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.

  • $\begingroup$ 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: $p/(1-p)$. $\endgroup$ – zoli Feb 25 '18 at 15:55
  • $\begingroup$ 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? $\endgroup$ – iknock Feb 25 '18 at 15:59

$$ \Pr(T>t) = \Pr(\text{No 6 appears in the first $t$ trials.}) = \left( \frac 5 6 \right)^t. $$ $$ \Pr(T=t) + \Pr(T>t) = \Pr(T>t-1). $$ $$ \Pr(T=t) + \left( \frac 5 6 \right)^t = \left( \frac 5 6 \right)^{t-1}. $$ $$ \Pr(T=t) = \left( \frac 5 6 \right)^{t-1} - \left( \frac 5 6 \right)^t = \left( \frac 5 6 \right)^{t-1}\left( 1 - \frac 5 6 \right). $$

  • $\begingroup$ sorry but i have difficulty of understanding your computation. are you saying the probability of obtaining 6 is 5/6? which definition are you saying is correct? $\endgroup$ – iknock Feb 25 '18 at 16:39
  • $\begingroup$ @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 $T>t. \qquad$ $\endgroup$ – Michael Hardy Feb 26 '18 at 2:10
  • $\begingroup$ i think you are considering the outcome as equally likely however in this case its specified that the probability of obtaining 6 is different from 1/6. in any case you are saying the first definition is the one you are considering. $\endgroup$ – iknock Feb 27 '18 at 2:02

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