Probability of equal amount of heads
$begingroup$
Assume two gamblers throw coins randomly, and we are interested in probability that both end up with the same amount of heads.
The answer to this question suggests that it is enough to count the events where "exactly 3 out of six coins show heads". Why is this the correct intepretation?
I know that the answer is correct, since
$$ frac{1}{2^{2n}}sum_{j=0}^{n}left(begin{array}{c}n\jend{array}right)^2=frac{1}{2^{2n}}left(begin{array}{c}2n\nend{array}right)$$
holds by straightforwardly summing all events with different possible numbers of heads.
However, why do we understand it as the same as "exactly 3 out of 6 heads", if there are events where all heads or none heads suffice?
probability combinatorics
asked Jan 20 at 19:00
NutleNutle
320110
$endgroup$
|
show 3 more comments
$begingroup$
Assume two gamblers throw coins randomly, and we are interested in probability that both end up with the same amount of heads.
The answer to this question suggests that it is enough to count the events where "exactly 3 out of six coins show heads". Why is this the correct intepretation?
I know that the answer is correct, since
$$ frac{1}{2^{2n}}sum_{j=0}^{n}left(begin{array}{c}n\jend{array}right)^2=frac{1}{2^{2n}}left(begin{array}{c}2n\nend{array}right)$$
holds by straightforwardly summing all events with different possible numbers of heads.
However, why do we understand it as the same as "exactly 3 out of 6 heads", if there are events where all heads or none heads suffice?
probability combinatorics
asked Jan 20 at 19:00
NutleNutle
320110
$endgroup$
$begingroup$
Why not ask the user who posted that solution?
$endgroup$
– lulu
Jan 20 at 19:07
$begingroup$
Note: I think you missed the part where one person turns all of their's over. That stage is critical.
$endgroup$
– lulu
Jan 20 at 19:08
$begingroup$
@lulu I think that answer is too old to expect a reply. Either way, yes, I didn't understand why the part you mention is critical for the interpretation.
$endgroup$
– Nutle
Jan 20 at 19:12
$begingroup$
That user is still active on the site. But, any way, if they both had $i$ Heads, then after turning them over one player has $3-i$ Heads, and $i+3-i=3$.
$endgroup$
– lulu
Jan 20 at 19:13
$begingroup$
with 9000+ answers I would expect his inbox being flooded constantly. :) Ping @hagen-von-eitzen for a comment? Anyway, this seems to be getting off the topic
$endgroup$
– Nutle
Jan 20 at 19:16
|
show 3 more comments
$begingroup$
Assume two gamblers throw coins randomly, and we are interested in probability that both end up with the same amount of heads.
The answer to this question suggests that it is enough to count the events where "exactly 3 out of six coins show heads". Why is this the correct intepretation?
I know that the answer is correct, since
$$ frac{1}{2^{2n}}sum_{j=0}^{n}left(begin{array}{c}n\jend{array}right)^2=frac{1}{2^{2n}}left(begin{array}{c}2n\nend{array}right)$$
holds by straightforwardly summing all events with different possible numbers of heads.
However, why do we understand it as the same as "exactly 3 out of 6 heads", if there are events where all heads or none heads suffice?
probability combinatorics
asked Jan 20 at 19:00
NutleNutle
320110
$endgroup$
Assume two gamblers throw coins randomly, and we are interested in probability that both end up with the same amount of heads.
The answer to this question suggests that it is enough to count the events where "exactly 3 out of six coins show heads". Why is this the correct intepretation?
I know that the answer is correct, since
$$ frac{1}{2^{2n}}sum_{j=0}^{n}left(begin{array}{c}n\jend{array}right)^2=frac{1}{2^{2n}}left(begin{array}{c}2n\nend{array}right)$$
holds by straightforwardly summing all events with different possible numbers of heads.
However, why do we understand it as the same as "exactly 3 out of 6 heads", if there are events where all heads or none heads suffice?
probability combinatorics
probability combinatorics
asked Jan 20 at 19:00
NutleNutle
320110
asked Jan 20 at 19:00
NutleNutle
320110
asked Jan 20 at 19:00
NutleNutle
320110
asked Jan 20 at 19:00
NutleNutle
320110
asked Jan 20 at 19:00
NutleNutle
320110
320110
$begingroup$
Why not ask the user who posted that solution?
$endgroup$
– lulu
Jan 20 at 19:07
$begingroup$
Note: I think you missed the part where one person turns all of their's over. That stage is critical.
$endgroup$
– lulu
Jan 20 at 19:08
$begingroup$
@lulu I think that answer is too old to expect a reply. Either way, yes, I didn't understand why the part you mention is critical for the interpretation.
$endgroup$
– Nutle
Jan 20 at 19:12
$begingroup$
That user is still active on the site. But, any way, if they both had $i$ Heads, then after turning them over one player has $3-i$ Heads, and $i+3-i=3$.
$endgroup$
– lulu
Jan 20 at 19:13
$begingroup$
with 9000+ answers I would expect his inbox being flooded constantly. :) Ping @hagen-von-eitzen for a comment? Anyway, this seems to be getting off the topic
$endgroup$
– Nutle
Jan 20 at 19:16
|
show 3 more comments
$begingroup$
Why not ask the user who posted that solution?
$endgroup$
– lulu
Jan 20 at 19:07
$begingroup$
Note: I think you missed the part where one person turns all of their's over. That stage is critical.
$endgroup$
– lulu
Jan 20 at 19:08
$begingroup$
@lulu I think that answer is too old to expect a reply. Either way, yes, I didn't understand why the part you mention is critical for the interpretation.
$endgroup$
– Nutle
Jan 20 at 19:12
$begingroup$
That user is still active on the site. But, any way, if they both had $i$ Heads, then after turning them over one player has $3-i$ Heads, and $i+3-i=3$.
$endgroup$
– lulu
Jan 20 at 19:13
$begingroup$
with 9000+ answers I would expect his inbox being flooded constantly. :) Ping @hagen-von-eitzen for a comment? Anyway, this seems to be getting off the topic
$endgroup$
– Nutle
Jan 20 at 19:16
$begingroup$
Why not ask the user who posted that solution?
$endgroup$
– lulu
Jan 20 at 19:07
$begingroup$
Why not ask the user who posted that solution?
$endgroup$
– lulu
Jan 20 at 19:07
$begingroup$
Note: I think you missed the part where one person turns all of their's over. That stage is critical.
$endgroup$
– lulu
Jan 20 at 19:08
$begingroup$
Note: I think you missed the part where one person turns all of their's over. That stage is critical.
$endgroup$
– lulu
Jan 20 at 19:08
$begingroup$
@lulu I think that answer is too old to expect a reply. Either way, yes, I didn't understand why the part you mention is critical for the interpretation.
$endgroup$
– Nutle
Jan 20 at 19:12
$begingroup$
@lulu I think that answer is too old to expect a reply. Either way, yes, I didn't understand why the part you mention is critical for the interpretation.
$endgroup$
– Nutle
Jan 20 at 19:12
$begingroup$
That user is still active on the site. But, any way, if they both had $i$ Heads, then after turning them over one player has $3-i$ Heads, and $i+3-i=3$.
$endgroup$
– lulu
Jan 20 at 19:13
$begingroup$
That user is still active on the site. But, any way, if they both had $i$ Heads, then after turning them over one player has $3-i$ Heads, and $i+3-i=3$.
$endgroup$
– lulu
Jan 20 at 19:13
$begingroup$
with 9000+ answers I would expect his inbox being flooded constantly. :) Ping @hagen-von-eitzen for a comment? Anyway, this seems to be getting off the topic
$endgroup$
– Nutle
Jan 20 at 19:16
$begingroup$
with 9000+ answers I would expect his inbox being flooded constantly. :) Ping @hagen-von-eitzen for a comment? Anyway, this seems to be getting off the topic
$endgroup$
– Nutle
Jan 20 at 19:16
|
show 3 more comments
1 Answer
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$begingroup$
Lets even say they threw n coins each other. For any i < n the odds for a guy to get i heads is equal to the odds of getting n-i heads. So saying "whats the odds for both getting i heads" is the same as "whats the odds for one getting i heads and the second getting n-i heads?" if you sum up all the i's that smaller than n, you get the question "whats the odds to get n time heads for the both of them".
answered Jan 20 at 19:12
ShaqShaq
2849
$endgroup$
add a comment |
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$begingroup$
Lets even say they threw n coins each other. For any i < n the odds for a guy to get i heads is equal to the odds of getting n-i heads. So saying "whats the odds for both getting i heads" is the same as "whats the odds for one getting i heads and the second getting n-i heads?" if you sum up all the i's that smaller than n, you get the question "whats the odds to get n time heads for the both of them".
answered Jan 20 at 19:12
ShaqShaq
2849
$endgroup$
add a comment |
$begingroup$
Lets even say they threw n coins each other. For any i < n the odds for a guy to get i heads is equal to the odds of getting n-i heads. So saying "whats the odds for both getting i heads" is the same as "whats the odds for one getting i heads and the second getting n-i heads?" if you sum up all the i's that smaller than n, you get the question "whats the odds to get n time heads for the both of them".
answered Jan 20 at 19:12
ShaqShaq
2849
$endgroup$
add a comment |
$begingroup$
Lets even say they threw n coins each other. For any i < n the odds for a guy to get i heads is equal to the odds of getting n-i heads. So saying "whats the odds for both getting i heads" is the same as "whats the odds for one getting i heads and the second getting n-i heads?" if you sum up all the i's that smaller than n, you get the question "whats the odds to get n time heads for the both of them".
answered Jan 20 at 19:12
ShaqShaq
2849
$endgroup$
Lets even say they threw n coins each other. For any i < n the odds for a guy to get i heads is equal to the odds of getting n-i heads. So saying "whats the odds for both getting i heads" is the same as "whats the odds for one getting i heads and the second getting n-i heads?" if you sum up all the i's that smaller than n, you get the question "whats the odds to get n time heads for the both of them".
answered Jan 20 at 19:12
ShaqShaq
2849
answered Jan 20 at 19:12
ShaqShaq
2849
answered Jan 20 at 19:12
ShaqShaq
2849
answered Jan 20 at 19:12
ShaqShaq
2849
2849
add a comment |
add a comment |
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$begingroup$
Why not ask the user who posted that solution?
$endgroup$
– lulu
Jan 20 at 19:07
$begingroup$
Note: I think you missed the part where one person turns all of their's over. That stage is critical.
$endgroup$
– lulu
Jan 20 at 19:08
$begingroup$
@lulu I think that answer is too old to expect a reply. Either way, yes, I didn't understand why the part you mention is critical for the interpretation.
$endgroup$
– Nutle
Jan 20 at 19:12
$begingroup$
That user is still active on the site. But, any way, if they both had $i$ Heads, then after turning them over one player has $3-i$ Heads, and $i+3-i=3$.
$endgroup$
– lulu
Jan 20 at 19:13
$begingroup$
with 9000+ answers I would expect his inbox being flooded constantly. :) Ping @hagen-von-eitzen for a comment? Anyway, this seems to be getting off the topic
$endgroup$
– Nutle
Jan 20 at 19:16