A box contains 2 biased coins, one with probability of 0.4 of head and other with probability 0.7 of head. A coim in randomly chose and tossed 10 times. Evaluate the conditional expectation of the number of heads given that the first 2 of the first 3 tosses were head.

My attempt:

Let $X$ denote the number of heads that appeared on those tosses. Define $$ X_i=1 \text{if a head appears}$$
$$X_i=0, otherwise$$

If a coin is randomly chosen, the probability of head should be

$$\frac{1}{2} 0.4 + \frac{1}{2} 0.7$$



Now, we can see that


Note that there are 4 cases where 2 heads (H) appears on the first 3 tosses:


If $ Y $is the event where 2 heads appear on the first 3 tosses, then, for example,

$$E(X_1|Y)=P(HHH)+P(HHT)+P(HTH) = (0.55)^3 + 2(0.55)^20.45 $$

We should have the same for $ X_2, X_3.$ But,$ X_4, ..., X_{10}$ are independent of Y (considering independent tosses), then

$$E[X_i|Y]=E[X_i]=0.55, i=4,...,10$$

Now, we shpuld sum and obtain the conditional expectation of $X.$ But I am getting 5.16, while the answer is 6.06. What is wrong?


  • $\begingroup$ The condition that "the first 2 of the first 3 tosses were heads" looks like a transcription error. Taken literally, it says the first 2 tosses (out of all 10) were heads; "of the first 3 tosses" is redundant information. Was it supposed to say "at least 2 of the first 3 tosses were heads," or perhaps even "exactly 2 of the first 3 tosses were heads"? $\endgroup$ – David K Oct 16 '14 at 18:52
  • $\begingroup$ First $2$ of first $3$ is well defined to me in this context. Suppose we had ABCDEFGHIJ as names to represent the $10$ coin tosses in order (A is first). The first $2$ of the first $3$ would be the first $2$ of ABC which are AB. Based on his example yes I agree he has it worded wrongly. $\endgroup$ – David Oct 16 '14 at 21:09

The question is a little ambiguous. When you say "the first 2 of the first 3 tosses are heads," that is clearly a typo; based on the answer you gave, I assume it should read "exactly 2 of the first 3 tosses are heads." (If it should instead read "at least 2 of the first 3 tosses are heads," the answer will differ.)

Your main error is treating the tosses as independent of each other: All the tosses depend on which coin you chose, and are independent given that choice, which isn't the same thing. So $$ P(2H \;|\; A)=3(0.4)^2(0.6) = 0.288, \\ P(2H \;|\; B)=3(0.7)^2(0.3) = 0.441, $$ where $A$ means you chose the tail-weighted coin and $B$ means you chose the head-weighted coin. In case $A$, the expected number of heads is clearly $2 + 0.4\times 7 = 4.8$; and in case $B$ it's $2 + 0.7\times 7=6.9$. So you can write $$ E[N_H\;|\;2H]=E[N_H\;|\;A]P(A\;|\;2H)+E[N_H\;|\;B]P(B\;|\;2H)\\ =4.8P(A \; | \; 2H)+6.9(1-P(A\;|\;2H))=6.9-2.1P(A\;|\;2H). $$ Moreover, $$ P(A\;|\;2H)=\frac{P(2H\;|\;A)P(A)}{P(2H)}=\frac{0.288 \cdot0.5}{0.288\cdot 0.5 + 0.441 \cdot 0.5}=\frac{32}{81}. $$ The result is then $$ E[N_H \;|\; 2H] = 6.9 - 2.1\times\frac{32}{81}=6.07. $$


The fact that you had more heads than tails in the first three tosses tells you that the coin with probability $0.7$ was more likely chosen. This affects the conditional expectation of all the tosses, not just the first three.

Judging from the result you were supposed to get, I think the "two of the first three" condition is supposed to mean that there were exactly two heads in the first three tosses. The prior probability of this is $p_1 = 3(0.7)^2(0.3)$ for the $0.7$ coin but $p_2 = 3(0.4)^2(0.6)$ for the $0.4$ coin. This makes the probability $\frac{p_1}{p_1 + p_2} = \frac{49}{81}$ that the $0.7$ coin is the one that was chosen. The expected number of heads is therefore $$2 + 7 \left(0.7 \frac{49}{81} + 0.4 \frac{32}{81} \right).$$ (I get approximately $6.0704$ when computing this value; I'm not sure why the answer key says $6.06.$ But other interpretations of the "two of three" condition seem to take us farther away from that value.)

  • $\begingroup$ How did you calculate those probabilities? $\endgroup$ – Giiovanna Oct 16 '14 at 19:10
  • $\begingroup$ My initial calculations (before editing the answer) actually were incorrect. I have corrected them and also given a bit more detail on how I got the conditional probability that the $0.7$ coin was chosen. $\endgroup$ – David K Oct 16 '14 at 19:24
  • $\begingroup$ How can anyone accurately tell that the head biased coin was more likely chosen just from the first $3$ coin flips? It seems to me as long as either coin can generate the same outcome, then there should be no bias as to which coin was chosen and therefore each one has to be equally "weighted" in the final answer as if it had an equal chance of being selected. $\endgroup$ – David Oct 16 '14 at 21:29
  • $\begingroup$ @David - This is a straightforward application of Bayes' Theorem. Even one flip of the coin tells you something about which coin you chose (that is, the probability you chose the $0.7$ coin is no longer $\frac 12$). $\endgroup$ – David K Oct 16 '14 at 23:12

As the question is written, you are saying that HH was gotten for the first $2$ coin tosses. I would say from only $2$ coin tosses that no conclusion can be drawn as to which biased coin was chosen because either one could have "easily" generated HH.

Thus my answer would be $2 + 8 * 0.55 = 6.4$

  • $\begingroup$ Compare this answer to the precisely-notated conditional probabilities computed in the answer by mjqxxxx. Using the notation of mjqxxxx, this answer asserts that $P(A\mid 2H) = \frac 12,$ which mjqxxxx showed is not true. $\endgroup$ – David K Oct 16 '14 at 23:21
  • $\begingroup$ It is debatable whether it is $50$% or not. With only $2$ heads observed, it could easily be either coin. For example, what if we didn't tell you what the outcome was of the first $2$ tosses, then what would be probability of each coin be? $50$%? So then how is telling you the outcome of the first $2$ tosses going to affect that? That would kinda be like saying if I had a fair coin (but didn't tell you it was a fair coin) and got $2$ heads on the first $2$ tosses that you would conclude that the coin is biased towards heads. $\endgroup$ – David Oct 16 '14 at 23:48
  • $\begingroup$ Why would I suspect that your coin is biased? If I had to assign a prior probability to that, it would be tiny, and the posterior probability also would be tiny after two heads. The OP, on the other hand, has very precise priors for everything. The only thing to debate is whether you want to use a standard model of probability or some kind of "fuzzy" model--which I suppose you can do, but the OP was working a problem whose answer was "supposed to" be precise to three digits, which implies to me they were looking for something more standard. $\endgroup$ – David K Oct 17 '14 at 2:39
  • $\begingroup$ Answer this question specifically. If I had a fair coin and didn't tell you it was fair and you saw $2$ heads come up out of the first $3$ tosses, would you be confident to say the coin is biased at that point? If not, then how can you say that seeing $2$ heads in $3$ tosses makes it more likely that the coin tossed was the one biased towards heads? My philosophy is that short term stuff is not indicative of long term stuff. For example, in tossing a coin that is $70$% biased towards heads, you may get a short string of tails to start but eventually heads will prevail. $\endgroup$ – David Oct 17 '14 at 4:23
  • $\begingroup$ Of course you can get a short-term trend opposite to the long-term trend. That's why the conditional probability after 3 flips is only slightly different from $\frac 12,$ and the only reason I'm willing to compute it is because we know that the prior probability was $\frac 12.$ Without that prior probability, it's a completely different question, with a completely different answer, and completely irrelevant to OP's question. $\endgroup$ – David K Oct 17 '14 at 12:29

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