# Odds of being correct X times in a row

Is there a simple way to know what the chances are of being correct for a given number of opportunities?

To keep this simple: I am either right or wrong with a 50/50 chance. What are the odds that I'll be correct 7 times in a row or 20 or simply X times?

... and can the answer be put in simple terms ;)

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The answer is different depending on if you mean "at least X times" or "exactly X times". The latter implies you're wrong at opportunity X+1, and this should be factored into the equation. See also youtube.com/watch?v=bFfSfzjhfC8 – Mark Jeronimus Jan 22 at 7:29

Assuming the outcome of "answering" any question is independent of the outcome of answering other questions:

Yes, you'll be right $n$ times in a row with probability: $\left(\dfrac{1}{2}\right)^n$

If $X$ and $Y$ are independent events, then the probability that both $X$ and $Y$ occur is the product of the probabilities of $X$ and $Y$. For each question, you have a $50/50$ chance of answering correctly, which translates to a probability of $\frac 12$: for "randomly guessing" one question, you'll have a probability of $\frac 12$ that you'll be correct.

So answering correctly two questions in a row has probability: $$\frac 12 \times \frac 12 = \left(\frac 12\right)^2 =\frac 14$$

The probability of answering correctly three questions in a row: $$\frac 12 \times \frac 12 \times \frac 12= \left(\frac 12\right)^3= \frac 18$$

$$\vdots$$

The probability of answering correctly $n$ questions in a row: $$\underbrace{\frac 12 \times \frac 12 \times \cdots \frac 12}_{\large n \;\text{factors}}= \left(\frac 12\right)^n$$

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Thank you, that is what I was looking for. – user84483 Jun 30 '13 at 15:36
You're welcome! – amWhy Jun 30 '13 at 15:36
@Amzoti Thanks for all the support! Yes, for me...it starts tonight, this week...gotta figure out to stay away, even if only to tackle a few other things on my long list of "things to do!" Baby steps! – amWhy Jul 1 '13 at 0:31

There is a simple formula if we assume independence. If $A$ and $B$ are independent events, then the probability that both $A$ and $B$ occur is the product of the probabilities of $A$ and $B$.

Let $A$ be the event I am right on my first guess, and let $B$ be the event I am right on my second guess. Then the probability I am right on both guesses is $(1/2)(1/2)$, or equivalently $(1/2)^2$.

Let $A$ be the event I am right on the first two guesses, and let $B$ be the event I am right on the third guess. Then the probability that both $A$ and $B$ happen, i.e. that I am right on the first $3$ guesses, is $(1.2)^2(1.2)=(1/2)^3$.

Let $A$ be the event I am right on the first three guesses, and let $B$ be the event I am right on the fourth guess. Then the probability that both $A$ and $B$ happen, i.e. that I am right on the first $4$ guesses, is $(1.2)^3(1.2)=(1/2)^4$.

We can keep on using the same reasoning, and conclude that the probability of being right for $X$ guesses in a row is $(1/2)^X$.

More generally, if my probability of being right on any guess is $p$, and I make $X$ independent guesses in a row, my probability of being right all $X$ times is $p^X$.

Remark: Your question used the word odds and we used the term probability. These are different but closely related notions. Probability is mathematically more useful, but odds are more directly useful to gamblers.

Roughly speaking, the odds against an event is the ratio of the probability the event happens to the probability it doesn't happen.

Take for example your $X=7$. The probability we guess right $7$ times in a row is $(1/2)^7$, which is $\frac{1}{128}$. The odds that this happens are $1$ to $127$.

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Thank you, this answer was also very useful. – user84483 Jun 30 '13 at 16:16

This is a geometric random variable described by the geometric distribution, specifically the first one defined in the article.

In terms of your question, "success" for the distribution means getting an answer wrong.

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