A thought experiment:

Hypothetically lets assume a machine can predict result of equally likely events correctly with probability 0.8(say a coin toss for example).

We know probability of getting a head/tail in a coin toss is 0.5, The machine beforehand predicts that head is the result of the next toss. Does this change the probability of getting a head/tail in the next toss? (because we know the machine predicts 8/10 times correctly and also assume the machine is intelligent and once event has occurred the machine predicts the result with 100% accuracy)

Does the first statement make sense? Is it even possible to call the events equally likely if the machine can predict the outcome with some accuracy?

Bonus discussion: Two cases I imagined: (Not sure if it is relevant to this forum)

  1. Either the machine can go into the future with some accuracy and return to the same timeline
  2. Or the machine knows every occurrence of tosses with some precision

Note: The machine cannot predict with 100% accuracy - This makes seeing the future with absolute precision not possible. (If not I think probability cannot exist right? we already know the future)

For me case 2 makes the event not equally likely because the machine always knows the future with some accuracy. Case 1 - Not really sure because the event has to happen normally, its just that the machine knows the result by traveling to the future but this might have to do with closed time loop and paradoxes

Edited first statement from: Hypothetically lets assume a machine can predict result of equally likely events correctly 8/10 times(say a coin toss for example).

  • $\begingroup$ The coin toss is independent of the device. $\endgroup$
    – AlvinL
    Aug 7, 2021 at 16:14
  • 2
    $\begingroup$ The probability of getting head/tail is would be $0.5$, since the event of tossing a coin is independent of the prediction by the machine. However if you are in future but don't know the outcome of tossed coin and you want to calculate the probability that head (or say, tail) had actually occurred when the machine shows head (or tail) is going to occur, then it's a simple question of Bayes' Theorem. $\endgroup$ Aug 7, 2021 at 16:32
  • $\begingroup$ @AmanKushwaha Can you explain the Bayes part a bit, I've not used probabilities in a while. It would help if you can point out what is A and B in this case Lets say A is head occurring B is Machine predicting head occurs so P(a) = 0.5 p(b) = machine predicting head (what will this be?) how to define this? p(b/a) = 1 I haven't mentioned this in the post if machine knows head happened it'll predict head 100% of times Am i being right with events here? $\endgroup$
    – Djock
    Aug 7, 2021 at 17:04
  • $\begingroup$ $P(\frac{B}{A})=$ "Probability that the machine predicts head given that head had actually occurred" which means the probability that machine's prediction is correct, that is equal to $\frac{8}{10}$ as you said. $\endgroup$ Aug 7, 2021 at 17:14
  • $\begingroup$ Also, let $A'$ be the event in which tail occured. Then, $P(\frac{B}{A'})$="Probability that the machine predicts head given that tail had occurred" which means the probability that machine's prediction is wrong, and that will be equal to $0.2$ . Now, can you apply Bayes' Theorem to calculate $P(\frac{A}{B})$? $\endgroup$ Aug 7, 2021 at 17:18

2 Answers 2


Let $M_i\in \{H,T\}$ be the machine's prediction for coin toss $i$ and $X_i\in\{H,T\}$ be the actual outcome of coin toss $i$.

What we want to study is the joint distribution $P_{MX}(M=m,X=x)$:

$$P_{MX}(M=H,X=H) = P_X(X=H)P_M(M=H|X=H)=0.5*0.8 = 0.4 = P_{MX}(M=T,X=T)$$

By symmetry, $$P_{MX}(M=H,X=T) = P_{MX}(M=T,X=H)=0.1$$

Therefore, the distribution of $(M,X)$ is given by:

$$P_{MX}(m,x) = 0.4 \mathbb{I}_{m=x} + 0.1\mathbb{I}_{m \neq x}$$

Note that the marginal probability $P_X(X=x) =P_{MX}(H,x) + P_{MX}(T,x) = 0.4+0.1=0.5$ as expected.

Also, $P_M(M=x) =P_{MX}(x,H) + P_{MX}(x,T) = 0.4+0.1=0.5$

So what is $P(X=H|M=H)$?:

$$P_{X|M}(X=H|M=H)=\frac{P_{XM}(X=H,M=H)}{P_M(H)} = \frac{0.4}{0.5} = 0.8 = P_{X|M}(X=T|M=T)$$

Again, not super surprising.

So you are much better off listening to the machine.


Provided the coin is fair, is it still possible to have a mechanism that predicts the outcomes correctly with some probability $>0.5$? Let $M_H$ denote event of machine predicting $H$ (heads).

Assume $P(M_H \mid H) = p$ for some $0< p\leqslant 1$. In your example $p=0.8$. Compute $P(H \mid M_H)$. By Bayes identity $$ P(H\mid M_H) P(M_H) = P(M_H \mid H) P(H) = 0.5p $$

By total probability $$P(M_H) = P(M_H\mid H)P(H) + P(M_H\mid T)P(T) = 0.5 $$

But then $P(H\mid M_H) = p$. This either contradicts fairness of the coin or $p=0.5$ is forced. Alternatively, the coin tossing could be Dependent on the machine, but that would defeat the purpose of "prediction".

But if $p=0.5$ is the best you can do, there's no point trying to build such a machine in the first place.

Notice what happens with $P(M_H)$. It only depends on the assumption that the coin is fair. So that would immediately tell you that in a fair game, there's no reason to have a bias toward any outcome. (duh!)


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