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Let's assume I have two coins for which I know the probabilities of heads - they are both fair coins of 0.5. So if I toss them both and get one heads and one tails, for me they are indistinguishable, so my probability of that event is 0.5.

Now let's assume that I know which one is slightly unfair with probability of heads 0.49999999999999. Once I toss them and get one heads and one tails, the probability of that event is approximately 0.25 because they are now distinguishable by their varying probability. So an infinitesimal variation in probability seems to change the problem categorically. Where am I making the error of judgment?

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  • $\begingroup$ Why is the probability of observing one head and one tail $\frac 12$? It seems to me that it should be at most $\frac 13$ since there are (at least) three possible outcomes that you might observe: (i) one head and one tail, (ii) two heads, and (iii) two tails. More rational people would argue that the coins are distinguishable, especially if you toss them in succession, or simultaneously toss one with the right hand and one with the left hand, etc. $\endgroup$
    – Dilip Sarwate
    Dec 4 '17 at 21:36
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    $\begingroup$ How do you observe a probability directly? Regardless, if we accept the coins are distinguishable, then yes: the event that the first is heads and the second is tails has a probability close to $0.25$. But so what? You are describing two different events: in the first example the event consists of two distinct outcomes (whether you can tell them apart is immaterial) whereas in the second example it consists of just one outcome. $\endgroup$
    – whuber ♦
    Dec 4 '17 at 21:38
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    $\begingroup$ @ Dilip, the fact there are three outcomes doesn't mean that probability is at most 1/3. @ whuber Let's say I have coins that have their probabilities written on them, and they are true. So I select a pair, toss the in front of you and ask you the probability of that event. Is it not the same event now? $\endgroup$
    – Cindy Almighty
    Dec 4 '17 at 21:52
  • $\begingroup$ If you know which one has which probability, what information can there possibly be in the result of the coin flips? You already know everything there is to know! $\endgroup$
    – jbowman
    Dec 4 '17 at 21:58
  • $\begingroup$ @jbowman The probability of the outcome that is the result of both coins. $\endgroup$
    – Cindy Almighty
    Dec 4 '17 at 22:19
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Ignoring the distinction between unconditional and conditional probabilities, and what are the effects of conditioning on the probabilities, is where the conceptual issue lies here.

Assume the two coins are visibly numbered, $c_1$ with known heads probability $p_1$, and $c_2$ with known heads probability $p_2$. Neither coins are assumed, or not assumed, "fair".

Then if we are told that the throw returned "one heads and one tails", but we have not actually looked at the coins, then the best we can say is the same thing we would have said prior to the throw:

$$P[\{HT,TH\}] = p_1(1-p_2) + (1-p_1)p_2$$

Assume now that we observe the coins, and them being visibly marked, we know which one came up heads and which one came up tails. But now, if we are to take into account what we have just observed, we are talking for the probability of the said event conditional on the additional information due to observing the coins. So here, we are trying to calculate the probability, say,

$$P\Big[\{HT,TH\} \mid \{c_1=H, c_2=T\}\Big]$$

It should be evident that now we are calculating a different probability than before, and so in any case, we should not confuse the first situation with the second. So let's focus on this second situation. Since we know which coin came up heads, we are no longer considering the event $\{HT,TH\}$, because it describes a situation where either of the coins came up heads, while we know which one actually came up heads. Therefore we conclude that

$$P\Big[\{HT,TH\} \mid \{c_1=H, c_2=T\}\Big] = P\Big[\{HT\}\Big] = p_1(1-p_2)$$

And this has nothing to do with what are the values $p_1$ and $p_2$, the "drop" in probability value does not happen due to their "varying probability", but because, by introducing additional information, we have restricted ourselves to consider a different event than without this information.

If this has been presented as some kind of conceptual "problem", "paradox" etc, it should not.

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  • $\begingroup$ Thank you for a detailed reply. I intuitively understand the distinction you outlined, and you have formalized it perfectly. The problem is conceptual in a sense that I am having a problem with setting up a good model for my problem. Let me elaborate further. $\endgroup$
    – Cindy Almighty
    Dec 4 '17 at 23:21
  • $\begingroup$ Let's say I've been calculating the probability of combinations of coin tosses assuming they are fair via a binomial distribution. Then someone gives me information that coins from 1994 have shifted probability 0.55:0.45. How can I include this info without converting the whole problem to the problem of distinct coins? I know how to modify the formula to use the information that, for example, 20% of coins are from 1994 and adjust the probability, but what happens after I see which coins are from which year? Is there no way to factor in that information without completely changing the approach? $\endgroup$
    – Cindy Almighty
    Dec 4 '17 at 23:24
  • $\begingroup$ But you do not "change completely the approach". My example shows that observing an outcome just re-directs us back to the same probability distribution, only to a different event. it is one thing to "incorporate a priori the knowledge that 20% of coins are from 1994". It is a different thing to observe an outcome and then go back and obtain what has been its a priori probability. $\endgroup$
    – Alecos Papadopoulos
    Dec 5 '17 at 0:02
  • $\begingroup$ I completely change the approach in conceptualizing the event as a physical thing. Let's say that I am trying to set a threshold on what I will accept, and I say I will not accept anything below a result with CDF 0.05 (in terms of number of heads). This can be calculated for any total number of coins. What if I want to use the individual outcomes because I just got them? So there is no way to incorporate that information and still use CDF? How do I compare results for 1000 coins and 3 coins when the probability of individual events keeps dropping with the number of coins? Thanks! $\endgroup$
    – Cindy Almighty
    Dec 5 '17 at 8:37
  • $\begingroup$ Another way to ask this is: Can I enrich the first approach with new info on individual tosses without changing the event in the statistical sense? $\endgroup$
    – Cindy Almighty
    Dec 5 '17 at 8:41
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What changed the problem is not the tiny change in probability but the change in your knowledge of the coins. Let's say that you have 2 coins that are precisely fair. Then the probability of [H,T] is 1/4 and the probability of [T,H] is also 1/4. This is regardless of whether the coins are distinguishable. But if they are distinguishable then you can tell the two outcomes apart (and that's regardless of whether they are precisely fair).

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