re: Solve this OT

i'm just reading OP and saw you were the last poster (and i respect your math skillz). 50%, right? 2 options for 2 boxes

No.

There are three possible opening moves:

1 - Take G1 from B1
2 - Take G2 from B1
3 - Take G1 From B2

That leaves three possible remaining moves:

1 - Take G2 from B1
2 - Take G1 from B1
3 - Take S1 from B2

your odds are 2/3

quote:

---

There are three possible opening moves:

---

no there aren't. there are only 2 (b/c we know we get a gold ball)

quote:

---

As always, it's time to get all Monte Carlo up in this bitch. This is in Python, and the leading dots on some of the lines just represent spaces that won't format correctly here any other way I've found. import random # Three boxes in the 'box' list, each box is # a list of two balls. box = [['G','G'],['G','S'],['S','S']] def choose_ball (): # We'll choose a box at random (and remember it) # and choose a ball from it. . this_box = random.randrange (3) . return (this_box, random.choice (box [this_box])) def other_ball (box_idx): # If we bother to look at the 'other' ball, we know # we have already chosen a 'G', so we must be in # either box [0] or box [1] # in box [0], the other ball has to be 'G', # in box [1], the other ball has to be 'S' . if box_idx == 0: .. return 'G' . if box_idx == 1: .. return 'S' trials = 0 golds = 0 # Now we do the test a million times for i in range (1000000): # Pick a ball and remember which box we're in . my_box, my_ball = choose_ball () . if my_ball == 'S': # Silver ball first -- don't count anything # Just do the next one .. continue . else: # Gold ball first, so we count this one # and look at the other ball in the box. .. trials = trials + 1 .. if other_ball (my_box) == 'G': ... golds = golds + 1 print (float (golds) / float (trials))

So, for each test, we pick a ball 1 million times. If we pick a silver ball, we go on to the next pick. If we pick a gold ball, we look at the other ball in the box we picked from. At the end we look at ONLY the times we picked a gold ball first, and we see the fraction of times that the other ball was also gold. It's fairly fast on a modern PC, so I ran it 10 times (remember it was a million picks for each run and another pick if we picked gold first).
Results are:
0.665746226428
0.667399396348
0.667402563641
0.667849064158
0.666473975721
0.665929128366
0.666613307294
0.666898288197
0.667379593472
0.666684691288

It's pretty safe to say that simulation agrees with the assertion that your chances are 2/3 to pick a second gold ball GIVEN that you've picked a gold ball first.

---

ETA: from tokenBoiler on page 10

quote:

---

Did you even read his post past your quote?

---

no

but i'm thinking about it

It's 67%, Slo.

You are overthinking this thing.

When the exercise started there were three options, but we're past the beginning.

Think of the contest this way.

There are three boxes in a grab bag. They are marked one, two and three.

Chicken goes first and pulls out box number three. We see it there next to the grab bag. We know it's not in the bag.

Slo goes next, and now what are the odds he picks box 1?

50/50

The different colored balls are really not relevant. The key part is that there's three boxes.

quote:

---

That leaves three possible remaining moves:

1 - Take G2 from B1
2 - Take G1 from B1
3 - Take S1 from B2

---

Those that think 50/50 are simply ignoring the fact that box 1 has two options. The brain wants to think that you either pulled G1 from B1 or G1 from B2 leaving only two second pull options, G2 from B1 or S1 from B2; 50/50. As the stats have proven, they would be wrong.

gotcha

shouldn't have posted witha keto headache