Jan 14 at 22:51
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%2f3073796%2fif-a-sports-team-is-down-by-1-2-in-a-best-of-7-series-what-is-their-chance-of-w%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
$begingroup$
The team that's behind needs to win three of the remaining four games.
One possibility is to explicitly write out all the possible scenarios: WWW, WWLW, WLWW, LWWW. These have probability 1/8, 1/16, 1/16, and 1/16, so the probability of winning is the sum of these, 5/16.
Alternatively, imagine that all four remaining games get played even if they don't need to be. Then the probability that the team that's behind wins at least three of its games is just the probability that a binomial(4, 1/2) random variable is at least 3 - that's
$$ {{4 choose 3} + {4 choose 4} over 2^4} = {4 + 1 over 16} = {5 over 16}$$
answered Jan 14 at 22:01
Michael LugoMichael Lugo
18.1k33576
$endgroup$
$begingroup$
for your second method, aren't you double counting some scenarios, such as WWW and WWWW?
$endgroup$
– Tim Weah
Jan 14 at 22:23
$begingroup$
with a tree diagram, the probability I got is 2/5. The list of scenarios are: WWW, WWLW, WWLL, WLWW, WLWL, WLL, LWWW, LWWL, LWL, LL, where the team wins on 4/10 of those.
$endgroup$
– Tim Weah
Jan 14 at 22:27
$begingroup$
@TimWeah In the second method, WWW doesn't appear, because they are looking at 4-game combinations, and that's only three games. For your calculation of probabilities, are you weighting each scenario by the probability of seeing that scenario?
$endgroup$
– Acccumulation
Jan 14 at 22:54
$begingroup$
@Acccumulation I just used a tree diagram (like the one below), and got 2/5. I'm just trying to figure out a way to show my solution numerically, but am having difficulty.
$endgroup$
– Tim Weah
Jan 14 at 22:57
add a comment |
$begingroup$
The team that's behind needs to win three of the remaining four games.
One possibility is to explicitly write out all the possible scenarios: WWW, WWLW, WLWW, LWWW. These have probability 1/8, 1/16, 1/16, and 1/16, so the probability of winning is the sum of these, 5/16.
Alternatively, imagine that all four remaining games get played even if they don't need to be. Then the probability that the team that's behind wins at least three of its games is just the probability that a binomial(4, 1/2) random variable is at least 3 - that's
$$ {{4 choose 3} + {4 choose 4} over 2^4} = {4 + 1 over 16} = {5 over 16}$$
answered Jan 14 at 22:01
Michael LugoMichael Lugo
18.1k33576
$endgroup$
$begingroup$
for your second method, aren't you double counting some scenarios, such as WWW and WWWW?
$endgroup$
– Tim Weah
Jan 14 at 22:23
$begingroup$
with a tree diagram, the probability I got is 2/5. The list of scenarios are: WWW, WWLW, WWLL, WLWW, WLWL, WLL, LWWW, LWWL, LWL, LL, where the team wins on 4/10 of those.
$endgroup$
– Tim Weah
Jan 14 at 22:27
$begingroup$
@TimWeah In the second method, WWW doesn't appear, because they are looking at 4-game combinations, and that's only three games. For your calculation of probabilities, are you weighting each scenario by the probability of seeing that scenario?
$endgroup$
– Acccumulation
Jan 14 at 22:54
$begingroup$
@Acccumulation I just used a tree diagram (like the one below), and got 2/5. I'm just trying to figure out a way to show my solution numerically, but am having difficulty.
$endgroup$
– Tim Weah
Jan 14 at 22:57
add a comment |
$begingroup$
The team that's behind needs to win three of the remaining four games.
One possibility is to explicitly write out all the possible scenarios: WWW, WWLW, WLWW, LWWW. These have probability 1/8, 1/16, 1/16, and 1/16, so the probability of winning is the sum of these, 5/16.
Alternatively, imagine that all four remaining games get played even if they don't need to be. Then the probability that the team that's behind wins at least three of its games is just the probability that a binomial(4, 1/2) random variable is at least 3 - that's
$$ {{4 choose 3} + {4 choose 4} over 2^4} = {4 + 1 over 16} = {5 over 16}$$
answered Jan 14 at 22:01
Michael LugoMichael Lugo
18.1k33576
$endgroup$
The team that's behind needs to win three of the remaining four games.
One possibility is to explicitly write out all the possible scenarios: WWW, WWLW, WLWW, LWWW. These have probability 1/8, 1/16, 1/16, and 1/16, so the probability of winning is the sum of these, 5/16.
Alternatively, imagine that all four remaining games get played even if they don't need to be. Then the probability that the team that's behind wins at least three of its games is just the probability that a binomial(4, 1/2) random variable is at least 3 - that's
$$ {{4 choose 3} + {4 choose 4} over 2^4} = {4 + 1 over 16} = {5 over 16}$$
answered Jan 14 at 22:01
Michael LugoMichael Lugo
18.1k33576
answered Jan 14 at 22:01
Michael LugoMichael Lugo
18.1k33576
answered Jan 14 at 22:01
Michael LugoMichael Lugo
18.1k33576
answered Jan 14 at 22:01
Michael LugoMichael Lugo
18.1k33576
18.1k33576
$begingroup$
for your second method, aren't you double counting some scenarios, such as WWW and WWWW?
$endgroup$
– Tim Weah
Jan 14 at 22:23
$begingroup$
with a tree diagram, the probability I got is 2/5. The list of scenarios are: WWW, WWLW, WWLL, WLWW, WLWL, WLL, LWWW, LWWL, LWL, LL, where the team wins on 4/10 of those.
$endgroup$
– Tim Weah
Jan 14 at 22:27
$begingroup$
@TimWeah In the second method, WWW doesn't appear, because they are looking at 4-game combinations, and that's only three games. For your calculation of probabilities, are you weighting each scenario by the probability of seeing that scenario?
$endgroup$
– Acccumulation
Jan 14 at 22:54
$begingroup$
@Acccumulation I just used a tree diagram (like the one below), and got 2/5. I'm just trying to figure out a way to show my solution numerically, but am having difficulty.
$endgroup$
– Tim Weah
Jan 14 at 22:57
add a comment |
$begingroup$
for your second method, aren't you double counting some scenarios, such as WWW and WWWW?
$endgroup$
– Tim Weah
Jan 14 at 22:23
$begingroup$
with a tree diagram, the probability I got is 2/5. The list of scenarios are: WWW, WWLW, WWLL, WLWW, WLWL, WLL, LWWW, LWWL, LWL, LL, where the team wins on 4/10 of those.
$endgroup$
– Tim Weah
Jan 14 at 22:27
$begingroup$
@TimWeah In the second method, WWW doesn't appear, because they are looking at 4-game combinations, and that's only three games. For your calculation of probabilities, are you weighting each scenario by the probability of seeing that scenario?
$endgroup$
– Acccumulation
Jan 14 at 22:54
$begingroup$
@Acccumulation I just used a tree diagram (like the one below), and got 2/5. I'm just trying to figure out a way to show my solution numerically, but am having difficulty.
$endgroup$
– Tim Weah
Jan 14 at 22:57
$begingroup$
for your second method, aren't you double counting some scenarios, such as WWW and WWWW?
$endgroup$
– Tim Weah
Jan 14 at 22:23
$begingroup$
for your second method, aren't you double counting some scenarios, such as WWW and WWWW?
$endgroup$
– Tim Weah
Jan 14 at 22:23
$begingroup$
with a tree diagram, the probability I got is 2/5. The list of scenarios are: WWW, WWLW, WWLL, WLWW, WLWL, WLL, LWWW, LWWL, LWL, LL, where the team wins on 4/10 of those.
$endgroup$
– Tim Weah
Jan 14 at 22:27
$begingroup$
with a tree diagram, the probability I got is 2/5. The list of scenarios are: WWW, WWLW, WWLL, WLWW, WLWL, WLL, LWWW, LWWL, LWL, LL, where the team wins on 4/10 of those.
$endgroup$
– Tim Weah
Jan 14 at 22:27
$begingroup$
@TimWeah In the second method, WWW doesn't appear, because they are looking at 4-game combinations, and that's only three games. For your calculation of probabilities, are you weighting each scenario by the probability of seeing that scenario?
$endgroup$
– Acccumulation
Jan 14 at 22:54
$begingroup$
@TimWeah In the second method, WWW doesn't appear, because they are looking at 4-game combinations, and that's only three games. For your calculation of probabilities, are you weighting each scenario by the probability of seeing that scenario?
$endgroup$
– Acccumulation
Jan 14 at 22:54
$begingroup$
@Acccumulation I just used a tree diagram (like the one below), and got 2/5. I'm just trying to figure out a way to show my solution numerically, but am having difficulty.
$endgroup$
– Tim Weah
Jan 14 at 22:57
$begingroup$
@Acccumulation I just used a tree diagram (like the one below), and got 2/5. I'm just trying to figure out a way to show my solution numerically, but am having difficulty.
$endgroup$
– Tim Weah
Jan 14 at 22:57
add a comment |
$begingroup$
Starting with a reply to the OP's request for a tree:
Calculate the probabilities of getting to a "winning" end $42$ and $43$, being sure to add the probabilities along all paths:
$underbrace{(1/2)^3}_{{rm path~to~} 42} + underbrace{3 (1/2)^4}_{{rm paths~to~} 43} = 5/16$
answered Jan 14 at 22:47
David G. StorkDavid G. Stork
11k41432
$endgroup$
$begingroup$
Thats exactly what I have, but (since a tree diagram is not sufficient enough for this course), I'm trying to figure out a way to show this numerically. (not sure if thats the correct term)
$endgroup$
– Tim Weah
Jan 14 at 22:51
add a comment |
$begingroup$
Starting with a reply to the OP's request for a tree:
Calculate the probabilities of getting to a "winning" end $42$ and $43$, being sure to add the probabilities along all paths:
$underbrace{(1/2)^3}_{{rm path~to~} 42} + underbrace{3 (1/2)^4}_{{rm paths~to~} 43} = 5/16$
answered Jan 14 at 22:47
David G. StorkDavid G. Stork
11k41432
$endgroup$
$begingroup$
Thats exactly what I have, but (since a tree diagram is not sufficient enough for this course), I'm trying to figure out a way to show this numerically. (not sure if thats the correct term)
$endgroup$
– Tim Weah
Jan 14 at 22:51
add a comment |
$begingroup$
Starting with a reply to the OP's request for a tree:
Calculate the probabilities of getting to a "winning" end $42$ and $43$, being sure to add the probabilities along all paths:
$underbrace{(1/2)^3}_{{rm path~to~} 42} + underbrace{3 (1/2)^4}_{{rm paths~to~} 43} = 5/16$
answered Jan 14 at 22:47
David G. StorkDavid G. Stork
11k41432
$endgroup$
Starting with a reply to the OP's request for a tree:
Calculate the probabilities of getting to a "winning" end $42$ and $43$, being sure to add the probabilities along all paths:
$underbrace{(1/2)^3}_{{rm path~to~} 42} + underbrace{3 (1/2)^4}_{{rm paths~to~} 43} = 5/16$
answered Jan 14 at 22:47
David G. StorkDavid G. Stork
11k41432
edited Jan 14 at 23:03
answered Jan 14 at 22:47
David G. StorkDavid G. Stork
11k41432
answered Jan 14 at 22:47
David G. StorkDavid G. Stork
11k41432
answered Jan 14 at 22:47
David G. StorkDavid G. Stork
11k41432
11k41432
$begingroup$
Thats exactly what I have, but (since a tree diagram is not sufficient enough for this course), I'm trying to figure out a way to show this numerically. (not sure if thats the correct term)
$endgroup$
– Tim Weah
Jan 14 at 22:51
add a comment |
$begingroup$
Thats exactly what I have, but (since a tree diagram is not sufficient enough for this course), I'm trying to figure out a way to show this numerically. (not sure if thats the correct term)
$endgroup$
– Tim Weah
Jan 14 at 22:51
$begingroup$
Thats exactly what I have, but (since a tree diagram is not sufficient enough for this course), I'm trying to figure out a way to show this numerically. (not sure if thats the correct term)
$endgroup$
– Tim Weah
Jan 14 at 22:51
$begingroup$
Thats exactly what I have, but (since a tree diagram is not sufficient enough for this course), I'm trying to figure out a way to show this numerically. (not sure if thats the correct term)
$endgroup$
– Tim Weah
Jan 14 at 22:51
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%2f3073796%2fif-a-sports-team-is-down-by-1-2-in-a-best-of-7-series-what-is-their-chance-of-w%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
2
$begingroup$
The winning scenarios are $WWW, WWLW, WLWW, LWWW$ so... If it helps, you can think of $WWW$ as short hand for the two scenarios $WWWW$ and $WWWL$.
$endgroup$
– lulu
Jan 14 at 21:48
$begingroup$
You can draw a tree diagram, if you're familiar with that.
$endgroup$
– saulspatz
Jan 14 at 21:50
$begingroup$
Note: your thought, $binom 43times frac 1{16}$, is close but not correct since your team might win the next three (in which case there is no "missing" slot).
$endgroup$
– lulu
Jan 14 at 21:51
$begingroup$
Drawing a tree diagram helped me, but is there a way to approach this problem without a diagram? I'm having difficulty excluding the probabilities that is impossible to occur, such as the team winning all of the remaining four games.
$endgroup$
– Tim Weah
Jan 14 at 21:53
$begingroup$
Either the series ends in exactly $2$ more games, or $3$ more games, or $4$ more games. For each case, calculate the probability each team wins in exactly that number of additional games. Then sum these probabilities for the given team.
$endgroup$
– David G. Stork
Jan 14 at 21:59