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Expected score of a player in a best-of match
The 2019 Stack Overflow Developer Survey Results Are InExpected value equals sum of probabilitiesAn application of probability in tennis gameProb question on gamesProbability of winning a match with different probability of independently winning a game in the match?Probability of winning a game in tennis?Backgammon game competition.Which option should $A$ prefer so that the probability of his winning the match is higher?Given Elo ratings, what is the expected number of points in an X games match?Probability of a chess gameConditional Expectation in a match of Chess with maximum 3 pointsTotal probability of a match, where first to $n$ games wins.
$begingroup$
Let $n>0$ and the probability for player $A$ to win a single game against $B$ be $p$.
A match consists of multiple independent games.
What is the expected score-line of $A$
if the match ends when $A$ or $B$ wins $n$ games? (best-of-$(2n-1)$)
By simulation, it looks like below for $n=10$ and varying $p$, but whats the formula?
$n=10$ as a function of $p$.">
Sounds easy, but...
Help appreciated! Thanks in advance.
probability
edited Apr 6 at 22:51
Eric Wofsey
192k14220352
asked Apr 6 at 20:00
fwgbfwgb
1284
$endgroup$
|
show 4 more comments
$begingroup$
Let $n>0$ and the probability for player $A$ to win a single game against $B$ be $p$.
A match consists of multiple independent games.
What is the expected score-line of $A$
if the match ends when $A$ or $B$ wins $n$ games? (best-of-$(2n-1)$)
By simulation, it looks like below for $n=10$ and varying $p$, but whats the formula?
$n=10$ as a function of $p$.">
Sounds easy, but...
Help appreciated! Thanks in advance.
probability
edited Apr 6 at 22:51
Eric Wofsey
192k14220352
asked Apr 6 at 20:00
fwgbfwgb
1284
$endgroup$
$begingroup$
What do you mean by "score line"? Are you just asking for the probability that $A$ wins the match?
$endgroup$
– lulu
Apr 6 at 20:11
$begingroup$
Score line means the result e.g. 10-9, 10-1, or 8-10. For $A$ : 10, 10, 8.
$endgroup$
– fwgb
Apr 6 at 20:15
$begingroup$
Ok, well...the probability that $A$ wins the first $n$ is $p^n$. To get the probability that $A$ wins in exactly $n+1$ games, $A$ must win game $n+1$ and lose exactly one of the first $n$, hence $ntimes (1-p)*p^n$. Can you finish from here?
$endgroup$
– lulu
Apr 6 at 20:17
$begingroup$
But also scores where $A$ lost should contribute to the required expectation.
$endgroup$
– fwgb
Apr 6 at 20:30
$begingroup$
Yes, of course. That's why my expression has terms involving $(1-p)^i$.
$endgroup$
– lulu
Apr 6 at 20:36
|
show 4 more comments
$begingroup$
Let $n>0$ and the probability for player $A$ to win a single game against $B$ be $p$.
A match consists of multiple independent games.
What is the expected score-line of $A$
if the match ends when $A$ or $B$ wins $n$ games? (best-of-$(2n-1)$)
By simulation, it looks like below for $n=10$ and varying $p$, but whats the formula?
$n=10$ as a function of $p$.">
Sounds easy, but...
Help appreciated! Thanks in advance.
probability
edited Apr 6 at 22:51
Eric Wofsey
192k14220352
asked Apr 6 at 20:00
fwgbfwgb
1284
$endgroup$
Let $n>0$ and the probability for player $A$ to win a single game against $B$ be $p$.
A match consists of multiple independent games.
What is the expected score-line of $A$
if the match ends when $A$ or $B$ wins $n$ games? (best-of-$(2n-1)$)
By simulation, it looks like below for $n=10$ and varying $p$, but whats the formula?
$n=10$ as a function of $p$.">
Sounds easy, but...
Help appreciated! Thanks in advance.
probability
probability
edited Apr 6 at 22:51
Eric Wofsey
192k14220352
asked Apr 6 at 20:00
fwgbfwgb
1284
edited Apr 6 at 22:51
Eric Wofsey
192k14220352
asked Apr 6 at 20:00
fwgbfwgb
1284
edited Apr 6 at 22:51
Eric Wofsey
192k14220352
edited Apr 6 at 22:51
Eric Wofsey
192k14220352
edited Apr 6 at 22:51
Eric Wofsey
192k14220352
192k14220352
asked Apr 6 at 20:00
fwgbfwgb
1284
asked Apr 6 at 20:00
fwgbfwgb
1284
asked Apr 6 at 20:00
fwgbfwgb
1284
1284
$begingroup$
What do you mean by "score line"? Are you just asking for the probability that $A$ wins the match?
$endgroup$
– lulu
Apr 6 at 20:11
$begingroup$
Score line means the result e.g. 10-9, 10-1, or 8-10. For $A$ : 10, 10, 8.
$endgroup$
– fwgb
Apr 6 at 20:15
$begingroup$
Ok, well...the probability that $A$ wins the first $n$ is $p^n$. To get the probability that $A$ wins in exactly $n+1$ games, $A$ must win game $n+1$ and lose exactly one of the first $n$, hence $ntimes (1-p)*p^n$. Can you finish from here?
$endgroup$
– lulu
Apr 6 at 20:17
$begingroup$
But also scores where $A$ lost should contribute to the required expectation.
$endgroup$
– fwgb
Apr 6 at 20:30
$begingroup$
Yes, of course. That's why my expression has terms involving $(1-p)^i$.
$endgroup$
– lulu
Apr 6 at 20:36
|
show 4 more comments
$begingroup$
What do you mean by "score line"? Are you just asking for the probability that $A$ wins the match?
$endgroup$
– lulu
Apr 6 at 20:11
$begingroup$
Score line means the result e.g. 10-9, 10-1, or 8-10. For $A$ : 10, 10, 8.
$endgroup$
– fwgb
Apr 6 at 20:15
$begingroup$
Ok, well...the probability that $A$ wins the first $n$ is $p^n$. To get the probability that $A$ wins in exactly $n+1$ games, $A$ must win game $n+1$ and lose exactly one of the first $n$, hence $ntimes (1-p)*p^n$. Can you finish from here?
$endgroup$
– lulu
Apr 6 at 20:17
$begingroup$
But also scores where $A$ lost should contribute to the required expectation.
$endgroup$
– fwgb
Apr 6 at 20:30
$begingroup$
Yes, of course. That's why my expression has terms involving $(1-p)^i$.
$endgroup$
– lulu
Apr 6 at 20:36
$begingroup$
What do you mean by "score line"? Are you just asking for the probability that $A$ wins the match?
$endgroup$
– lulu
Apr 6 at 20:11
$begingroup$
What do you mean by "score line"? Are you just asking for the probability that $A$ wins the match?
$endgroup$
– lulu
Apr 6 at 20:11
$begingroup$
Score line means the result e.g. 10-9, 10-1, or 8-10. For $A$ : 10, 10, 8.
$endgroup$
– fwgb
Apr 6 at 20:15
$begingroup$
Score line means the result e.g. 10-9, 10-1, or 8-10. For $A$ : 10, 10, 8.
$endgroup$
– fwgb
Apr 6 at 20:15
$begingroup$
Ok, well...the probability that $A$ wins the first $n$ is $p^n$. To get the probability that $A$ wins in exactly $n+1$ games, $A$ must win game $n+1$ and lose exactly one of the first $n$, hence $ntimes (1-p)*p^n$. Can you finish from here?
$endgroup$
– lulu
Apr 6 at 20:17
$begingroup$
Ok, well...the probability that $A$ wins the first $n$ is $p^n$. To get the probability that $A$ wins in exactly $n+1$ games, $A$ must win game $n+1$ and lose exactly one of the first $n$, hence $ntimes (1-p)*p^n$. Can you finish from here?
$endgroup$
– lulu
Apr 6 at 20:17
$begingroup$
But also scores where $A$ lost should contribute to the required expectation.
$endgroup$
– fwgb
Apr 6 at 20:30
$begingroup$
But also scores where $A$ lost should contribute to the required expectation.
$endgroup$
– fwgb
Apr 6 at 20:30
$begingroup$
Yes, of course. That's why my expression has terms involving $(1-p)^i$.
$endgroup$
– lulu
Apr 6 at 20:36
$begingroup$
Yes, of course. That's why my expression has terms involving $(1-p)^i$.
$endgroup$
– lulu
Apr 6 at 20:36
|
show 4 more comments
1 Answer
1
active
oldest
votes
$begingroup$
Using this it follows
$E = sum_k=1^n P(A$ wins $k$ games before $B$ wins $n)$
which gives
$E = sum _k=1^n sum _j=k^n+k-1 tbinomk+n-1j p^j (1-p)^k+n-1-j$.
This coincides with the simualations. Can we simplify that?
answered Apr 6 at 21:02
fwgbfwgb
1284
$endgroup$
add a comment |
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1 Answer
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1 Answer
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$begingroup$
Using this it follows
$E = sum_k=1^n P(A$ wins $k$ games before $B$ wins $n)$
which gives
$E = sum _k=1^n sum _j=k^n+k-1 tbinomk+n-1j p^j (1-p)^k+n-1-j$.
This coincides with the simualations. Can we simplify that?
answered Apr 6 at 21:02
fwgbfwgb
1284
$endgroup$
add a comment |
$begingroup$
Using this it follows
$E = sum_k=1^n P(A$ wins $k$ games before $B$ wins $n)$
which gives
$E = sum _k=1^n sum _j=k^n+k-1 tbinomk+n-1j p^j (1-p)^k+n-1-j$.
This coincides with the simualations. Can we simplify that?
answered Apr 6 at 21:02
fwgbfwgb
1284
$endgroup$
add a comment |
$begingroup$
Using this it follows
$E = sum_k=1^n P(A$ wins $k$ games before $B$ wins $n)$
which gives
$E = sum _k=1^n sum _j=k^n+k-1 tbinomk+n-1j p^j (1-p)^k+n-1-j$.
This coincides with the simualations. Can we simplify that?
answered Apr 6 at 21:02
fwgbfwgb
1284
$endgroup$
Using this it follows
$E = sum_k=1^n P(A$ wins $k$ games before $B$ wins $n)$
which gives
$E = sum _k=1^n sum _j=k^n+k-1 tbinomk+n-1j p^j (1-p)^k+n-1-j$.
This coincides with the simualations. Can we simplify that?
answered Apr 6 at 21:02
fwgbfwgb
1284
edited Apr 7 at 19:48
answered Apr 6 at 21:02
fwgbfwgb
1284
answered Apr 6 at 21:02
fwgbfwgb
1284
answered Apr 6 at 21:02
fwgbfwgb
1284
1284
add a comment |
add a comment |
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$begingroup$
What do you mean by "score line"? Are you just asking for the probability that $A$ wins the match?
$endgroup$
– lulu
Apr 6 at 20:11
$begingroup$
Score line means the result e.g. 10-9, 10-1, or 8-10. For $A$ : 10, 10, 8.
$endgroup$
– fwgb
Apr 6 at 20:15
$begingroup$
Ok, well...the probability that $A$ wins the first $n$ is $p^n$. To get the probability that $A$ wins in exactly $n+1$ games, $A$ must win game $n+1$ and lose exactly one of the first $n$, hence $ntimes (1-p)*p^n$. Can you finish from here?
$endgroup$
– lulu
Apr 6 at 20:17
$begingroup$
But also scores where $A$ lost should contribute to the required expectation.
$endgroup$
– fwgb
Apr 6 at 20:30
$begingroup$
Yes, of course. That's why my expression has terms involving $(1-p)^i$.
$endgroup$
– lulu
Apr 6 at 20:36