Guessing a coin toss correctly

[member 428835]

Given $n$ coin tosses of a biased coin with heads probability $p$, what is the probability of guessing correctly $k\leq n$ or more? I think this is $\sum_{i=k}^{n}C^n_{n-i}p^i(1-p)^{n-i}$ but this can't be correct since it is not symmetric for $p$ vs $1-p$. Also, what's the expected number of correct guesses we would predict the guesser to guess correctly? Sure feels something like $np(1-p)$, but I dunno.

But now let's suppose the guesser knows there will be exactly $k$ heads tossed ($n-k$ tails). In this case what would the odds of a correct guess be? We know there are $C^n_k$ different ways to arrange the heads/tails.

Thanks for any insight!

 

[Hornbein]

Given $n$ coin tosses of a biased coin with heads probability $p$, what is the probability of guessing correctly $k\leq n$ or more? I think this is $\sum_{i=k}^{n}nC{n-i}p^i(1-p)^{n-i}$ but this can't be correct since it is not symmetric for $p$ vs $. Also, what's the expected number of correct guesses we would predict the guesser to guess correctly? Sure feels something like$np(1-p)$, but I dunno. But now let's suppose the guesser knows there will be exactly$k$heads tossed ($n-k$tails). In this case what would the odds of a correct guess be? We know there are$nCk## different ways to arrange the heads/tails.

Thanks for any insight!

It depends on what the guesser does. To make progress I'll assume that if p > 0.5 then to maximize the score the guesser guesses heads every time. The average correct guesses will then be np.

You seem to assume that the guesser must guess heads exactly half the time. Then the average is np/2 + n(1-p)/2 = (np+n-np)/2 = n/2. Kind of surprising.

 

[PeroK]

But now let's suppose the guesser knows there will be exactly $k$ heads tossed ($n-k$ tails). In this case what would the odds of a correct guess be? We know there are $C^n_k$ different ways to arrange the heads/tails.

You have an unknown arrangement of $k$ heads and $n-k$ tails. Assuming all such arrangements are equally likely (e.g. if you toss a fair or biased coin $n$ times and happen to get $k$ heads), then the best you can do is guess one of these possible arrangements.

 

[PeroK]

Given $n$ coin tosses of a biased coin with heads probability $p$, what is the probability of guessing correctly $k\leq n$ or more? I think this is $\sum_{i=k}^{n}C^n_{n-i}p^i(1-p)^{n-i}$ but this can't be correct since it is not symmetric for $p$ vs $1-p$.

Take one biased coin toss, where the guesser does not know it's biased and is equally likely to guess heads or tails. The probability of guessing correctly is $\frac 1 2 p + \frac 1 2 (1-p) = \frac 1 2$. Kind of amusing!

 

[pbuk]

Given $n$ coin tosses of a biased coin with heads probability $p$, what is the probability of guessing correctly $k\leq n$ or more?

• What does this mean:
  • Tossing a biased coin $n$ times and guessing heads or tails correctly at least $k$ times?
  • Predicting correctly that there are at least $k$ heads in $n$ tosses?
  • Something else?
• Does the guesser know the bias?
• Why do you think any answer would be interesting?

But now let's suppose the guesser knows there will be exactly $k$ heads tossed ($n-k$ tails). In this case what would the odds of a correct guess be?

This is not very well thought through: there is no single "odds of a correct guess", for instance the guesser can always "guess" the $n$th toss with 100% certainty (if there have already been $k$ heads it must be tails).

 

[FactChecker]

Take one biased coin toss, where the guesser does not know it's biased and is equally likely to guess heads or tails. The probability of guessing correctly is $\frac 1 2 p + \frac 1 2 (1-p) = \frac 1 2$. Kind of amusing!

It's not so surprising, for this reason:
Suppose you have already tossed a coin (fair or not) and got a head. At that point P(head)=1 and P(tail)=0, so it is about as unfair as it can be. Yet the probability of guessing correctly (head) is 1/2, simply because that is assumed to be how the guessing is done.

 

[WWGD]

Standard odds a:b definition is related to probabability by $$ a:b \rightarrow \frac{a}{a+b} $$, so that,
e.g., 1:1 odds correspond to a probability of 1/2; odds of 2:1 correspond to 1/(1+2)=1/3, etc.

In your case, the probability is the sum of the probabilities of all such events.