# probability, geometric distribution definitions

Consider an unfair die, where the probability of obtaining $6$ is $p ≠ 1/6.$ The die is thrown several times. Call $T$ the RV that counts the number of throws before a 6 appears for the first time. What will be the distribution of $T.$

I consider this as the number of trial before the first success and hence the distribution is geometric having $p(T=t) = p(1-p)^{t-1}$ however im also has a hesitation if this is the case of the number of failures before the first success having $p(T=t) = p(1-p)^t$

which one is correct and why we choose one from the other? or is it correct to model this case in one of the two definition.

• The first formula is certainly wrong. Think of $t=0$, that is, the case when the first roll is immediately a $6$. The probability is $p$. As opposed to what the first formula would tell: $p/(1-p)$. – zoli Feb 25 '18 at 15:55
• but since the first is to count number of trials it has a support starting from 1 which is the first trial. and if we find 6 on the first roll we get the probability p as you said. what do you think? – iknock Feb 25 '18 at 15:59

$$\Pr(T>t) = \Pr(\text{No 6 appears in the first t trials.}) = \left( \frac 5 6 \right)^t.$$ $$\Pr(T=t) + \Pr(T>t) = \Pr(T>t-1).$$ $$\Pr(T=t) + \left( \frac 5 6 \right)^t = \left( \frac 5 6 \right)^{t-1}.$$ $$\Pr(T=t) = \left( \frac 5 6 \right)^{t-1} - \left( \frac 5 6 \right)^t = \left( \frac 5 6 \right)^{t-1}\left( 1 - \frac 5 6 \right).$$
• @iknock : The probability of NOT getting a $6$ on any trial is $5/6.$ So the probability of getting a $\text{non-}6$ on all of the first $t$ trials is $(5/6)^t.$ Therefore that is the probability that $T>t. \qquad$ – Michael Hardy Feb 26 '18 at 2:10