I’m currently trying to do the Raylong quest involving winning
higher or lower sixteen times. I’ve been using the strategy of if
it’s over five, choose lower and if it’s under five, choose higher
but have not had any luck. Is there a better strategy to winning or
is it just luck? Thanks :p
I actually normally do higher if it is over 2 and then lower is
really depending on what kinds of numbers that were under it.
Like once I had a 3 and it was one.
So its really depending!!
Audi famam illius.
Solus in hostes ruit
et patriam servavit.
Audi famam illius.
Cucurrit quaeque tetigit destruens.
Audi famam illius.
Audi famam illius.
Spes omnibus, mihi quoque.
Terror omnibus, mihi quoque.
Ille
iuxta me.
Ille iuxta me.
Socii sunt mihi,
qui olim viri fortes
rivalesque erant.
Saeve certando pugnandoque
splendor crescit.
The game is based on probability as far as I can tell. If he
chooses the number 9, you have a 9/10 probability (or 90% chance)
of getting the answer right if you choose "lower" and 1/10 (or 10%)
chance of getting it right when you choose "higher." At five, you
have a 50% - 50% chance of getting it right no matter which one you
chose so it is mainly luck when he picks numbers around the middle.
(Of course, with it being probability, sometimes he picks the
number 10 when his first number is 9 at about a 10% rate.)
Ehhh... not true. He never picks a number twice. With this in mind,
it's a chance out of nine. At 9 you have an 8/9 chance with lower,
about 88.9% chance, while higher is 1/9, about 11.1%. At 5, you
have a 4/9 percent chance at lower to 5/9 chance at higher.
Another way of thinking about it is it's always a 50% chance. This
way of thinking is only if you believe that he already has two
numbers. If he already has two numbers, it's the probability that
he picks the smaller one first if you pick higher, or if he picks
the larger one and you pick lower.