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There is a unfair coin and you want to find out its actual probability, to find the exact probability is actually not an easy task, you need to toss it many many times to get the close to actual result.

Gladly, someone has already calculated that and he is observing you while you are tossing. You toss the coin 4 times at the beginning, and get 2 tails and 2 heads as THTH. Then all of a sudden the observer said:

The chance of getting $2$ tails and $2$ heads with this coin is actually half of the probability of getting a tail in a single toss with it.

By this information,

Can you find the most probable probability of getting a tail in a single toss with this coin?

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  • $\begingroup$ The chance of getting 2 tails and 2 heads - in the THTH pattern or any pattern? $\endgroup$ – boboquack Nov 10 '17 at 9:49
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    $\begingroup$ @boboquack...oh...so many dirty jokes can be spawned from your comment. $\endgroup$ – Marius Nov 10 '17 at 9:59
  • $\begingroup$ @boboquack: since "2 tails and 2 heads" already clashes with the THTH pattern; I would infer that the order of outcomes is not relevant. (and if the statement is not related to having flipped the THTH pattern, that means the THTH serves no narrative purpose whatsoever, which seems counterintuitive) $\endgroup$ – Flater Nov 10 '17 at 10:51
  • $\begingroup$ I find this to be very amusing, since it's impossible to bias a coin: stat.berkeley.edu/~nolan/Papers/dice.pdf $\endgroup$ – MechMK1 Nov 10 '17 at 11:21
  • $\begingroup$ @MechMK1 yes you cant find it but get close to the actual probability as I stated in the question as "you need to toss it many many times to get close to actual result." $\endgroup$ – Oray Nov 10 '17 at 11:22
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Let $p$ be the probability of getting tails.

We are told that

$\frac{1}{2}p=p^2(1-p)^2\binom 42\implies 12p-24p^2+12p^3-1=0\text{ or }p=0$ by rearranging.

This expression doesn't have a nice factorisation, so plugging it into the cubic formula, we get (given that $0\leq p\leq1$)

$p = \frac{1}{12} \left(8 + \frac{-4 - 4 i\sqrt3}{\sqrt[3]{1 + 3 i \sqrt7}} + i (\sqrt3 + i) \sqrt[3]{1 + 3 i \sqrt7}\right)$
or
$p=\frac{1}{12} \left(8 + \frac{4i(\sqrt3+i)}{\sqrt[3]{1 + 3 i \sqrt7}} + (-i\sqrt3 - 1) \sqrt[3]{1 + 3 i \sqrt7}\right)$

Now, these aren't very nice numbers. But they are approximately:

$0.1037$ and $0.6388$

In reality, I think that:

the solution from the first spoiler, $p=0$, is most likely as such a finely calibrated coin would be impossible, or at least $p$ is infinitesimal (since supposedly we did flip tails)

even though

from the data given, that we did flip two heads and tails, the root at $p\approx0.6388$ is most likely.

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  • $\begingroup$ Nice work. Now, which one of these two seems more reasonable ;) $\endgroup$ – ABcDexter Nov 10 '17 at 10:10
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    $\begingroup$ In my opinion, in reality, you cannot have two tails with p is close to 0. I would go for 0.6388 since I got two tails in 4 rolls. I wish I would have asked in that way :) $\endgroup$ – Oray Nov 10 '17 at 10:22
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    $\begingroup$ Since you did get a tail, it can't be p=0. $\endgroup$ – Francis Davey Nov 10 '17 at 13:22
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    $\begingroup$ I'd say the 0.6388 is the most likely. This is the one that is most likely to produce the two tails and two heads that we saw as we flipped our coins. $\endgroup$ – Chris Nov 10 '17 at 15:40
  • $\begingroup$ Regardless of whether you consider p=0 possible or not, the answer is clearly No, you can't tell, since there are at least two possibilities that can't be ruled out. $\endgroup$ – trentcl Nov 10 '17 at 16:18
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Disclaimer As per the comment of Jaap Scherphuis the equation of this answer equation is incorrect. Thread carefully

Gave wolfram alpha:

1/x*1/(1-x)*1/x*1/(1-x) = 2/x

And got back:

x = 1.5652

Which gives

Chance of tails ~64%

Thoughts:

hmm i remember seeing the numberphile video about TH ordering and how it mattered!, I reject this and stand by my solution! I'd be interested to have flaws pointed out though. :)

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    $\begingroup$ The $1/(1-x)$ factors should be $1-1/x$ because if $p=1/x$ is the probability of tails then the probability of heads is $q=1-p=1-1/x$. Oddly enough, the answer from your incorrect equation turns out to be exactly the same as the correct answer for 2H and 2T in any order! $\endgroup$ – Jaap Scherphuis Nov 10 '17 at 10:10
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    $\begingroup$ Ofcourse that's correct. Classic, correct result even when doing significant mistakes! That's maths for you ^^ $\endgroup$ – Adam Nov 10 '17 at 10:15

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