You flip a fair coin twice resulting in the sample space {HH, HT, TH, TT}. How can we show that each point in the sample space has probability 1/4 without invoking independence?

Work so far

If the coin flips are independent, we get the answer immediately: by fairness, we know p(H) = p(T) = 1/2 and by independence we have p({HT}) = p(H) * p(T) = 1/4 (with the same result for any other point in the sample space).

But how do we do it without the independence assumption?

Here's the path I've been working on:

Let event A = Heads on first flip = {HH, HT} and let event B = Tails on first flip = {TT, TH}.

By disjointness and fairness we have p(A) = P({HH}) + p({HT}) = 1/2 and p(B) = P({TT}) + p({TH}) = 1/2

This gives us 2 equations in 4 unknowns. We can add to this system the fact that p(sample space) = p{HH, HT, TH, TT} = p({HH}) + p({HT}) + P({TT}) + p({TH}) = 1.

The idea would be to define another event that does not assume independence that would provide a fourth equation and allow us to solve this system for the component point-probabilities.

Can anyone provide guidance?

  • $\begingroup$ What do you mean by a "fair coin"? The usual thought would include independence ("each toss comes up H with probability $\frac 12$ " or something like that). Or do you have a different notion of "fair" in mind? $\endgroup$ – lulu Sep 7 '15 at 2:10
  • $\begingroup$ Here, for example: en.wikipedia.org/wiki/Fair_coin . Independence is built into the definition. $\endgroup$ – lulu Sep 7 '15 at 2:12
  • $\begingroup$ Appreciated. In this context, a fair coin means the probabilities of heads and tails are equal. I'd like to understand how to proceed without independence although I understand that is part of the typical understanding of a 'fair coin.' $\endgroup$ – Chris Sep 7 '15 at 2:24
  • $\begingroup$ You mean, "on each toss the probabilities are equal"? That's independence. $\endgroup$ – lulu Sep 7 '15 at 2:25
  • $\begingroup$ I see what you are saying, but assuming independence from the start makes problems like the following trivial: 2 identical and balanced coins are tossed once. Let H(1) be the event the first coin lands heads and H(2) be the event the second coin lands heads. Show that these are independent events. Do you have any idea on how to proceed without assuming what we're supposed to show? $\endgroup$ – Chris Sep 7 '15 at 2:35

If "fairness" is defined to mean "$P(H)=P(T)=\frac{1}{2}$," then independance follows from fairness logically, so we can avoid saying "by independance" by using the word "fairness" a bunch. However, if the only property that we know of a coin is that it is fair, then we can't show what you wish without using something at least as strong as independence.


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