Puzzle : Draw The Maximum Sum

Assume you are blindfolded and placed in front of a large bowl containing currency in $50, $20, $10, and $5 denominations. You are allowed to reach in and remove bills, one bill at a time. The drawing stops as soon as you have selected four-of-a-kind— four bills of the same denomination. 

What is the maximum sum of money you could accumulate before the drawing ends?

THIS should be the maximum sum! 

Source

The Maximum That You Can Get!

What was the question?

The quick response to the question by any body would be 3 x 50 + 1 x 20 = $170 would be the maximum as next pick of highest currency of $50 will stop drawing of currencies. But that 5th attempt may not be of $50 & could be of $20,$10 or $5. So this case doesn't count those possibilities. This is worst case scenario where 3 out of 4 picks resulted into accumulation of 3 currencies of same denomination of $50.

So ideally, after drawing three $50 s, three $20 s, three $10 s, three $5 s and finally drawing one anything will stop drawing attempts as that will accumulate 4 bills of same denomination (of $50/$20/$10/$5). The best that can be picked is $50 to get the maximum.

That is total maximum of 50 x 3 + 20 x 3 + 10 x 3 + 5 x 3 + 50 = $305 can be accumulated in 12 attempts & stopping after 13th attempt. And this will be the best case.

Maximize The Chance of White Ball

There are two empty bowls in a room. You have 50 white balls and 50 black balls. After you place the balls in the bowls, a random ball will be picked from a random bowl. Distribute the balls (all of them) into the bowls to maximize the chance of picking a white ball. 

This is the way to maximize the chances!

Way to Maximize White Ball Probability

What was the task given?

Let's distribute 50 black ball in one bowl & other 50 white ball in another bowl.

Then,

Probability (White Ball) = (1/2)(0/50) + (1/2)(50/50) = 0.5.

Now, if 1 white ball is kept in 1 bowl and other 49 white + 50 black = 99 balls in other bowl, then


Probability (White Ball) = (1/2)(1/1) + (1/2)(49/99) = 0.747.

That's nearly equal to 3/4 which is certainly higher than the previous case. And that's the way of maximizing the probability of white ball.