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Coin flipping and Bayes' theorem… but where does binomial theorem come in?

Bayes' Law and Coin Flipshi, for an independent event, like flipping a fair coin does Pr(Amid B) always equal to Pr(Bmid A)?Bayes factor for fair and biased coininterpretting prior and posteriorBayes Rule in 2 Fair and 1 Biased coinCoin toss how fair is the coin?If a coin toss is observed to come up as heads many times, does that affect the probability of the next toss?Using Bayes Theorem to calculate the probability of a coin being biased given a specific test resultIs it best to approach solving this question using Binomial Distribution and Conditional Probability in the following way?Bayes Theorem to find probability of getting a sum given a coin toss

0

$begingroup$

Consider the following question:

You are face with two identical coins. One is fair, and the
other comes up Heads 90% of the time. You flip coins, which
results in THHHTHHHTH (seven heads, three tails). What is the probability that
the coin you’ve been flipping is the unfair one?

Is it safe to assume order is irrelevant and to look for probability of seven heads and three tails with binomial distribution:

Let F be event that coin is fair, B be event that coin is biased, T be the observed toss. Then, using Bayes':
$$
P(B|T)=frac{P(T|B)times P(B)} {P(T)}
$$

$
P(T|B)=binom{10}{7} times 0.9^7times 0.1^3 =0.0574
$

$
P(B)=0.5
$

$
P(T)=P(T|F)cdot P(F) + P(T|B)cdot P(B) = 0.1172
$

Thus the probability that the coin flipped is biased is 24.5%

Is this reasoning correct?

probability conditional-probability binomial-distribution bayes-theorem

share|cite|improve this question

edited Mar 13 at 9:50

Lawrence

asked Mar 12 at 13:56

LawrenceLawrence

32

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I get $P(T) = 0.087$. Are you certain that you remembered to multiply by both $P(F) = 0.5$ and $P(B) = 0.5$?
  $endgroup$
  – Arthur
  Mar 12 at 14:13
  
  
  

  
  
  
  

add a comment | 

0

$begingroup$

Consider the following question:

You are face with two identical coins. One is fair, and the
other comes up Heads 90% of the time. You flip coins, which
results in THHHTHHHTH (seven heads, three tails). What is the probability that
the coin you’ve been flipping is the unfair one?

Is it safe to assume order is irrelevant and to look for probability of seven heads and three tails with binomial distribution:

Let F be event that coin is fair, B be event that coin is biased, T be the observed toss. Then, using Bayes':
$$
P(B|T)=frac{P(T|B)times P(B)} {P(T)}
$$

$
P(T|B)=binom{10}{7} times 0.9^7times 0.1^3 =0.0574
$

$
P(B)=0.5
$

$
P(T)=P(T|F)cdot P(F) + P(T|B)cdot P(B) = 0.1172
$

Thus the probability that the coin flipped is biased is 24.5%

Is this reasoning correct?

probability conditional-probability binomial-distribution bayes-theorem

share|cite|improve this question

edited Mar 13 at 9:50

Lawrence

asked Mar 12 at 13:56

LawrenceLawrence

32

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I get $P(T) = 0.087$. Are you certain that you remembered to multiply by both $P(F) = 0.5$ and $P(B) = 0.5$?
  $endgroup$
  – Arthur
  Mar 12 at 14:13
  
  
  

  
  
  
  

add a comment | 

$begingroup$

Consider the following question:

You are face with two identical coins. One is fair, and the
other comes up Heads 90% of the time. You flip coins, which
results in THHHTHHHTH (seven heads, three tails). What is the probability that
the coin you’ve been flipping is the unfair one?

Is it safe to assume order is irrelevant and to look for probability of seven heads and three tails with binomial distribution:

Let F be event that coin is fair, B be event that coin is biased, T be the observed toss. Then, using Bayes':
$$
P(B|T)=frac{P(T|B)times P(B)} {P(T)}
$$

$
P(T|B)=binom{10}{7} times 0.9^7times 0.1^3 =0.0574
$

$
P(B)=0.5
$

$
P(T)=P(T|F)cdot P(F) + P(T|B)cdot P(B) = 0.1172
$

Thus the probability that the coin flipped is biased is 24.5%

Is this reasoning correct?

probability conditional-probability binomial-distribution bayes-theorem

share|cite|improve this question

edited Mar 13 at 9:50

Lawrence

asked Mar 12 at 13:56

LawrenceLawrence

32

$endgroup$

share|cite|improve this question

edited Mar 13 at 9:50

Lawrence

asked Mar 12 at 13:56

LawrenceLawrence

32

share|cite|improve this question

edited Mar 13 at 9:50

Lawrence

asked Mar 12 at 13:56

LawrenceLawrence

32

asked Mar 12 at 13:56

LawrenceLawrence

32

asked Mar 12 at 13:56

LawrenceLawrence

32

2 Answers
2

active

oldest

votes

1

$begingroup$

Apart from a minor possible mistake I pointed out in a comment above, this looks good.

Is it safe to assume order is irrelevant and to look for probability of seven heads and three tails with binomial distribution

Let's see: With the binomial distribution, the final formula you get is
$$
frac{binom{10}3cdot 0.9^7cdot 0.1^3cdotfrac12}{frac12cdotbinom{10}3cdot 0.9^7cdot 0.1^3 + frac12cdotbinom{10}3cdot0.5^{10}}
$$
What happens if we don't use the binomial distribution and instead care about order? We get
$$
frac{ 0.9^7cdot 0.1^3cdotfrac12}{frac12cdot 0.9^7cdot 0.1^3 + frac12cdot0.5^{10}}
$$
which, you may notice, has exactly the same value because algebraically we have only simplified the fraction by $binom{10}3$. So not only is it safe to assume; it doesn't matter one way or the other.

One final notational nitpick: Is there a specific reason you use both $B$ and $L$ rather than, say, $B$ and $overline B$?

share|cite|improve this answer

answered Mar 12 at 14:20

ArthurArthur

118k7118202

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Of course it seems so obvious now! Thanks. only used B and F because the notation didn't occur to me.
  $endgroup$
  – Lawrence
  Mar 13 at 9:48
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @Lawrence The only real harm is the mental overhead of remembering one extra letter. On the other hand, applying $overline{phantom B}$ isn't completely without mental overhead either. So it's not really that important. However, once you start having more events that need to be negated, the difference becomes clearer. If you have a coin and a die, both of which can be either fair or biased, and you also wonder whether it's raining outside, having three letters with overlines starts becoming a lot easier than having six letters.
  $endgroup$
  – Arthur
  Mar 13 at 9:59
  

  
  
  
  

add a comment | 

0

$begingroup$

Although it does not matter in calculation as demonstrated by Arthur, yet the condition indicates the specific order of the $10$ tosses. So ($F-$ fair, $U-$ unfair):
$$P(Fcap HHHHTTHHTH)=frac12cdot frac{1}{2^{10}};\
P(Ucap HHHHTTHHTH)=\
frac12cdot left(frac{9}{10}right)^7cdot left(frac1{10}right)^3;\
P(U|HHHHTTHHTH)=frac{P(Ucap HHHHTTHHTH)}{P(Fcap HHHHTTHHTH)+P(Ucap HHHHTTHHTH)}=\
frac{frac{9^7}{2cdot 10^{10}}}{frac{9^7}{2cdot 10^{10}}+frac{1}{2^{11}}}approx 0.3288.$$

share|cite|improve this answer

answered Mar 12 at 15:22

farruhotafarruhota

21.3k2841

$endgroup$

add a comment | 

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1

$begingroup$

Apart from a minor possible mistake I pointed out in a comment above, this looks good.

Is it safe to assume order is irrelevant and to look for probability of seven heads and three tails with binomial distribution

Let's see: With the binomial distribution, the final formula you get is
$$
frac{binom{10}3cdot 0.9^7cdot 0.1^3cdotfrac12}{frac12cdotbinom{10}3cdot 0.9^7cdot 0.1^3 + frac12cdotbinom{10}3cdot0.5^{10}}
$$
What happens if we don't use the binomial distribution and instead care about order? We get
$$
frac{ 0.9^7cdot 0.1^3cdotfrac12}{frac12cdot 0.9^7cdot 0.1^3 + frac12cdot0.5^{10}}
$$
which, you may notice, has exactly the same value because algebraically we have only simplified the fraction by $binom{10}3$. So not only is it safe to assume; it doesn't matter one way or the other.

One final notational nitpick: Is there a specific reason you use both $B$ and $L$ rather than, say, $B$ and $overline B$?

share|cite|improve this answer

answered Mar 12 at 14:20

ArthurArthur

118k7118202

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Of course it seems so obvious now! Thanks. only used B and F because the notation didn't occur to me.
  $endgroup$
  – Lawrence
  Mar 13 at 9:48
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @Lawrence The only real harm is the mental overhead of remembering one extra letter. On the other hand, applying $overline{phantom B}$ isn't completely without mental overhead either. So it's not really that important. However, once you start having more events that need to be negated, the difference becomes clearer. If you have a coin and a die, both of which can be either fair or biased, and you also wonder whether it's raining outside, having three letters with overlines starts becoming a lot easier than having six letters.
  $endgroup$
  – Arthur
  Mar 13 at 9:59
  

  
  
  
  

add a comment | 

1

$begingroup$

Apart from a minor possible mistake I pointed out in a comment above, this looks good.

Is it safe to assume order is irrelevant and to look for probability of seven heads and three tails with binomial distribution

Let's see: With the binomial distribution, the final formula you get is
$$
frac{binom{10}3cdot 0.9^7cdot 0.1^3cdotfrac12}{frac12cdotbinom{10}3cdot 0.9^7cdot 0.1^3 + frac12cdotbinom{10}3cdot0.5^{10}}
$$
What happens if we don't use the binomial distribution and instead care about order? We get
$$
frac{ 0.9^7cdot 0.1^3cdotfrac12}{frac12cdot 0.9^7cdot 0.1^3 + frac12cdot0.5^{10}}
$$
which, you may notice, has exactly the same value because algebraically we have only simplified the fraction by $binom{10}3$. So not only is it safe to assume; it doesn't matter one way or the other.

One final notational nitpick: Is there a specific reason you use both $B$ and $L$ rather than, say, $B$ and $overline B$?

share|cite|improve this answer

answered Mar 12 at 14:20

ArthurArthur

118k7118202

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Of course it seems so obvious now! Thanks. only used B and F because the notation didn't occur to me.
  $endgroup$
  – Lawrence
  Mar 13 at 9:48
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @Lawrence The only real harm is the mental overhead of remembering one extra letter. On the other hand, applying $overline{phantom B}$ isn't completely without mental overhead either. So it's not really that important. However, once you start having more events that need to be negated, the difference becomes clearer. If you have a coin and a die, both of which can be either fair or biased, and you also wonder whether it's raining outside, having three letters with overlines starts becoming a lot easier than having six letters.
  $endgroup$
  – Arthur
  Mar 13 at 9:59
  

  
  
  
  

add a comment | 

$begingroup$

Apart from a minor possible mistake I pointed out in a comment above, this looks good.

Is it safe to assume order is irrelevant and to look for probability of seven heads and three tails with binomial distribution

Let's see: With the binomial distribution, the final formula you get is
$$
frac{binom{10}3cdot 0.9^7cdot 0.1^3cdotfrac12}{frac12cdotbinom{10}3cdot 0.9^7cdot 0.1^3 + frac12cdotbinom{10}3cdot0.5^{10}}
$$
What happens if we don't use the binomial distribution and instead care about order? We get
$$
frac{ 0.9^7cdot 0.1^3cdotfrac12}{frac12cdot 0.9^7cdot 0.1^3 + frac12cdot0.5^{10}}
$$
which, you may notice, has exactly the same value because algebraically we have only simplified the fraction by $binom{10}3$. So not only is it safe to assume; it doesn't matter one way or the other.

One final notational nitpick: Is there a specific reason you use both $B$ and $L$ rather than, say, $B$ and $overline B$?

share|cite|improve this answer

answered Mar 12 at 14:20

ArthurArthur

118k7118202

$endgroup$

share|cite|improve this answer

answered Mar 12 at 14:20

ArthurArthur

118k7118202

answered Mar 12 at 14:20

ArthurArthur

118k7118202

answered Mar 12 at 14:20

ArthurArthur

118k7118202

0

$begingroup$

Although it does not matter in calculation as demonstrated by Arthur, yet the condition indicates the specific order of the $10$ tosses. So ($F-$ fair, $U-$ unfair):
$$P(Fcap HHHHTTHHTH)=frac12cdot frac{1}{2^{10}};\
P(Ucap HHHHTTHHTH)=\
frac12cdot left(frac{9}{10}right)^7cdot left(frac1{10}right)^3;\
P(U|HHHHTTHHTH)=frac{P(Ucap HHHHTTHHTH)}{P(Fcap HHHHTTHHTH)+P(Ucap HHHHTTHHTH)}=\
frac{frac{9^7}{2cdot 10^{10}}}{frac{9^7}{2cdot 10^{10}}+frac{1}{2^{11}}}approx 0.3288.$$

share|cite|improve this answer

answered Mar 12 at 15:22

farruhotafarruhota

21.3k2841

$endgroup$

add a comment | 

0

$begingroup$

Although it does not matter in calculation as demonstrated by Arthur, yet the condition indicates the specific order of the $10$ tosses. So ($F-$ fair, $U-$ unfair):
$$P(Fcap HHHHTTHHTH)=frac12cdot frac{1}{2^{10}};\
P(Ucap HHHHTTHHTH)=\
frac12cdot left(frac{9}{10}right)^7cdot left(frac1{10}right)^3;\
P(U|HHHHTTHHTH)=frac{P(Ucap HHHHTTHHTH)}{P(Fcap HHHHTTHHTH)+P(Ucap HHHHTTHHTH)}=\
frac{frac{9^7}{2cdot 10^{10}}}{frac{9^7}{2cdot 10^{10}}+frac{1}{2^{11}}}approx 0.3288.$$

share|cite|improve this answer

answered Mar 12 at 15:22

farruhotafarruhota

21.3k2841

$endgroup$

add a comment | 

$begingroup$

Although it does not matter in calculation as demonstrated by Arthur, yet the condition indicates the specific order of the $10$ tosses. So ($F-$ fair, $U-$ unfair):
$$P(Fcap HHHHTTHHTH)=frac12cdot frac{1}{2^{10}};\
P(Ucap HHHHTTHHTH)=\
frac12cdot left(frac{9}{10}right)^7cdot left(frac1{10}right)^3;\
P(U|HHHHTTHHTH)=frac{P(Ucap HHHHTTHHTH)}{P(Fcap HHHHTTHHTH)+P(Ucap HHHHTTHHTH)}=\
frac{frac{9^7}{2cdot 10^{10}}}{frac{9^7}{2cdot 10^{10}}+frac{1}{2^{11}}}approx 0.3288.$$

share|cite|improve this answer

answered Mar 12 at 15:22

farruhotafarruhota

21.3k2841

$endgroup$

share|cite|improve this answer

answered Mar 12 at 15:22

farruhotafarruhota

21.3k2841

answered Mar 12 at 15:22

farruhotafarruhota

21.3k2841

answered Mar 12 at 15:22

farruhotafarruhota

21.3k2841