Question: Chance of Arrival in Unit Reinforcement

[zovs66]

I like simple math and was never called to be a mathematician, in ASL there are folks that can recite the probability of rolling 2d6 for various things, in those days (and even to this day) I just squeeze my dice a litter tighter and whisper a prayer and roll the bones. Probably why I am not a gambler or risk taker in real life...lol

[scorpio_rocks]

Han Solo (and probably zovs66) should look away now...

For those that want the maths:

Use the “not” trick. If probability of a unit arriving is 20%, then the probability of not arriving is 80% (0.2 and 0.8 respectively).
Not arriving(X) in a number of turns(N) is X ^ N. Therefore The chance they arrive by a certain turn is 1 - X ^ N.

e.g.:
The chance of appearance in 1 turn is 20%, in 2 turns is 1 - (0.8 ^ 2) = 36% , in 3 turns is 1 - (0.8 ^ 3) = 48.8%, in 4 turns is 1-(0.8 ^ 4) = 59%, in 5 turns is 1-(0.8 ^ 5) = 67%, etc.

It never becomes a certainty, but there is a 99% chance that they will have arrived by turn 21!

[zovs66]

You lost me on

Not arriving(X) in a number of turns(N) is X ^ N. Therefore The chance they arrive by a certain turn is 1 - X ^ N.

The rest was just some numbers wrapped in pretty characters...

[shahadi]

Originally Posted by scorpio_rocks

Han Solo (and probably zovs66) should look away now...

LoL... Han Solo and his lucky dice.

Thanks for all contributions. Informative.

[scorpio_rocks]