1
$\begingroup$

I am trying to calculate the probability of reaching n successes on a trial, knowing that a failure will increase success chance.

In simpler term, what is the mean number of trials before we reach $x$ successes with $p = min(y + wF, 1)$? (with $F$ being the number of failures)

This does include the fact that after a certain number of failures, success is guaranteed ($p>1$).

I've tried to inspire myself from this question, but I don't want $F$ to reset upon success.

$\endgroup$
7
  • $\begingroup$ Confusing. Looks like you intend that if $wy = q$ and $F$ is the number of failures that have already occurred, then chance of failure on next trial is $qF$. If this is wrong, please advise. If this is right, then what is the chance of failure on the very 1st trial? $\endgroup$ Commented Jun 15, 2021 at 7:19
  • $\begingroup$ Re previous comment - perhaps I am misinterpreting your mathematical intent. Please edit your question, using MathJax to show math. Also, re previous comment, I question whether my interpretation is tenable. Besides the problem of determining chance of failure on 1st trial, chance of failure on any trial must be $\leq 1$. However, the expression $qF$ may exceed $1$. $\endgroup$ Commented Jun 15, 2021 at 7:23
  • $\begingroup$ @user2661923 Thanks for your comment, I'll try to update my question to make it more clear. Also, correcting my typo: $p$ is the chance of success, defined by a base probability $y$ plus $wF$. Not multiplied. $\endgroup$
    – Docteur
    Commented Jun 15, 2021 at 7:27
  • $\begingroup$ Then, how do you prevent $y + wF$ from exceeding $1$? $\endgroup$ Commented Jun 15, 2021 at 7:28
  • 1
    $\begingroup$ You could simply specify $p = \min(y + wF, 1)$, if that accurately represents your intent. $\endgroup$ Commented Jun 15, 2021 at 8:13

0

You must log in to answer this question.

Browse other questions tagged .