Mixing
Why does the conditional probability of having chosen a biased coin given an unusual result decrease with a...
$begingroup$
Let's say $1$ out of three coins is biased and lands on tails with probability $p>1/2$
We choose a coin randomly from the three coins and throw that coin $10$ times. Given that it lands on tails $8$ out of $10$ times the probability of having chosen the biased coin is $$P(B|8,2)=frac{P(8,2|B)P(B)}{P(8,2|B)P(B)+P(8,2|B^c)P(B^c)}=frac{{10choose8}p^8(1-p)^2cdot{1over3}}{{10choose8}p^8(1-p)^2cdot{1over3}+{10choose8}({1over2})^{10}cdot{2over3}}$$
And this is equal to $0.7746$ if $p=0.8$ or $0.5791$ if $p=0.6$ etc
I would expect that value to increase if we had more than $3$ coins and only one of them was biased with $p=0.8$ but it's not the case.
probability binomial-distribution
asked Jan 28 at 16:54
H. WalterH. Walter
1047
$endgroup$
add a comment |
$begingroup$
Let's say $1$ out of three coins is biased and lands on tails with probability $p>1/2$
We choose a coin randomly from the three coins and throw that coin $10$ times. Given that it lands on tails $8$ out of $10$ times the probability of having chosen the biased coin is $$P(B|8,2)=frac{P(8,2|B)P(B)}{P(8,2|B)P(B)+P(8,2|B^c)P(B^c)}=frac{{10choose8}p^8(1-p)^2cdot{1over3}}{{10choose8}p^8(1-p)^2cdot{1over3}+{10choose8}({1over2})^{10}cdot{2over3}}$$
And this is equal to $0.7746$ if $p=0.8$ or $0.5791$ if $p=0.6$ etc
I would expect that value to increase if we had more than $3$ coins and only one of them was biased with $p=0.8$ but it's not the case.
probability binomial-distribution
asked Jan 28 at 16:54
H. WalterH. Walter
1047
$endgroup$
$begingroup$
"I would expect that value to increase if we had more than $10$ coins" Do you mean 10 coins instead of the three in your first paragraph?
$endgroup$
– Arthur
Jan 28 at 16:57
$begingroup$
Please don't post the actual question in the title.
$endgroup$
– Mohammad Zuhair Khan
Jan 28 at 16:59
add a comment |
$begingroup$
Let's say $1$ out of three coins is biased and lands on tails with probability $p>1/2$
We choose a coin randomly from the three coins and throw that coin $10$ times. Given that it lands on tails $8$ out of $10$ times the probability of having chosen the biased coin is $$P(B|8,2)=frac{P(8,2|B)P(B)}{P(8,2|B)P(B)+P(8,2|B^c)P(B^c)}=frac{{10choose8}p^8(1-p)^2cdot{1over3}}{{10choose8}p^8(1-p)^2cdot{1over3}+{10choose8}({1over2})^{10}cdot{2over3}}$$
And this is equal to $0.7746$ if $p=0.8$ or $0.5791$ if $p=0.6$ etc
I would expect that value to increase if we had more than $3$ coins and only one of them was biased with $p=0.8$ but it's not the case.
probability binomial-distribution
asked Jan 28 at 16:54
H. WalterH. Walter
1047
$endgroup$
Let's say $1$ out of three coins is biased and lands on tails with probability $p>1/2$
We choose a coin randomly from the three coins and throw that coin $10$ times. Given that it lands on tails $8$ out of $10$ times the probability of having chosen the biased coin is $$P(B|8,2)=frac{P(8,2|B)P(B)}{P(8,2|B)P(B)+P(8,2|B^c)P(B^c)}=frac{{10choose8}p^8(1-p)^2cdot{1over3}}{{10choose8}p^8(1-p)^2cdot{1over3}+{10choose8}({1over2})^{10}cdot{2over3}}$$
And this is equal to $0.7746$ if $p=0.8$ or $0.5791$ if $p=0.6$ etc
I would expect that value to increase if we had more than $3$ coins and only one of them was biased with $p=0.8$ but it's not the case.
probability binomial-distribution
probability binomial-distribution
asked Jan 28 at 16:54
H. WalterH. Walter
1047
asked Jan 28 at 16:54
H. WalterH. Walter
1047
edited Jan 28 at 17:04
H. Walter
asked Jan 28 at 16:54
H. WalterH. Walter
1047
asked Jan 28 at 16:54
H. WalterH. Walter
1047
asked Jan 28 at 16:54
H. WalterH. Walter
1047
1047
$begingroup$
"I would expect that value to increase if we had more than $10$ coins" Do you mean 10 coins instead of the three in your first paragraph?
$endgroup$
– Arthur
Jan 28 at 16:57
$begingroup$
Please don't post the actual question in the title.
$endgroup$
– Mohammad Zuhair Khan
Jan 28 at 16:59
add a comment |
$begingroup$
"I would expect that value to increase if we had more than $10$ coins" Do you mean 10 coins instead of the three in your first paragraph?
$endgroup$
– Arthur
Jan 28 at 16:57
$begingroup$
Please don't post the actual question in the title.
$endgroup$
– Mohammad Zuhair Khan
Jan 28 at 16:59
$begingroup$
"I would expect that value to increase if we had more than $10$ coins" Do you mean 10 coins instead of the three in your first paragraph?
$endgroup$
– Arthur
Jan 28 at 16:57
$begingroup$
"I would expect that value to increase if we had more than $10$ coins" Do you mean 10 coins instead of the three in your first paragraph?
$endgroup$
– Arthur
Jan 28 at 16:57
$begingroup$
Please don't post the actual question in the title.
$endgroup$
– Mohammad Zuhair Khan
Jan 28 at 16:59
$begingroup$
Please don't post the actual question in the title.
$endgroup$
– Mohammad Zuhair Khan
Jan 28 at 16:59
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
If you had more than 3 coins, in the fraction
$$
P(B mid 8, 2) = frac{P(8,2|B)P(B)}{P(8,2|B)P(B)+P(8,2|B^c)P(B^c)} = frac{P(8,2|B)}{P(8,2|B)+P(8,2|B^c)frac{P(B^c)}{P(B)}}
$$
the only numbers that change are $P(B)$ and $P(B^c)$. Namely, $P(B)$ goes down, and $P(B^c)$ goes up. That means that the denominator goes up, which means your resulting probability goes down.
In Bayesian terms: your evidence $P(8, 2 mid B)$ has not changed, your prior $P(B)$ has gone down, so your posterior $P(B mid 8, 2)$ has also gone down.
answered Jan 28 at 17:02
Mees de VriesMees de Vries
17.6k13059
$endgroup$
1
$begingroup$
+1. In intuitive terms, as the total number of coins increases but all else is kept equal, you get the same amount of additional evidence that you picked the biased coin, but less reason to believe that you picked the biased coin in the first place. So that second term in the denominator, which is sort of your "prior skepticism" that you picked a biased coin, increases.
$endgroup$
– Ian
Jan 28 at 17:11
add a comment |
$begingroup$
Here is an appeal to intuition. Think of these two scenarios:
- You have ten coins, one of which is biased. You pick one of them and toss it ten times, getting 8 tails.
- You have ten coins, one of which is biased. You eliminate seven of them using some completely certain method. Then you pick up one of the remaining three and toss it ten times, getting tails 8 times.
Which of these scenarios makes you more likely to hold the biased coin, you think? Note that the second case is exactly the same as just starting with three coins.
answered Jan 28 at 17:09
ArthurArthur
120k7121206
$endgroup$
add a comment |
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2 Answers
2
active
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2 Answers
2
active
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active
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active
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$begingroup$
If you had more than 3 coins, in the fraction
$$
P(B mid 8, 2) = frac{P(8,2|B)P(B)}{P(8,2|B)P(B)+P(8,2|B^c)P(B^c)} = frac{P(8,2|B)}{P(8,2|B)+P(8,2|B^c)frac{P(B^c)}{P(B)}}
$$
the only numbers that change are $P(B)$ and $P(B^c)$. Namely, $P(B)$ goes down, and $P(B^c)$ goes up. That means that the denominator goes up, which means your resulting probability goes down.
In Bayesian terms: your evidence $P(8, 2 mid B)$ has not changed, your prior $P(B)$ has gone down, so your posterior $P(B mid 8, 2)$ has also gone down.
answered Jan 28 at 17:02
Mees de VriesMees de Vries
17.6k13059
$endgroup$
1
$begingroup$
+1. In intuitive terms, as the total number of coins increases but all else is kept equal, you get the same amount of additional evidence that you picked the biased coin, but less reason to believe that you picked the biased coin in the first place. So that second term in the denominator, which is sort of your "prior skepticism" that you picked a biased coin, increases.
$endgroup$
– Ian
Jan 28 at 17:11
add a comment |
$begingroup$
If you had more than 3 coins, in the fraction
$$
P(B mid 8, 2) = frac{P(8,2|B)P(B)}{P(8,2|B)P(B)+P(8,2|B^c)P(B^c)} = frac{P(8,2|B)}{P(8,2|B)+P(8,2|B^c)frac{P(B^c)}{P(B)}}
$$
the only numbers that change are $P(B)$ and $P(B^c)$. Namely, $P(B)$ goes down, and $P(B^c)$ goes up. That means that the denominator goes up, which means your resulting probability goes down.
In Bayesian terms: your evidence $P(8, 2 mid B)$ has not changed, your prior $P(B)$ has gone down, so your posterior $P(B mid 8, 2)$ has also gone down.
answered Jan 28 at 17:02
Mees de VriesMees de Vries
17.6k13059
$endgroup$
1
$begingroup$
+1. In intuitive terms, as the total number of coins increases but all else is kept equal, you get the same amount of additional evidence that you picked the biased coin, but less reason to believe that you picked the biased coin in the first place. So that second term in the denominator, which is sort of your "prior skepticism" that you picked a biased coin, increases.
$endgroup$
– Ian
Jan 28 at 17:11
add a comment |
$begingroup$
If you had more than 3 coins, in the fraction
$$
P(B mid 8, 2) = frac{P(8,2|B)P(B)}{P(8,2|B)P(B)+P(8,2|B^c)P(B^c)} = frac{P(8,2|B)}{P(8,2|B)+P(8,2|B^c)frac{P(B^c)}{P(B)}}
$$
the only numbers that change are $P(B)$ and $P(B^c)$. Namely, $P(B)$ goes down, and $P(B^c)$ goes up. That means that the denominator goes up, which means your resulting probability goes down.
In Bayesian terms: your evidence $P(8, 2 mid B)$ has not changed, your prior $P(B)$ has gone down, so your posterior $P(B mid 8, 2)$ has also gone down.
answered Jan 28 at 17:02
Mees de VriesMees de Vries
17.6k13059
$endgroup$
If you had more than 3 coins, in the fraction
$$
P(B mid 8, 2) = frac{P(8,2|B)P(B)}{P(8,2|B)P(B)+P(8,2|B^c)P(B^c)} = frac{P(8,2|B)}{P(8,2|B)+P(8,2|B^c)frac{P(B^c)}{P(B)}}
$$
the only numbers that change are $P(B)$ and $P(B^c)$. Namely, $P(B)$ goes down, and $P(B^c)$ goes up. That means that the denominator goes up, which means your resulting probability goes down.
In Bayesian terms: your evidence $P(8, 2 mid B)$ has not changed, your prior $P(B)$ has gone down, so your posterior $P(B mid 8, 2)$ has also gone down.
answered Jan 28 at 17:02
Mees de VriesMees de Vries
17.6k13059
answered Jan 28 at 17:02
Mees de VriesMees de Vries
17.6k13059
answered Jan 28 at 17:02
Mees de VriesMees de Vries
17.6k13059
answered Jan 28 at 17:02
Mees de VriesMees de Vries
17.6k13059
17.6k13059
1
$begingroup$
+1. In intuitive terms, as the total number of coins increases but all else is kept equal, you get the same amount of additional evidence that you picked the biased coin, but less reason to believe that you picked the biased coin in the first place. So that second term in the denominator, which is sort of your "prior skepticism" that you picked a biased coin, increases.
$endgroup$
– Ian
Jan 28 at 17:11
add a comment |
1
$begingroup$
+1. In intuitive terms, as the total number of coins increases but all else is kept equal, you get the same amount of additional evidence that you picked the biased coin, but less reason to believe that you picked the biased coin in the first place. So that second term in the denominator, which is sort of your "prior skepticism" that you picked a biased coin, increases.
$endgroup$
– Ian
Jan 28 at 17:11
1
1
$begingroup$
+1. In intuitive terms, as the total number of coins increases but all else is kept equal, you get the same amount of additional evidence that you picked the biased coin, but less reason to believe that you picked the biased coin in the first place. So that second term in the denominator, which is sort of your "prior skepticism" that you picked a biased coin, increases.
$endgroup$
– Ian
Jan 28 at 17:11
$begingroup$
+1. In intuitive terms, as the total number of coins increases but all else is kept equal, you get the same amount of additional evidence that you picked the biased coin, but less reason to believe that you picked the biased coin in the first place. So that second term in the denominator, which is sort of your "prior skepticism" that you picked a biased coin, increases.
$endgroup$
– Ian
Jan 28 at 17:11
add a comment |
$begingroup$
Here is an appeal to intuition. Think of these two scenarios:
- You have ten coins, one of which is biased. You pick one of them and toss it ten times, getting 8 tails.
- You have ten coins, one of which is biased. You eliminate seven of them using some completely certain method. Then you pick up one of the remaining three and toss it ten times, getting tails 8 times.
Which of these scenarios makes you more likely to hold the biased coin, you think? Note that the second case is exactly the same as just starting with three coins.
answered Jan 28 at 17:09
ArthurArthur
120k7121206
$endgroup$
add a comment |
$begingroup$
Here is an appeal to intuition. Think of these two scenarios:
- You have ten coins, one of which is biased. You pick one of them and toss it ten times, getting 8 tails.
- You have ten coins, one of which is biased. You eliminate seven of them using some completely certain method. Then you pick up one of the remaining three and toss it ten times, getting tails 8 times.
Which of these scenarios makes you more likely to hold the biased coin, you think? Note that the second case is exactly the same as just starting with three coins.
answered Jan 28 at 17:09
ArthurArthur
120k7121206
$endgroup$
add a comment |
$begingroup$
Here is an appeal to intuition. Think of these two scenarios:
- You have ten coins, one of which is biased. You pick one of them and toss it ten times, getting 8 tails.
- You have ten coins, one of which is biased. You eliminate seven of them using some completely certain method. Then you pick up one of the remaining three and toss it ten times, getting tails 8 times.
Which of these scenarios makes you more likely to hold the biased coin, you think? Note that the second case is exactly the same as just starting with three coins.
answered Jan 28 at 17:09
ArthurArthur
120k7121206
$endgroup$
Here is an appeal to intuition. Think of these two scenarios:
- You have ten coins, one of which is biased. You pick one of them and toss it ten times, getting 8 tails.
- You have ten coins, one of which is biased. You eliminate seven of them using some completely certain method. Then you pick up one of the remaining three and toss it ten times, getting tails 8 times.
Which of these scenarios makes you more likely to hold the biased coin, you think? Note that the second case is exactly the same as just starting with three coins.
answered Jan 28 at 17:09
ArthurArthur
120k7121206
answered Jan 28 at 17:09
ArthurArthur
120k7121206
answered Jan 28 at 17:09
ArthurArthur
120k7121206
answered Jan 28 at 17:09
ArthurArthur
120k7121206
120k7121206
add a comment |
add a comment |
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$begingroup$
"I would expect that value to increase if we had more than $10$ coins" Do you mean 10 coins instead of the three in your first paragraph?
$endgroup$
– Arthur
Jan 28 at 16:57
$begingroup$
Please don't post the actual question in the title.
$endgroup$
– Mohammad Zuhair Khan
Jan 28 at 16:59