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Show that given n coins, if the probability of getting heads an even number of times is 1/2 then there is at least one fair coin

If nine coins are tossed, what is the probability that the number of heads is even?If you toss $1000$ fair coins $10$ times each, what is the probability the *some* coin will get $10$ heads?If one of $n$ coins is fair, then find the probability that the total number of heads is evenProbability of getting 499000–501000 heads if a fair coin is flipped $10^6$ timesProbability of fair coin given 20 heads observedWhat is the connection between probability of getting $m$ heads from $n$ coins and getting $km$ heads from $kn$ coins?Two coins tossed $5$ times with different probabilites for heads on each coin. What's the probability of getting at least $3$ heads?Probability of flipping 10,000 coins and getting at least one 10 consecutive heads?1 Fair Coin, 1 Biased Coin. Keep getting heads. How many flips until probability of it being fair < 0.1Showing probability that $A$ and $B$ flip the same number of heads is equal to a total of $k$ heads.If I flip two fair coins, and then tell you that one is heads, what is the probability that the other coin is also heads?

4

1

$begingroup$

So the set up is as follows:
We have n coins being flipped independently, not necessarily all fair. I know that if there is at least one fair coin then the probability of getting an even number of heads after flipping is 12.
I want to show the converse, that if the probability is 1/2(of getting an even number of heads) then there is at least one fair coin.

probability conditional-probability

share|cite|improve this question

asked Mar 10 at 19:32

physicsPphysicsP

213

New contributor

physicsP is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  My guess is that there is some reasonably elegant argument by induction.
  $endgroup$
  – Brian Tung
  Mar 10 at 19:47
  

  
  
  
  

add a comment | 

4

1

$begingroup$

So the set up is as follows:
We have n coins being flipped independently, not necessarily all fair. I know that if there is at least one fair coin then the probability of getting an even number of heads after flipping is 12.
I want to show the converse, that if the probability is 1/2(of getting an even number of heads) then there is at least one fair coin.

probability conditional-probability

share|cite|improve this question

asked Mar 10 at 19:32

physicsPphysicsP

213

New contributor

physicsP is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  My guess is that there is some reasonably elegant argument by induction.
  $endgroup$
  – Brian Tung
  Mar 10 at 19:47
  

  
  
  
  

add a comment | 

$begingroup$

So the set up is as follows:
We have n coins being flipped independently, not necessarily all fair. I know that if there is at least one fair coin then the probability of getting an even number of heads after flipping is 12.
I want to show the converse, that if the probability is 1/2(of getting an even number of heads) then there is at least one fair coin.

probability conditional-probability

share|cite|improve this question

asked Mar 10 at 19:32

physicsPphysicsP

213

New contributor

physicsP is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

$endgroup$

share|cite|improve this question

asked Mar 10 at 19:32

physicsPphysicsP

213

New contributor

physicsP is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

share|cite|improve this question

asked Mar 10 at 19:32

physicsPphysicsP

213

New contributor

physicsP is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

asked Mar 10 at 19:32

physicsPphysicsP

213

New contributor

physicsP is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

asked Mar 10 at 19:32

physicsPphysicsP

213

2 Answers
2

active

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4

$begingroup$

Not as Elegant a Basic Approach as I Had Foreseen. We show this by induction.

First, for $n = 1$, a single coin. Obviously, then the probability of an even number of heads is simply the probability that this coin flips tails. If this coin is unfair, this probability is clearly not equal to $1/2$. Therefore the coin must be fair. This establishes the basis step.

Now, suppose that the proposition is true for some $n > 0$. Let us now show it for $n+1$. The antecedent is that the probability of an even number of heads in these $n+1$ flips is $1/2$. If (at least) one of the first $n$ coins is fair, then the consequent is true.

If, on the other hand, none of the first $n$ coins is fair, we already know that such a circumstance does not permit the probability of an even number of heads in the first $n$ tosses to be $1/2$. Let us say therefore that this probability is instead $P_n not= 1/2$, and let the $n+1$th coin have a probability of heads of $q$. Then the probability that the number of heads is even after all $n+1$ tosses is

$$
P_n+1 = P_n(1-q) + (1-P_n)q = P_n + q(1-2P_n)
$$

But we know, by hypothesis, that $P_n+1 = 1/2$, so we write

$$
frac12 = P_n + q(1-2P_n)
$$

which gives us, after some simple algebra,

$$
q = frac1/2-P_n1-2P_n = frac12
$$

This establishes the induction step and the proposition is shown.

share|cite|improve this answer

edited Mar 10 at 20:58

answered Mar 10 at 20:27

Brian TungBrian Tung

26.3k32555

$endgroup$

• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  For $n=1$, shouldn't the probability of an even number of heads be the probability to get tails? Since that is the probability of getting 0 heads, which is even, rather than 1 heads, which is odd.
  $endgroup$
  – Dasherman
  Mar 10 at 20:57
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @Dasherman: Yes, it should! In fact, when I read your comment, I thought I had in fact written "tails" (as is correct), and was dismayed to notice that I actually wrote "heads." Thanks for the catch!
  $endgroup$
  – Brian Tung
  Mar 10 at 20:58
  
  

  
  
  
  

add a comment | 

0

$begingroup$

This is really the same as the answer by @BrianTung but the presentation is a tad shorter. :)

Assume a set of $n$ coins has that property. Partition this set into two arbitrary non-empty subsets $X, Y$ and let $p_X = (1 over 2 + x), p_Y = (1over 2 + y)$ be the respective probabilities of each set to have an even number of heads. Then:

$$ 1 over 2 = p_X p_Y + (1 - p_X) (1 - p_Y) = (1 over 2 + x) (1 over 2 + y) + (1 over 2 - x) (1 over 2 - y) = 1 over 2 + 2xy$$

after you expand and realize the cross-terms cancel. Thus either $x$ or $y$ (or both) must be $0$, i.e. one (or both) of the subsets must have this property. As you recur downward you eventually reach a single coin which must be fair.

share|cite|improve this answer

answered 2 days ago

antkamantkam

2,067212

$endgroup$

add a comment | 

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4

$begingroup$

Not as Elegant a Basic Approach as I Had Foreseen. We show this by induction.

First, for $n = 1$, a single coin. Obviously, then the probability of an even number of heads is simply the probability that this coin flips tails. If this coin is unfair, this probability is clearly not equal to $1/2$. Therefore the coin must be fair. This establishes the basis step.

Now, suppose that the proposition is true for some $n > 0$. Let us now show it for $n+1$. The antecedent is that the probability of an even number of heads in these $n+1$ flips is $1/2$. If (at least) one of the first $n$ coins is fair, then the consequent is true.

If, on the other hand, none of the first $n$ coins is fair, we already know that such a circumstance does not permit the probability of an even number of heads in the first $n$ tosses to be $1/2$. Let us say therefore that this probability is instead $P_n not= 1/2$, and let the $n+1$th coin have a probability of heads of $q$. Then the probability that the number of heads is even after all $n+1$ tosses is

$$
P_n+1 = P_n(1-q) + (1-P_n)q = P_n + q(1-2P_n)
$$

But we know, by hypothesis, that $P_n+1 = 1/2$, so we write

$$
frac12 = P_n + q(1-2P_n)
$$

which gives us, after some simple algebra,

$$
q = frac1/2-P_n1-2P_n = frac12
$$

This establishes the induction step and the proposition is shown.

share|cite|improve this answer

edited Mar 10 at 20:58

answered Mar 10 at 20:27

Brian TungBrian Tung

26.3k32555

$endgroup$

• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  For $n=1$, shouldn't the probability of an even number of heads be the probability to get tails? Since that is the probability of getting 0 heads, which is even, rather than 1 heads, which is odd.
  $endgroup$
  – Dasherman
  Mar 10 at 20:57
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @Dasherman: Yes, it should! In fact, when I read your comment, I thought I had in fact written "tails" (as is correct), and was dismayed to notice that I actually wrote "heads." Thanks for the catch!
  $endgroup$
  – Brian Tung
  Mar 10 at 20:58
  
  

  
  
  
  

add a comment | 

4

$begingroup$

Not as Elegant a Basic Approach as I Had Foreseen. We show this by induction.

First, for $n = 1$, a single coin. Obviously, then the probability of an even number of heads is simply the probability that this coin flips tails. If this coin is unfair, this probability is clearly not equal to $1/2$. Therefore the coin must be fair. This establishes the basis step.

Now, suppose that the proposition is true for some $n > 0$. Let us now show it for $n+1$. The antecedent is that the probability of an even number of heads in these $n+1$ flips is $1/2$. If (at least) one of the first $n$ coins is fair, then the consequent is true.

If, on the other hand, none of the first $n$ coins is fair, we already know that such a circumstance does not permit the probability of an even number of heads in the first $n$ tosses to be $1/2$. Let us say therefore that this probability is instead $P_n not= 1/2$, and let the $n+1$th coin have a probability of heads of $q$. Then the probability that the number of heads is even after all $n+1$ tosses is

$$
P_n+1 = P_n(1-q) + (1-P_n)q = P_n + q(1-2P_n)
$$

But we know, by hypothesis, that $P_n+1 = 1/2$, so we write

$$
frac12 = P_n + q(1-2P_n)
$$

which gives us, after some simple algebra,

$$
q = frac1/2-P_n1-2P_n = frac12
$$

This establishes the induction step and the proposition is shown.

share|cite|improve this answer

edited Mar 10 at 20:58

answered Mar 10 at 20:27

Brian TungBrian Tung

26.3k32555

$endgroup$

• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  For $n=1$, shouldn't the probability of an even number of heads be the probability to get tails? Since that is the probability of getting 0 heads, which is even, rather than 1 heads, which is odd.
  $endgroup$
  – Dasherman
  Mar 10 at 20:57
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @Dasherman: Yes, it should! In fact, when I read your comment, I thought I had in fact written "tails" (as is correct), and was dismayed to notice that I actually wrote "heads." Thanks for the catch!
  $endgroup$
  – Brian Tung
  Mar 10 at 20:58
  
  

  
  
  
  

add a comment | 

$begingroup$

Not as Elegant a Basic Approach as I Had Foreseen. We show this by induction.

First, for $n = 1$, a single coin. Obviously, then the probability of an even number of heads is simply the probability that this coin flips tails. If this coin is unfair, this probability is clearly not equal to $1/2$. Therefore the coin must be fair. This establishes the basis step.

Now, suppose that the proposition is true for some $n > 0$. Let us now show it for $n+1$. The antecedent is that the probability of an even number of heads in these $n+1$ flips is $1/2$. If (at least) one of the first $n$ coins is fair, then the consequent is true.

If, on the other hand, none of the first $n$ coins is fair, we already know that such a circumstance does not permit the probability of an even number of heads in the first $n$ tosses to be $1/2$. Let us say therefore that this probability is instead $P_n not= 1/2$, and let the $n+1$th coin have a probability of heads of $q$. Then the probability that the number of heads is even after all $n+1$ tosses is

$$
P_n+1 = P_n(1-q) + (1-P_n)q = P_n + q(1-2P_n)
$$

But we know, by hypothesis, that $P_n+1 = 1/2$, so we write

$$
frac12 = P_n + q(1-2P_n)
$$

which gives us, after some simple algebra,

$$
q = frac1/2-P_n1-2P_n = frac12
$$

This establishes the induction step and the proposition is shown.

share|cite|improve this answer

edited Mar 10 at 20:58

answered Mar 10 at 20:27

Brian TungBrian Tung

26.3k32555

$endgroup$

share|cite|improve this answer

edited Mar 10 at 20:58

answered Mar 10 at 20:27

Brian TungBrian Tung

26.3k32555

answered Mar 10 at 20:27

Brian TungBrian Tung

26.3k32555

answered Mar 10 at 20:27

Brian TungBrian Tung

26.3k32555

0

$begingroup$

This is really the same as the answer by @BrianTung but the presentation is a tad shorter. :)

Assume a set of $n$ coins has that property. Partition this set into two arbitrary non-empty subsets $X, Y$ and let $p_X = (1 over 2 + x), p_Y = (1over 2 + y)$ be the respective probabilities of each set to have an even number of heads. Then:

$$ 1 over 2 = p_X p_Y + (1 - p_X) (1 - p_Y) = (1 over 2 + x) (1 over 2 + y) + (1 over 2 - x) (1 over 2 - y) = 1 over 2 + 2xy$$

after you expand and realize the cross-terms cancel. Thus either $x$ or $y$ (or both) must be $0$, i.e. one (or both) of the subsets must have this property. As you recur downward you eventually reach a single coin which must be fair.

share|cite|improve this answer

answered 2 days ago

antkamantkam

2,067212

$endgroup$

add a comment | 

0

$begingroup$

This is really the same as the answer by @BrianTung but the presentation is a tad shorter. :)

Assume a set of $n$ coins has that property. Partition this set into two arbitrary non-empty subsets $X, Y$ and let $p_X = (1 over 2 + x), p_Y = (1over 2 + y)$ be the respective probabilities of each set to have an even number of heads. Then:

$$ 1 over 2 = p_X p_Y + (1 - p_X) (1 - p_Y) = (1 over 2 + x) (1 over 2 + y) + (1 over 2 - x) (1 over 2 - y) = 1 over 2 + 2xy$$

after you expand and realize the cross-terms cancel. Thus either $x$ or $y$ (or both) must be $0$, i.e. one (or both) of the subsets must have this property. As you recur downward you eventually reach a single coin which must be fair.

share|cite|improve this answer

answered 2 days ago

antkamantkam

2,067212

$endgroup$

add a comment | 

$begingroup$

This is really the same as the answer by @BrianTung but the presentation is a tad shorter. :)

Assume a set of $n$ coins has that property. Partition this set into two arbitrary non-empty subsets $X, Y$ and let $p_X = (1 over 2 + x), p_Y = (1over 2 + y)$ be the respective probabilities of each set to have an even number of heads. Then:

$$ 1 over 2 = p_X p_Y + (1 - p_X) (1 - p_Y) = (1 over 2 + x) (1 over 2 + y) + (1 over 2 - x) (1 over 2 - y) = 1 over 2 + 2xy$$

after you expand and realize the cross-terms cancel. Thus either $x$ or $y$ (or both) must be $0$, i.e. one (or both) of the subsets must have this property. As you recur downward you eventually reach a single coin which must be fair.

share|cite|improve this answer

answered 2 days ago

antkamantkam

2,067212

$endgroup$

share|cite|improve this answer

answered 2 days ago

antkamantkam

2,067212

answered 2 days ago

antkamantkam

2,067212

answered 2 days ago

antkamantkam

2,067212