For a sequence of 10 coin flips, the probability of all heads is 1 in 1024 i.e. $$1/2^{10}.$$

And the probability for an individual coin flip to be heads is $1$ in $2$.

After $9$ coins have been flipped, is the probability of the last coin being heads still $\frac12$ for that last coin?

I ask with respect to the "Monty Hall Problem" -- https://en.wikipedia.org/wiki/Monty_Hall_problem -- in which the probability of an event changes in an unintuitive way.

Does anything like that apply in this coin flip example?

  • 5
    $\begingroup$ Nope. Each individual toss is a $\frac 12$ affair. $\endgroup$
    – lulu
    Jan 23, 2017 at 21:40
  • $\begingroup$ If you don't trust someone just saying "its 1/2 just because", you can approach via a more tedious and laborious approach: Each sequence of flips is equally likely to occur. Any sequence of ten flips which ends with a head can be described via a sequence of choices: is first coin head or tail? is second coin head or tail? ... is ninth coin head or tail? Applying multiplication principle there are $2^9$ sequences ending with a head. Taking the ratio of number ending with head over number total regardless you have probability is $2^9/2^{10}=\frac{1}{2}$ $\endgroup$
    – JMoravitz
    Jan 23, 2017 at 21:46
  • $\begingroup$ @commenters: Thanks. I posted this question before I re-read the details of the Monty Hall Problem. After reading that, though, it struck me that that probably does not apply in this case. But I posted anyway, just in case my question would be of interest to someone. Thanks! $\endgroup$
    – JDS
    Jan 23, 2017 at 21:51
  • $\begingroup$ Just to elaborate: I would say that this is definitional. We define a "fair coin" to be one for which each trial independently comes up $H$ or $T$ with probability $\frac 12$. It's a harder question if you talk about real coins. For real coins, you might reasonably say that the $\frac 12$ might be slightly off and you could use the observed frequency to recalibrate your estimate. But...I don't think you were asking about that, were you? $\endgroup$
    – lulu
    Jan 23, 2017 at 21:52
  • 2
    $\begingroup$ @JDS In the Monty Hall problem, the probability of choosing the car at the very beginning is $1/3$, and that probability doesn't change after one of the other doors is opened. What changes is that you're given a new opportunity, and the probability of success is different for this event. $\endgroup$
    – Théophile
    Jan 23, 2017 at 21:53

1 Answer 1


The Monty Hall problem is about dependency and changing knowledge.

You choose a door.   There is a probability of 1/3 that there is a prize behind it.   One of the remaining two doors is then opened revealing that there is no prize behind it.   You are given the opportunity to change to the third door or stay with the initial choice, now that you know the identity of one of the doors without a prize.

There is still a probability of 1/3 that there was a prize behind the door you choose before that information was generated.   But, what of the remaining door?

Because of dependency (since there is only one prize), if you were right first time there is certainly no prize behind the third door, but if you were wrong, there certainly is.   So the probability of there being a prize behind the third door is $2/3$.   ("Third door" being "neither the door you chose nor the door that was revealed".)

None of which has anything to do with your coin toss problem as the coins are independently tossed and you gain no knowledge of the final toss by knowing about any of the previous.

(It would be different if, say, you were told that the coin being tossed was either fair or had a specified bias, with equal likelihood, and a certain number of heads had shown in the first nine tosses.   Then you are in the realm of Bayesian probability.)


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