Probability of each runner winning
$begingroup$
All runners times are independent of each other.
Each Runner runs five trials and the times are measured in seconds.
A (31 28 24 33 29)
B (38 35 36 35 39)
C (33 40 41 35 38)
D (29 29 26 30 31)
E (35 34 31 30 28)
F (34 33 38 33 31)
If they were all to run again in a time trial and their times are again independent of each other, what would the probability of each runner winning outright be?
Is there a quick way to work this out because I have tried using excel and may have found a correct answer using grids etc but it is a very long winded way.
probability
asked Dec 15 '18 at 21:53
JJ317JJ317
11
$endgroup$
add a comment |
$begingroup$
All runners times are independent of each other.
Each Runner runs five trials and the times are measured in seconds.
A (31 28 24 33 29)
B (38 35 36 35 39)
C (33 40 41 35 38)
D (29 29 26 30 31)
E (35 34 31 30 28)
F (34 33 38 33 31)
If they were all to run again in a time trial and their times are again independent of each other, what would the probability of each runner winning outright be?
Is there a quick way to work this out because I have tried using excel and may have found a correct answer using grids etc but it is a very long winded way.
probability
asked Dec 15 '18 at 21:53
JJ317JJ317
11
$endgroup$
$begingroup$
There be a tie for first place? That is, would you have won the race if you tie for first place?
$endgroup$
– Bram28
Dec 15 '18 at 21:56
$begingroup$
Oh no sorry, what would the probability of each runner winning outright be should be the question
$endgroup$
– JJ317
Dec 15 '18 at 22:03
$begingroup$
Ok, that's what I assumed in my Answer below
$endgroup$
– Bram28
Dec 15 '18 at 22:11
$begingroup$
Thank you for the reply, but I may have not worded the question correctly. Is it possible to work out the probability of each runner winning without running to the exact time it has run before. As in A doesn't have to run to exactly 24,28 or 29. "A" could theoretically run below or above his best and worst times but is likely to run to around his mean of 29 with a relatively small variance either side.
$endgroup$
– JJ317
Dec 15 '18 at 22:21
$begingroup$
Ah! Well, that certainly makes it a lot more realistic .... but unfortunately also mich more of a pain in the ass to compute. ... you'd have to assume normal distributions for the times for each of those runners, and then ... frankly, I don't know what then ... kind of out of my league here, sorry!
$endgroup$
– Bram28
Dec 15 '18 at 22:35
add a comment |
$begingroup$
All runners times are independent of each other.
Each Runner runs five trials and the times are measured in seconds.
A (31 28 24 33 29)
B (38 35 36 35 39)
C (33 40 41 35 38)
D (29 29 26 30 31)
E (35 34 31 30 28)
F (34 33 38 33 31)
If they were all to run again in a time trial and their times are again independent of each other, what would the probability of each runner winning outright be?
Is there a quick way to work this out because I have tried using excel and may have found a correct answer using grids etc but it is a very long winded way.
probability
asked Dec 15 '18 at 21:53
JJ317JJ317
11
$endgroup$
All runners times are independent of each other.
Each Runner runs five trials and the times are measured in seconds.
A (31 28 24 33 29)
B (38 35 36 35 39)
C (33 40 41 35 38)
D (29 29 26 30 31)
E (35 34 31 30 28)
F (34 33 38 33 31)
If they were all to run again in a time trial and their times are again independent of each other, what would the probability of each runner winning outright be?
Is there a quick way to work this out because I have tried using excel and may have found a correct answer using grids etc but it is a very long winded way.
probability
probability
asked Dec 15 '18 at 21:53
JJ317JJ317
11
asked Dec 15 '18 at 21:53
JJ317JJ317
11
edited Dec 15 '18 at 22:04
JJ317
asked Dec 15 '18 at 21:53
JJ317JJ317
11
asked Dec 15 '18 at 21:53
JJ317JJ317
11
asked Dec 15 '18 at 21:53
JJ317JJ317
11
11
$begingroup$
There be a tie for first place? That is, would you have won the race if you tie for first place?
$endgroup$
– Bram28
Dec 15 '18 at 21:56
$begingroup$
Oh no sorry, what would the probability of each runner winning outright be should be the question
$endgroup$
– JJ317
Dec 15 '18 at 22:03
$begingroup$
Ok, that's what I assumed in my Answer below
$endgroup$
– Bram28
Dec 15 '18 at 22:11
$begingroup$
Thank you for the reply, but I may have not worded the question correctly. Is it possible to work out the probability of each runner winning without running to the exact time it has run before. As in A doesn't have to run to exactly 24,28 or 29. "A" could theoretically run below or above his best and worst times but is likely to run to around his mean of 29 with a relatively small variance either side.
$endgroup$
– JJ317
Dec 15 '18 at 22:21
$begingroup$
Ah! Well, that certainly makes it a lot more realistic .... but unfortunately also mich more of a pain in the ass to compute. ... you'd have to assume normal distributions for the times for each of those runners, and then ... frankly, I don't know what then ... kind of out of my league here, sorry!
$endgroup$
– Bram28
Dec 15 '18 at 22:35
add a comment |
$begingroup$
There be a tie for first place? That is, would you have won the race if you tie for first place?
$endgroup$
– Bram28
Dec 15 '18 at 21:56
$begingroup$
Oh no sorry, what would the probability of each runner winning outright be should be the question
$endgroup$
– JJ317
Dec 15 '18 at 22:03
$begingroup$
Ok, that's what I assumed in my Answer below
$endgroup$
– Bram28
Dec 15 '18 at 22:11
$begingroup$
Thank you for the reply, but I may have not worded the question correctly. Is it possible to work out the probability of each runner winning without running to the exact time it has run before. As in A doesn't have to run to exactly 24,28 or 29. "A" could theoretically run below or above his best and worst times but is likely to run to around his mean of 29 with a relatively small variance either side.
$endgroup$
– JJ317
Dec 15 '18 at 22:21
$begingroup$
Ah! Well, that certainly makes it a lot more realistic .... but unfortunately also mich more of a pain in the ass to compute. ... you'd have to assume normal distributions for the times for each of those runners, and then ... frankly, I don't know what then ... kind of out of my league here, sorry!
$endgroup$
– Bram28
Dec 15 '18 at 22:35
$begingroup$
There be a tie for first place? That is, would you have won the race if you tie for first place?
$endgroup$
– Bram28
Dec 15 '18 at 21:56
$begingroup$
There be a tie for first place? That is, would you have won the race if you tie for first place?
$endgroup$
– Bram28
Dec 15 '18 at 21:56
$begingroup$
Oh no sorry, what would the probability of each runner winning outright be should be the question
$endgroup$
– JJ317
Dec 15 '18 at 22:03
$begingroup$
Oh no sorry, what would the probability of each runner winning outright be should be the question
$endgroup$
– JJ317
Dec 15 '18 at 22:03
$begingroup$
Ok, that's what I assumed in my Answer below
$endgroup$
– Bram28
Dec 15 '18 at 22:11
$begingroup$
Ok, that's what I assumed in my Answer below
$endgroup$
– Bram28
Dec 15 '18 at 22:11
$begingroup$
Thank you for the reply, but I may have not worded the question correctly. Is it possible to work out the probability of each runner winning without running to the exact time it has run before. As in A doesn't have to run to exactly 24,28 or 29. "A" could theoretically run below or above his best and worst times but is likely to run to around his mean of 29 with a relatively small variance either side.
$endgroup$
– JJ317
Dec 15 '18 at 22:21
$begingroup$
Thank you for the reply, but I may have not worded the question correctly. Is it possible to work out the probability of each runner winning without running to the exact time it has run before. As in A doesn't have to run to exactly 24,28 or 29. "A" could theoretically run below or above his best and worst times but is likely to run to around his mean of 29 with a relatively small variance either side.
$endgroup$
– JJ317
Dec 15 '18 at 22:21
$begingroup$
Ah! Well, that certainly makes it a lot more realistic .... but unfortunately also mich more of a pain in the ass to compute. ... you'd have to assume normal distributions for the times for each of those runners, and then ... frankly, I don't know what then ... kind of out of my league here, sorry!
$endgroup$
– Bram28
Dec 15 '18 at 22:35
$begingroup$
Ah! Well, that certainly makes it a lot more realistic .... but unfortunately also mich more of a pain in the ass to compute. ... you'd have to assume normal distributions for the times for each of those runners, and then ... frankly, I don't know what then ... kind of out of my league here, sorry!
$endgroup$
– Bram28
Dec 15 '18 at 22:35
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Note that $D$ will always beat $B$ and $C$, so
$$P(B) =P(C)=0$$
Also, if one can be a 'winner' only by being an outright winner (i.e. When there is a tie for first place there is no winner at all), then $F $can never win against $D$ either, and hence
$$P(F)=0$$
So, that leaves $A$, $D$, and $E$
Now, $E$ can only win against $D$ when running in $28$ seconds, and in that case $D$ has to run one of its $4$ slower times, and $A$ one of $3$ so the probability of that is $$P(E) = frac{1}{5}cdot frac{4}{5}cdot frac{3}{5}=frac{12}{125}$$
$A$ can win in several ways:
$A$ runs $24$: $D$ and $E$ are always slower
$A$ runs $28$: $D$ runs slower in $4$ cases, and $E$ runs slower in $4$ cases
$A$ runs $29$ : $D$ runs slower in $2$ cases, and $E$ runs slower in $4$ cases
This gives a total of $5cdot 5 + 4 cdot 4 + 2 cdot 4= 49$ cases, out of $125$ possible times between $A$, $D$, and $E$, so:
$$P(A)=frac{49}{125}$$
$D$ wins if:
$D$ runs $26$: $A$ is slower in $4$ cases, E in all $5$
$D$ runs $29$ ($2$ cases for $D$): $A$ slower in $2$ cases, $E$ in $4$
$D$ runs $30$: $A$ slower in $2$ cases, $E$ in $3$
$D$ runs $31$ $A$ slower in $1$ case,$E$ in $2$
Total: $20+16+6+2=44$ cases ... out of $125$
So: $$P(D)=frac{44}{125}$$
Sanity check:
In how many cases can we get a tie for first place?
28: A and E: D slower in $4$ cases
29: A and D (2 cases): E slower in $4$: $8$ cases
30: D and E: A slower in $2$ cases
31: just A and E: no cases, since d will be faster or tie as well
31: just A and D: E slower in 2 cases
31: just D and E: A slower in 1 case
Total: 17 out of 125 cases
Total winner or tie: 122 out of 125 ... darn! Made a mistake somewhere!
OK, gotta pick up and eat my pizza...will be back in an hour...
answered Dec 15 '18 at 22:11
Bram28Bram28
63.3k44793
$endgroup$
add a comment |
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$begingroup$
Note that $D$ will always beat $B$ and $C$, so
$$P(B) =P(C)=0$$
Also, if one can be a 'winner' only by being an outright winner (i.e. When there is a tie for first place there is no winner at all), then $F $can never win against $D$ either, and hence
$$P(F)=0$$
So, that leaves $A$, $D$, and $E$
Now, $E$ can only win against $D$ when running in $28$ seconds, and in that case $D$ has to run one of its $4$ slower times, and $A$ one of $3$ so the probability of that is $$P(E) = frac{1}{5}cdot frac{4}{5}cdot frac{3}{5}=frac{12}{125}$$
$A$ can win in several ways:
$A$ runs $24$: $D$ and $E$ are always slower
$A$ runs $28$: $D$ runs slower in $4$ cases, and $E$ runs slower in $4$ cases
$A$ runs $29$ : $D$ runs slower in $2$ cases, and $E$ runs slower in $4$ cases
This gives a total of $5cdot 5 + 4 cdot 4 + 2 cdot 4= 49$ cases, out of $125$ possible times between $A$, $D$, and $E$, so:
$$P(A)=frac{49}{125}$$
$D$ wins if:
$D$ runs $26$: $A$ is slower in $4$ cases, E in all $5$
$D$ runs $29$ ($2$ cases for $D$): $A$ slower in $2$ cases, $E$ in $4$
$D$ runs $30$: $A$ slower in $2$ cases, $E$ in $3$
$D$ runs $31$ $A$ slower in $1$ case,$E$ in $2$
Total: $20+16+6+2=44$ cases ... out of $125$
So: $$P(D)=frac{44}{125}$$
Sanity check:
In how many cases can we get a tie for first place?
28: A and E: D slower in $4$ cases
29: A and D (2 cases): E slower in $4$: $8$ cases
30: D and E: A slower in $2$ cases
31: just A and E: no cases, since d will be faster or tie as well
31: just A and D: E slower in 2 cases
31: just D and E: A slower in 1 case
Total: 17 out of 125 cases
Total winner or tie: 122 out of 125 ... darn! Made a mistake somewhere!
OK, gotta pick up and eat my pizza...will be back in an hour...
answered Dec 15 '18 at 22:11
Bram28Bram28
63.3k44793
$endgroup$
add a comment |
$begingroup$
Note that $D$ will always beat $B$ and $C$, so
$$P(B) =P(C)=0$$
Also, if one can be a 'winner' only by being an outright winner (i.e. When there is a tie for first place there is no winner at all), then $F $can never win against $D$ either, and hence
$$P(F)=0$$
So, that leaves $A$, $D$, and $E$
Now, $E$ can only win against $D$ when running in $28$ seconds, and in that case $D$ has to run one of its $4$ slower times, and $A$ one of $3$ so the probability of that is $$P(E) = frac{1}{5}cdot frac{4}{5}cdot frac{3}{5}=frac{12}{125}$$
$A$ can win in several ways:
$A$ runs $24$: $D$ and $E$ are always slower
$A$ runs $28$: $D$ runs slower in $4$ cases, and $E$ runs slower in $4$ cases
$A$ runs $29$ : $D$ runs slower in $2$ cases, and $E$ runs slower in $4$ cases
This gives a total of $5cdot 5 + 4 cdot 4 + 2 cdot 4= 49$ cases, out of $125$ possible times between $A$, $D$, and $E$, so:
$$P(A)=frac{49}{125}$$
$D$ wins if:
$D$ runs $26$: $A$ is slower in $4$ cases, E in all $5$
$D$ runs $29$ ($2$ cases for $D$): $A$ slower in $2$ cases, $E$ in $4$
$D$ runs $30$: $A$ slower in $2$ cases, $E$ in $3$
$D$ runs $31$ $A$ slower in $1$ case,$E$ in $2$
Total: $20+16+6+2=44$ cases ... out of $125$
So: $$P(D)=frac{44}{125}$$
Sanity check:
In how many cases can we get a tie for first place?
28: A and E: D slower in $4$ cases
29: A and D (2 cases): E slower in $4$: $8$ cases
30: D and E: A slower in $2$ cases
31: just A and E: no cases, since d will be faster or tie as well
31: just A and D: E slower in 2 cases
31: just D and E: A slower in 1 case
Total: 17 out of 125 cases
Total winner or tie: 122 out of 125 ... darn! Made a mistake somewhere!
OK, gotta pick up and eat my pizza...will be back in an hour...
answered Dec 15 '18 at 22:11
Bram28Bram28
63.3k44793
$endgroup$
add a comment |
$begingroup$
Note that $D$ will always beat $B$ and $C$, so
$$P(B) =P(C)=0$$
Also, if one can be a 'winner' only by being an outright winner (i.e. When there is a tie for first place there is no winner at all), then $F $can never win against $D$ either, and hence
$$P(F)=0$$
So, that leaves $A$, $D$, and $E$
Now, $E$ can only win against $D$ when running in $28$ seconds, and in that case $D$ has to run one of its $4$ slower times, and $A$ one of $3$ so the probability of that is $$P(E) = frac{1}{5}cdot frac{4}{5}cdot frac{3}{5}=frac{12}{125}$$
$A$ can win in several ways:
$A$ runs $24$: $D$ and $E$ are always slower
$A$ runs $28$: $D$ runs slower in $4$ cases, and $E$ runs slower in $4$ cases
$A$ runs $29$ : $D$ runs slower in $2$ cases, and $E$ runs slower in $4$ cases
This gives a total of $5cdot 5 + 4 cdot 4 + 2 cdot 4= 49$ cases, out of $125$ possible times between $A$, $D$, and $E$, so:
$$P(A)=frac{49}{125}$$
$D$ wins if:
$D$ runs $26$: $A$ is slower in $4$ cases, E in all $5$
$D$ runs $29$ ($2$ cases for $D$): $A$ slower in $2$ cases, $E$ in $4$
$D$ runs $30$: $A$ slower in $2$ cases, $E$ in $3$
$D$ runs $31$ $A$ slower in $1$ case,$E$ in $2$
Total: $20+16+6+2=44$ cases ... out of $125$
So: $$P(D)=frac{44}{125}$$
Sanity check:
In how many cases can we get a tie for first place?
28: A and E: D slower in $4$ cases
29: A and D (2 cases): E slower in $4$: $8$ cases
30: D and E: A slower in $2$ cases
31: just A and E: no cases, since d will be faster or tie as well
31: just A and D: E slower in 2 cases
31: just D and E: A slower in 1 case
Total: 17 out of 125 cases
Total winner or tie: 122 out of 125 ... darn! Made a mistake somewhere!
OK, gotta pick up and eat my pizza...will be back in an hour...
answered Dec 15 '18 at 22:11
Bram28Bram28
63.3k44793
$endgroup$
Note that $D$ will always beat $B$ and $C$, so
$$P(B) =P(C)=0$$
Also, if one can be a 'winner' only by being an outright winner (i.e. When there is a tie for first place there is no winner at all), then $F $can never win against $D$ either, and hence
$$P(F)=0$$
So, that leaves $A$, $D$, and $E$
Now, $E$ can only win against $D$ when running in $28$ seconds, and in that case $D$ has to run one of its $4$ slower times, and $A$ one of $3$ so the probability of that is $$P(E) = frac{1}{5}cdot frac{4}{5}cdot frac{3}{5}=frac{12}{125}$$
$A$ can win in several ways:
$A$ runs $24$: $D$ and $E$ are always slower
$A$ runs $28$: $D$ runs slower in $4$ cases, and $E$ runs slower in $4$ cases
$A$ runs $29$ : $D$ runs slower in $2$ cases, and $E$ runs slower in $4$ cases
This gives a total of $5cdot 5 + 4 cdot 4 + 2 cdot 4= 49$ cases, out of $125$ possible times between $A$, $D$, and $E$, so:
$$P(A)=frac{49}{125}$$
$D$ wins if:
$D$ runs $26$: $A$ is slower in $4$ cases, E in all $5$
$D$ runs $29$ ($2$ cases for $D$): $A$ slower in $2$ cases, $E$ in $4$
$D$ runs $30$: $A$ slower in $2$ cases, $E$ in $3$
$D$ runs $31$ $A$ slower in $1$ case,$E$ in $2$
Total: $20+16+6+2=44$ cases ... out of $125$
So: $$P(D)=frac{44}{125}$$
Sanity check:
In how many cases can we get a tie for first place?
28: A and E: D slower in $4$ cases
29: A and D (2 cases): E slower in $4$: $8$ cases
30: D and E: A slower in $2$ cases
31: just A and E: no cases, since d will be faster or tie as well
31: just A and D: E slower in 2 cases
31: just D and E: A slower in 1 case
Total: 17 out of 125 cases
Total winner or tie: 122 out of 125 ... darn! Made a mistake somewhere!
OK, gotta pick up and eat my pizza...will be back in an hour...
answered Dec 15 '18 at 22:11
Bram28Bram28
63.3k44793
edited Dec 15 '18 at 23:00
answered Dec 15 '18 at 22:11
Bram28Bram28
63.3k44793
answered Dec 15 '18 at 22:11
Bram28Bram28
63.3k44793
answered Dec 15 '18 at 22:11
Bram28Bram28
63.3k44793
63.3k44793
add a comment |
add a comment |
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$begingroup$
There be a tie for first place? That is, would you have won the race if you tie for first place?
$endgroup$
– Bram28
Dec 15 '18 at 21:56
$begingroup$
Oh no sorry, what would the probability of each runner winning outright be should be the question
$endgroup$
– JJ317
Dec 15 '18 at 22:03
$begingroup$
Ok, that's what I assumed in my Answer below
$endgroup$
– Bram28
Dec 15 '18 at 22:11
$begingroup$
Thank you for the reply, but I may have not worded the question correctly. Is it possible to work out the probability of each runner winning without running to the exact time it has run before. As in A doesn't have to run to exactly 24,28 or 29. "A" could theoretically run below or above his best and worst times but is likely to run to around his mean of 29 with a relatively small variance either side.
$endgroup$
– JJ317
Dec 15 '18 at 22:21
$begingroup$
Ah! Well, that certainly makes it a lot more realistic .... but unfortunately also mich more of a pain in the ass to compute. ... you'd have to assume normal distributions for the times for each of those runners, and then ... frankly, I don't know what then ... kind of out of my league here, sorry!
$endgroup$
– Bram28
Dec 15 '18 at 22:35