numbers question

[sagefr0g]

If you have a bet that has 766 possible outcomes, where each outcome can occur randomly, how many trials of a simulation would be needed to insure that each possible outcome occurs at least once?

and would that number of trials be considered equivalent to the long run for the bet postulated?

i thought maybe I could use the method where one figures out how many coin flips is required to expect to come up with at least one heads & one tails which is

= (1+1/p)(1−p)+(1 + (1/(1−p)))p

but I’m not sure the proper way to do so.

 

[Meistro]

you would need to run an infinite number of simulations to ensure that all numbers are hit.

 

[sagefr0g]

you would need to run an infinite number of simulations to ensure that all numbers are hit.

YOIKS
so i take it that would also be the case for the coin flip case where one desires to ensure at least one heads and one tails presents?
obviously expectation and certainty are far apart concepts.
uhmm so, like for blackjack, the combinations of what can happen are (i dunno) huge. how does one determine when is enough far as running some number of rounds in a simulation.

 

[Meistro]

"so i take it that would also be the case for the coin flip case where one desires to ensure at least one heads and one tails presents?"

Yes that is correct as far as my understanding of probability is concerned. There is a chance that you flop tails 5 million times in a row. that chance can be expressed as .5 to the 5 millionth power. of course after a certain point you get to a situation where the outcome is so unlikely you can basically ignore it.

 

[DSchles]

If you have a bet that has 766 possible outcomes, where each outcome can occur randomly, how many trials of a simulation would be needed to insure that each possible outcome occurs at least once?

and would that number of trials be considered equivalent to the long run for the bet postulated?

i thought maybe I could use the method where one figures out how many coin flips is required to expect to come up with at least one heads & one tails which is

= (1+1/p)(1−p)+(1 + (1/(1−p)))p

but I’m not sure the proper way to do so.

There's never 100% certainty that an uncertain event will occur. The best you can do is to state that, if you perform x trials, there is a y probability that the event will occur. And you can make y as close to 100% as you like.

There are online calculators that will do this for binomial events, such as coin flips.

Don