3 ms·
I'm not sure what you mean by 'expectation of one case' here, but just to clarify: rolling a ten-sided dice, 10 times, gives a 65% chance of rolling at least on
by feral 9y ago
I'm not sure what you mean by 'expectation of one case' here, but just to clarify: rolling a ten-sided dice, 10 times, gives a 65% chance of rolling at least one 10. So you still have a 35% chance of zero 'wins'.
I can't comment on the rest of your model, but its worth keeping in mind how the binomials shake out.
- ryandrake 9y agoAck, you're absolutely right. Don't know where I was going with the hypothetical example but my math was totally wrong.
- waterhouse 9y agoActually, your statement was entirely correct. "Let's say 10% of start-ups succeed ... You would need to join 10 of these start-ups ... to have an expectation of one case [of success]." It is a true statement that the expected number of successes after 10 tries would be 1. It is also a true statement that your chance of zero successes after 10 tries is 34.8%. (The balance comes from the outcomes that include multiple successes.) It may be that the latter is the more relevant and interesting statement to make, but your math was totally right. ;-)