I Guessing a coin toss correctly

  • I
  • Thread starter Thread starter member 428835
  • Start date Start date
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!
 
Last edited by a moderator:
Physics news on Phys.org
joshmccraney said:
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.
 
  • Like
Likes PeroK and member 428835
joshmccraney said:
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.
 
  • Like
Likes Hornbein
joshmccraney said:
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!
 
joshmccraney said:
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?
joshmccraney said:
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).
 
PeroK said:
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.
 
  • Like
Likes PeroK
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.
 

Similar threads

Replies
21
Views
2K
Replies
57
Views
6K
Replies
41
Views
7K
Replies
4
Views
2K
Replies
5
Views
1K
Back
Top