Current Events > What is the probability that two coins land on heads given that at least one did

(+) Yeah! (+)

Let's see how smart CE is...
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50/50

It either happens or it doesnt
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Two_Dee posted...

50/50

It either happens or it doesnt

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I'm giving any CE topic on stats and probabilities no more than twenty posts until someone gets really angry at either the Monty Hall Problem or, better, the Birthday Paradox:

In a room of just 23 people theres a 50-50 chance of two people having the same birthday. In a room of 75 theres a 99.9% chance of two people matching.

prettyprincess posted...

Two_Dee posted...

50/50

It either happens or it doesnt

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prettyprincess posted...

Two_Dee posted...

50/50

It either happens or it doesnt

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So far CE's not that smart...
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ArchiePeck posted...

or, better, the Birthday Paradox:

In a room of just 23 people theres a 50-50 chance of two people having the same birthday. In a room of 75 theres a 99.9% chance of two people matching.

first one seems right
but the second one would also be 50-50
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Tmaster148 posted...

75% (3/4)

nope

RedWhiteBlue posted...

Not gonna do the math but 25%?

surprisingly no
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I actually misread the question.

It's 1/3.

HH
HT
TH

You don't count TT.
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Tmaster148 posted...

I actually misread the question.

It's 1/3.

HH
HT
TH

You don't count TT.

This
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Bayesian formula hijinks?
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prettyprincess posted...

ArchiePeck posted...

or, better, the Birthday Paradox:

In a room of just 23 people theres a 50-50 chance of two people having the same birthday. In a room of 75 theres a 99.9% chance of two people matching.

first one seems right
but the second one would also be 50-50

nope, it's 99.9% at 70 people.
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getting pretty sick of the "I'm proud to not know math" meme.
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50/50*
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People always trust intuition over logic

It's why we're so dumb
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TC comes from r/iamverysmart
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The odds of anything at all are always about tree fiddy
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https://www.youtube.com/watch?v=hI-j9TItxtU
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You got a 33 1/3 chance of landing on heads. But I got a 66 2/3 chance of landing on heads.

So you take your 33 1/3 chance, minus my 25% chance, and you got an 8 1/3 chance of landing on heads.

But then you take my 75% chance of landing on heads, and then add 66 2/3 percents, I got 141 2/3 chance of landing on heads.

The numbers don't lie!
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RedWhiteBlue posted...

Tmaster148 posted...

I actually misread the question.

It's 1/3.

HH
HT
TH

You don't count TT.

Actually you forgot the other HH. First is the free heads, second HH is where the second H is the free heads.

So 50%

I feel like that one doesnt count because either way there is one free heads and one earned heads
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Alkaloid posted...

People always trust intuition over logic

It's why we're so dumb

I live reasonably close to Atlantic City so inevitably I have friends who like to gamble. Listening to their rationale can be mind boggling. "Well if I'm lucky that night, I raise my bets." Take that AC money and spend it on a night course in statistics...
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Its 50%.

If one of them is guaranteed to land on heads, then the only relevant coin in the calculation is the other one, which has a 50% chance to land on heads or tails.

Ic3Bullet posted...

Its 50%.

If one of them is guaranteed to land on heads, then the only relevant coin in the calculation is the other one, which has a 50% chance to land on heads or tails.

I'm pretty sure this is correct, and an application of Bayes rule will get you there.
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COVxy posted...

Ic3Bullet posted...

Its 50%.

If one of them is guaranteed to land on heads, then the only relevant coin in the calculation is the other one, which has a 50% chance to land on heads or tails.

I'm pretty sure this is correct, and an application of Bayes rule will get you there.

That would only be true if the situation was that the first coin toss was heads. It's not. You can either get a Heads on the first toss or the second toss.
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Ic3Bullet posted...

Its 50%.

If one of them is guaranteed to land on heads, then the only relevant coin in the calculation is the other one, which has a 50% chance to land on heads or tails.

If you knew that a particular coin had landed heads, this would be correct. You don't. You only know that at least one coin is heads.

tiornys posted...

You only know that at least one coin is heads.

Exactly so you only worry about one since you know one will be heads.
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dave_is_slick posted...

tiornys posted...

You only know that at least one coin is heads.

Exactly so you only worry about one since you know one will be heads.

Let's try presenting this rigorously. The probability of event A happening given that event B has happened is P(A and B)/P(B). In this case, event A is "both coins are heads" and event B is "at least one coin is heads". P(A and B) = P(A) = 1/4. If both coins are heads, then it must be the case that at least one coin is heads, and the odds of getting two heads with no other knowledge is very straightforward. P(B) = 3/4. Namely, if the two flips are anything other than TT, then at least one coin is heads. (1/4)/(3/4) = 1/3.

You can't eliminate one of the coins based on the information that "at least one coin is heads". You would need to know "the first coin is heads" or similar.

Two_Dee posted...

50/50

It either happens or it doesnt

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what if it lands perfectly on it's side

checkmate liberals
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Tmaster148 posted...

That would only be true if the situation was that the first coin toss was heads. It's not. You can either get a Heads on the first toss or the second toss.

Ah, the wording is a bit artificial. In that case, 1*.25/.75 = 1/3

It is a very artificially constructed question, as there are no real circumstances where both coins are flipped in order and you'd have information regarding only one but not both. The only way to really construct this experiment in the mind is that you flip one, check, and then flip the other.
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COVxy posted...

Ic3Bullet posted...

Its 50%.

If one of them is guaranteed to land on heads, then the only relevant coin in the calculation is the other one, which has a 50% chance to land on heads or tails.

I'm pretty sure this is correct, and an application of Bayes rule will get you there.

Though I'm inclined to side with the person with the expected value signature,

Tmaster148 posted...

I actually misread the question.

It's 1/3.

HH
HT
TH

You don't count TT.

This is the least confusing interpretation to me. Probability has some fucked up results and if the problem space is small enough to write down, I'm writing it down.
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COVxy posted...

Ah, the wording is a bit artificial. In that case, 1*.25/.75 = 1/3

It is a very artificially constructed question, as there are no real circumstances where both coins are flipped in order and you'd have information regarding only one but not both. The only way to really construct this experiment in the mind is that you flip one, check, and then flip the other.

One way to construct this scenario with coins is as follows:

--I flip two coins behind a curtain
--I tell you that at least one coin is heads
--You can now bet on whether or not both coins are heads

Which bet you should make depends on the betting odds offered. If you're being offered even money, you should bet against both coins being heads.

For a more realistic scenario, we need to move away from coin flips. Say I have a quantum process whereby some particle decays into some other particle 50% of the time. Say I also have a detector that can sense the presence of a given particle but not the amount of that particle. If I start with two particles of the first type, then this exact calculation gives me the odds that both particles have decayed if my detector tells me that at least one has decayed.

If you have two detectors, one for each particle, given that the sensor is selective to a given particle, the problem breaks down into a different problem.

Both detectors go off:
either both decayed or neither decayed
If only one goes off:
you have specific information about which particle decayed into the other.

It seems like this is still very artificially constructed. Though, I may be simply misreading your realistic scenario.

EDIT: Oh, letting off say two particles of the same type. Above regards two particles each different. But you can work out the reasonable inferences given the other situation too. Now if you had a general decay detector, and the particles necessarily decayed at the same time, and the detector only has the sensitivity to detect occurrence, the analogy works.
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COVxy posted...

Now if you had a general decay detector, and the particles necessarily decayed at the same time, and the detector only has the sensitivity to detect occurrence, the analogy works.

Yes, this is what I was trying to say.

A real-life example (the puzzle is commonly told in these terms): consider all families with two children, of which at least one is a boy. How many of those families have two boys?
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Tmaster148 posted...

COVxy posted...

Ic3Bullet posted...

Its 50%.

If one of them is guaranteed to land on heads, then the only relevant coin in the calculation is the other one, which has a 50% chance to land on heads or tails.

I'm pretty sure this is correct, and an application of Bayes rule will get you there.

That would only be true if the situation was that the first coin toss was heads. It's not. You can either get a Heads on the first toss or the second toss.

There are still only two possible outcomes. Either both will be heads, or one will be tails. Therefore 50%.

Ic3Bullet posted...

Tmaster148 posted...

COVxy posted...

Ic3Bullet posted...

Its 50%.

If one of them is guaranteed to land on heads, then the only relevant coin in the calculation is the other one, which has a 50% chance to land on heads or tails.

I'm pretty sure this is correct, and an application of Bayes rule will get you there.

That would only be true if the situation was that the first coin toss was heads. It's not. You can either get a Heads on the first toss or the second toss.

There are still only two possible outcomes. Either both will be heads, or one will be tails. Therefore 50%.

For an individual coin yes. But not for the set of possible outcomes of 2 coin flips.
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Ic3Bullet posted...

There are still only two possible outcomes. Either both will be heads, or one will be tails. Therefore 50%.

HT and TH are not the same thing.
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H-known H-Unknown
H-Unknown H-Known
H-known T-Unknown
T-Unknown H-Known

The two tails outcomes don't exist.

It's 50/50
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Sex-Machine posted...

H-known H-Unknown
H-Unknown H-Known

These are not separate cases, so you're double counting HH.

tiornys posted...

Sex-Machine posted...

H-known H-Unknown
H-Unknown H-Known

These are not separate cases, so you're double counting HH.

Then why are the two tails outcomes different
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Sex-Machine posted...

Then why are the two tails outcomes different

Because if you were to experience those events they would physically be distinguishable. Not so much with HH.
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Sex-Machine posted...

tiornys posted...

Sex-Machine posted...

H-known H-Unknown
H-Unknown H-Known

These are not separate cases, so you're double counting HH.

Then why are the two tails outcomes different

Because order matters.

HH is the same as HH. HT is not the same as TH.
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Sex-Machine posted...

tiornys posted...

Sex-Machine posted...

H-known H-Unknown
H-Unknown H-Known

These are not separate cases, so you're double counting HH.

Then why are the two tails outcomes different

treat the coins as penny and dime.

you know one coin is heads, but not which. each coin has 2 possible outcomes, so in total there are 4 possible outcomes:

PennyHead + DimeHead
PennyHead + DimeTail
PennyTail + DimeHead
PennyTail + DimeTail.

from the setup, you know the 4th outcome is impossible. and of the three others, only the first is a success. therefore, 1/3.

it is functionally the same as if the coins are the same, it's just a lttle harder to see.
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Sex-Machine posted...

tiornys posted...

Sex-Machine posted...

H-known H-Unknown
H-Unknown H-Known

These are not separate cases, so you're double counting HH.

Then why are the two tails outcomes different

Because ordering matters before the choice in that case.

If you want, you can frame the problem in a way such that H-Known H-Unknown really is a separate case from H-unknown H-Known, but if you do it that way, then the total sum of the probability of the two H-H cases equals the probability of the H-T case equals the probability of the T-H case. Here's how that would work:

I flip two coins behind a curtain. With equal probability, I get H-H, or H-T, or T-H, or T-T. Because of the problem parameters, I didn't get T-T. The remaining three possibilities still have equal probability. I now tell you that at least one coin is heads. In the H-T or T-H cases, I am forced to pick the single heads result as the "known" result. In the H-H case, I can choose. Let's say I pick randomly between them with equal chance for each. Then H(k)-H occurs in half of H-H cases, and H-H(k) occurs in the other half of H-H cases.

So now the probability break down looks like this:
P(H(k)-H) = 1/2 * 1/3 = 1/6
P(H-H(k)) = 1/2 * 1/3 = 1/6
P(H(k)-T) = 1 * 1/3 = 1/3
P(T-H(k)) = 1 * 1/3 = 1/3

If I use some method to pick that gives unequal chances, the individual probabilities of P(H(k)-H) and P(H-H(k)) will change but their combined probability will not. For example, if I pick the first coin 90% of the time, then P(H(k)-H) = 9/10 * 1/3 = 3/10 and P(H-H(k)) = 1/10 * 1/3 = 1/30 for a combined probability of 10/30 = 1/3.

Butterfiles posted...

what if it lands perfectly on it's side

checkmate liberals

This is why I posted the video. It is super rare, but it can and does happen. I personally have had it happen twice in my life.
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