Question: Chance of Arrival in Unit Reinforcement

[zovs66]

John, not sure where your rough guess and maths come from, but it pretty clear i the manual how this works (unless I am missing something).

Set reinforcement turn for this unit, at the current hex. Set reinforcement chance, this is a percentage chance of arrival per turn, NOT a one shot chance. Therefore a reinforcement with a 10 per cent chance which does not arrive on the due turn will test again on each subsequent turn after its due turn as and until it makes the chance roll or the game ends, whichever comes first.

[geoff]

This is probability theory. If there's a 10% chance of the reinforcement arriving, then in theory after five turns there is a 50% chance that the reinforcement will have arrived at some point in turns 1 through 5, although the chance of it arriving on any specific turn is still 10%.

[Imp]

Originally Posted by zovs66

John, not sure where your rough guess and maths come from, but it pretty clear i the manual how this works (unless I am missing something).

It is probability its not as simple as Geoff quotes 5 tries at 10% does not give 50%, wish it did I would place loads of bets then.
Roulette bet on black / red 50% chance second try would be 100% I cannot lose unless 0 comes up.

Chances of it turning up each individual turn remains the same at 20% but the probability that will occur increases with each turn is what I am saying.

The chance for it to appear on a certain turn has to take into account the chances it failed to appear on the previous turns.
At 10% Turn 2 only has a 90% chance of happening because 10% of the time it will appear on turn 1 etc etc.

Because of this each turn the probability of it turning up increases not doing the maths but at 20% the chances of it not turning up by turn 10 is remote, by turn 27 you probably have more chance of getting hit by a car.
That's 26 times it failed to turn up with a 1 in 5 (20%) chance every turn. Law of averages says it should have appeared over 5 times by then.