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 ;)
$\endgroup$ 14 Answers
$\begingroup$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: $$\underbrace{\frac 12 \times \frac 12 \times \cdots \frac 12}_{\large n \;\text{factors}}= \left(\frac 12\right)^n$$
$\endgroup$ 3 $\begingroup$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 $127$.
$\endgroup$ 1 $\begingroup$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.
$\endgroup$ $\begingroup$You guys are making it too complicated. It goes like this:
double your last...
1st time: 1 in 2 - (win a single coin toss) 2nd time: 1 in 4 - (win 2 consecutive coin tosses) . . .etc. 3rd time: 1 in 8 4th time: 1 in 16 5th time: 1 in 32 6th time: 1 in 64 . . . etc . . . 1 in 2 to the nth. If you wanna get complicated with it then if 2 events have the same odds, there's no such thing as "independence" it's all in your head. Just add number of attempts together (nth)<--(this is an exponent). If the odds aren't 50/50 or if there are multiple events multiply the odds together. 1 in 4 should be represented as the fraction 1/4, 2 in 17 as 2/17 and so on... An example would be: "What would the odds of winning both a 1 in 3 bet and a 1 in 6 bet?" 1/3 X 1/6 = 1/18 or just multiply the 3 and 6 to make it easy. If you wanna know the odds of winning one and losing another, you don't do anything different, just flip the losses upside-down (inverse fractions) Make sense? "What are the odds of winning a 1 in 4 bet and a 1 in 6 bet, then losing a 1 in 4 bet?" 1/4 X 1/6 X 4/1(or just 4)= 1/6. I'm not going to bother with probability vs odds syntax since both boil down to the same thing. Just make sure that a preposition is never the word that you end your sentence in.
$\endgroup$
