[Darts at Morty's]

When Davey and Ian playing together:

Combined probability of hitting a bull's eye will be -

90% + 10% x 80% = 98% (If Davey fails to hit, when going first)

or

80% + 20% x 90% = 98% (If Ian fails to hit, when going first)

Clearly, more than 97% probability by Alex, so Alex definitely made a BAD bet.

Easier to just calculate odds of neither Davey or Ian hitting bullseye: .1 * .2 = .02 or 2%

[Darts at Morty's]

Curr3nt did a lot of figuring to come up with the same conclusion I got by multiplying .1 by .2 and comparing it to .03. On 100 throws Alex loses a net 1 time.

[Darts at Morty's]

Seems pretty simple. The odds that Davey and Ian both miss is 2% (.2*.1). The odds that Alex will miss is 3%, so he has made a bad bet. Is there something more complicated that I am missing?

[Two kids, with a nod to Teanchi and Beanchi]

If Ned said it, probability is 1 that it is GG;

If Red said it, it is either BG, GB, or GG, probability is 1/3 it is GG

If Ted said it, you know the older child is G, probability is 1/2 it is GG

If Zed said it, you know the coin selected child is G. He mentions sex of other child (G) so probability of GG is 1

1/4*1+1/4*1/3+1/4*1/2+1/4*1=17/24 = probability that it is GG.

[A Child and a Staircase]

I have a clarification question. If there are 10 steps and a landing at the bottom and the top, the child would have to take 11 steps if taken one at a time, right?

[Going Postal]

Nice going Prismatic. Is there logic or math behind this solution or trial and error?