# Binomial Distribution Question (Exactly/At Least $x$ Trials for Success)

I am a little embarrassed to ask this question, but for the life of me I cannot figure it out.

A man pays \$1 a throw to win a \$3 doll. His probability of landing a throw is 0.1. What is the probability that two throws will be required to win the doll? What is the probability that $x$ throws is needed?

The book said the probability that $x$ throws is required is $0.1(0.9)^{x-1}$. How did they get this?

I tried using a binomial distribution, but I don't know the number of trials

EDIT:

I have another similar problem, but i am having problems with determining the success rate

Three men toss coins to see who pays for coffee. If all three match, they toss again. Otherwise the "odd man" pays for the coffee. What is the probability that they will need to do this more than once? And at most twice?

I basically thought that outcomes like HHH is equal to TTT and likewise any combination of HHT = TTH.

So essentially we either stop after the first toss or we toss again and that makes the success probability equal to 1/2 and the failure is equal to 1/2.

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You do know the number of trials: $x$. But the distribution can not be binomial, since in this case, you stop playing once you land a throw. Whereas in a "binomial game", you'd play a fixed amount of times, regardless if you already won from the first or any preceding trial. – Raskolnikov Oct 14 '12 at 19:13
– joriki Oct 14 '12 at 19:14
"So essentially we either stop after the first toss or we toss again and that makes the success probability equal to 1/2 and the failure is equal to 1/2." Absolutely not! Either a meteor destroys earth tomorrow or not, but it is far from 50/50; one of the events is extremely unlikely. The chance of getting HHH is $(1/2)^2 = 1/8$, and the same for TTT. Thus, the probability of "success" (i.e. getting one of those two outcomes) is $1/4$. See if you can work it out from there (it is like the doll problem). – Austin Mohr Oct 14 '12 at 21:29

To win the doll in exactly two throws means that the first throw was a failure (90% likely) and the second throw a success (10% likely). Assuming each throw is independent of the others, you can just multiply the probabilities together to get $$0.9 \cdot 0.1 = 0.09.$$

More generally, if exactly $x$ throws are required, then the first $x-1$ throws were failures and the final throw a success. Multiplying it all out gives $$0.9^{x-1}\cdot 0.1.$$

The probability of requiring more than three throws to win can be seen as the complement of requiring three or fewer throws to win. That is, $$\Pr(\text{more than three throws}) = 1 - \Pr(\text{three or fewer throws}).$$ As is common with problems using "at most" or "at least", it is useful to break the event into several instances of "exactly": \begin{align*} & \Pr(\text{3 or fewer throws})\\ = & \Pr(\text{exactly 1 throw}) + \Pr(\text{exactly 2 throws}) + \Pr(\text{exactly 3 throws})\\ = & 0.9^0 \cdot 0.1 + 0.9^1 \cdot 0.1 + 0.9^2 \cdot 0.1\\ = & 0.271. \end{align*} Finally, \begin{align*} \Pr(\text{more than three throws}) &= 1 - \Pr(\text{three or fewer throws})\\ &= 1 - 0.271\\ &= 0.729. \end{align*}

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What about the probablity of more than three throws to win? I tried doing $1 - (0.9)^2 0.1 = 0.919$ and it was wrong – Hawk Oct 14 '12 at 20:22
What you've computed is the probability of "not exactly three throws to win", which is not the same as "more than three throws to win. See my updated answer. – Austin Mohr Oct 14 '12 at 20:26
Oh okay, so I essentially do $0.1 \sum_{x = 4}^{\infty} (0.9)^{x-1}$ – Hawk Oct 14 '12 at 20:33
@jak Your method works also. It seems you understand this more fully than you think. :) – Austin Mohr Oct 14 '12 at 20:36
Do you mind if I throw in another similiar problem? – Hawk Oct 14 '12 at 21:06