Bryan visits a carnival booth where Carl shows him 10 boxes. Exactly one of the boxes contains a gold coin; the other boxes are empty. Bryan randomly takes one of the boxes, but he doesn't open it. Carl then opens five other boxes that he knows are empty and shows Bryan that they are empty. Carl then tells Bryan he can either keep his initially chosen box or return it and choose one of the remaining closed boxes instead. If Bryan chooses to return his box and choose another one instead, what is the probability Bryan will choose the box with the gold coin? Express your answer as a common fraction.
1 - When Bryan initially picked one of the 10 boxes, his chances were: 1/10 and the chances of the other 9 boxes were: 1/10 for each box.
2 - When Carl opens 5 empty boxes, the chances of your box is still 1/10 but the chances of the remaining boxes suddenly jumped to 1/5 for each box including yours.
3 - So, if you keep your original box, your chances are still 1/10 because you picked it out of 10 boxes. But, Carl is giving you the chance to increase your chances from 1/10 or 10% to 1/5 or 20%. Therefore, you should switch your original box to one of remaining 4 boxes, therby doubling your chances of winning the gold coin, from 10% to 20%.
+2440 Guest’s solution looks like Mr. BB’s logic. 
This statement cannot be true: 2- When Carl opens 5 empty boxes, the chances of your box is still 1/10 but the chances of the remaining boxes suddenly jumped to 1/5 for each box including yours
The reason is Bryan’s box cannot have a simultaneous probability of (1/10) and (1/5).
Solution:
Bryan’s box has (1/10) probability of having the gold coin, and a (9/10) probability of not having it. The (9/10) probability is distributed across the (9) unchosen boxes.
After Carl opens (5) empty boxes, this (9/10) probability is now distributed across (4) boxes.
By choosing one of the four boxes, the probability changes to
(1/4) * (9/10) = (9/40)
= (22.5%)
GA
