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Ask The Wizard #425

Your task is to cut the triangle below into two pieces equal in area with one vertical cut (meaning parallel to side a). Where should you make the cut?

 

anonymous

The cut should be made b/sqrt(2) from the far right side of the triangle. Note that the height does not matter.

 

Here is my solution (PDF).

This question is asked and discussed in my forum at Wizard of Vegas.

In Ask the Wizard #424 someone asked, "A one-meter stick is cut in two random places. What is the expected area of the smallest of the three pieces created?" My question is what would the answer of c random cuts?

anonymous

1/(c+1)^2

 

Here is my solution (PDF).

This problem is asked and discussed in my forum at Wizard of Vegas.

How many ways can you place six distinct balls into three identical boxes?

anonymous

The answer is 122

 

The tricky part of this problem is that the boxes are identical. If they were different, the answer could be easily found as 36 = 729.

Let's label the balls A to F. Begin by placing ball A in any box.

First, let's say you put ball B into one of the two empty boxes left. From there on the three boxes are different because they have different balls in them, including one still empty. The number of ways to place the other four balls is 34 = 81.

Second, let's say you put ball B into the same box as ball A and C into one of the empty boxes. From there on the three boxes are different because they have different balls in them, including one still empty. The number of ways to place the other three balls is 33 = 27.

Third, let's say you put balls B and C into the same box as ball A and D into one of the empty boxes. From there on the three boxes are different because they have different balls in them, including one still empty. The number of ways to place the other two balls is 32 = 9.

Fourth, let's say you put balls B, C and D into the same box as ball A and E into one of the empty boxes. From there on the three boxes are different because they have different balls in them, including one still empty. The number of ways to place the other ball is 31 = 3.

Fifth, let's say you put balls B to E into the same box as ball A and F into one of the empty boxes. There are no balls left, so there is just one way balls A-E are in one box and F in another.

Sixth and finally, there is one way only to all the balls in the same box.

So the answer is 34 + 33 + 32 + 31 + 2 = 122.

This question is asked and discussed in my forum at Wizard of Vegas.

What is the average dealer final total in blackjack, assuming the dealer neither gets a blackjack nor busts?

anonymous

 

Decks Stand
Soft
17		 Hit
Soft
17
1 18.840370 18.880098
2 18.842675 18.882895
6 18.844207 18.884750
8 18.844399 18.884981
Infinite 18.848634 18.895356