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If I flip two fair coins, and then tell you that one is heads, what is the probability the other coin is also heads?


The answer given in my text is $1/3$ since we have the options HT, TH, HH remaining from the original sample space. But I'm just not sure how to reconcile that with reality. Let's say I flipped a coin in the air really high and my friend called me on the phone and told me that he just flipped a coin and it was heads. As my coin falls to the ground do I surmise the probability that it lands on heads is $1/3$? Seems like it should be $1/2$ to me—how can the outcome of one coin be dependent on the other? I'm perfectly comfortable with say, the fact that the odds of getting a head and tails is $1/2$ given the sample space, but this seems different.

Is there probabilistic difference between the posture of the original question and my fictitious scenario?

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  • $\begingroup$ Events are not considered as independent. $\endgroup$
    – Creator
    Feb 5 '19 at 2:14
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The difference is

  • in the first scenario, all you're told is that one of the coins is heads, but you don't know which one. This only rules out TT, so you still have three possibilities: HH, HT or TH.
  • in the second scenario, you're told that a particular coin is heads, namely, your friend's. This rules out both TT and HT, leaving only two possibilities, HH and TH.
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If I flip two fair coins, and then tell you that one is heads, what is the probability the unknown coin is heads?

Let's say I flipped a coin in the air really high and my friend called me on the phone and told me that he just flipped a coin and it was heads.

"I flip two coins and tell you that one is heads" is a different situation than "We flip two coins and you tell me that yours is heads."

In the first situation, you don'y know which of the coins I am referring to, only that at least one among the two shows heads.

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  • $\begingroup$ So if someone doesn't tell you which coin is heads, the sample space remains large, but if they tell you which coin it is, the sample space collapses? That's hard to swallow. (1) What if they flipped the coins and said, "the one on the left is heads"? (2) What if after the flips they said, "the one on this side is heads", but you missed seeing them gesture to said side? (3) Are there experimental results on this? $\endgroup$
    – Zduff
    Feb 5 '19 at 2:34
  • $\begingroup$ If you have more information about a condition, then you may narrow down the possible outcomes. $\endgroup$ Feb 5 '19 at 2:49
  • $\begingroup$ @Zduff People may have done experiments, but I'd be surprised if they published it because it proves nothing. The law of large numbers in a sense proves that the experiment will give you the approximately right ratio. $\endgroup$ Feb 5 '19 at 2:50
  • $\begingroup$ Maybe it would prove nothing, but it would strongly suggest that a specific interpretation of a problem is wrong. $\endgroup$
    – Zduff
    Feb 5 '19 at 2:59
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    $\begingroup$ @Zduff You may be interested in this related puzzle that illustrates how having more information can change the probability: jesperjuul.net/ludologist/2010/06/08/…. $\endgroup$
    – kccu
    Feb 5 '19 at 3:15
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Most "story problems" in probability are badly formulated, with unstated or unclear assumptions. This one is no exception. What is the "unknown" coin? If you haven't told me which one is heads, then as far as I'm concerned they are both unknown. I will assume that you are asking about the probability that both coins are heads.

The probability that both coins are heads is $1/4$ under the usual assumptions of fairness and independence. That's the answer to the question you asked. However, I think you meant to ask about some conditional probability, i.e., the con;ditional probability that both coins are heads, given that you told me "one is heads". However, in order to determine that conditional probability, I have to know, under what conditions will you tell me "one is heads"? Since that information is not given in the problem, I will consider a few of the many possible cases. I will assume that the coins are a dime and a quarter (or your coin and your phone friend's coin).

I. You will tell me "one is heads" iff at least one coin is heads. In that case the conditional probability of "both heads" is $1/3$.

II. You will tell me "one is heads" iff the dime is heads. In that case the conditional probability of "both heads" is $1/2$.

III. You will tell me "one is heads" iff the quarter is heads. In that case the conditional probability of "both heads" is $1/2$.

IV. You will tell me "one is heads" iff exactly one coin is heads. In that case the conditional probability of "both heads" is $0$.

V. You will tell me "one is heads" iff both coins are heads. In that case the conditional probability of "both heads" is $1$.

VI. You will always tell me "one is heads" regardless of what happens with the coins. In that case the conditional probability of "both heads" is $1/4$, the same as the unconditional probability.

VII. You will never tell me "one is heads". In that case the conditional probability is undefined.

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  • $\begingroup$ Okay I see. I weirdly imagined it as this scenario: I will tell you "one is heads" iff I look at one but don't observe (or focus on the other) the other, and the one I do look at is heads. Which now, reflecting on your breakdown, is a very unnatural way of looking at it. So... I have no idea why I thought of it like that, but thanks for detecting and focusing on my confusion. $\endgroup$
    – Zduff
    Feb 5 '19 at 3:11
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    $\begingroup$ @Zduff If you randomly look at one of the coins and it is heads, then the probability of two heads is 1/2. These sorts of ambiguously worded conditional probability scenarios are the most annoying story problems in elementary math, and I wish they would just go away 😁. bof's answer is very good in that it shows the probabilistic ambiguity in the original wording. $\endgroup$
    – Ned
    Feb 5 '19 at 15:19
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This is ambiguous and the language can be interpreted in at least three ways.

1) you flipped two coins. One of them was heads. Let's mark that one with red tape. Let's mark the other with green tape. What is the probability of the one with green tape is also heads.

Answer: 1/2.

This is basically how you interpreted the problem. It seems a linguistically fair way to interpret it.

2) You flip two coins. I call you up and say. "Hey, is at least one of the coins heads?" You look and correctly say "yes". I ask "is the other one heads?" You answer correctly.

What is the probability you said yes.

The answer is 1/3 for the reason the book gave.[1]

This is what the book meant. It represents a basic lesson in conditional probability.

But it's a linguistic nightmare. "One lands head" ="at least one is heads" is fine but then "the other one" forces us to inaccurately declare we had one coin in mind when we said "one was heads" when we didn't

3) you flip two coins. One lands heads. It's not that zero landed heads. It's not that two landed heads. One landed heads.

What is the probability the other is heads?

Answer: 0, of course.

......

Perhaps this could be better stated as:

You flip two coins. You get at least one head. What is the probability that both are heads.

.....

I leave you with a joke.

In American currency we have the following coins: pennies worth $1$ cent, nickels worth $5$ cents, dimes worth $10$ cents, and quarters worth $25 $ cents.

I have two (American) coins. They add up two 30 cents. One of them is not a nickel. How is that possible?

Answer: The one that is not a nickel is a quarter. The one that is a nickel is a nickel.

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[1] there are four ways a coin can land.

HH, HT, TH, TT. But one of them is not possible.

The three possible ways are HH,HT,TH and they are equally likely. Of those three ways in only one of them is "the other coin" heads.

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