March 10th, 2020 at 7:33:52 PM
permalink

A casino opens a game that works as follows:

There are 2020 balls, numbered 1, 2, 3, ..., 2019, 2020

1000 of them are drawn, and you win whatever the lowest number drawn is.

What is the expected result of the game?

There are 2020 balls, numbered 1, 2, 3, ..., 2019, 2020

1000 of them are drawn, and you win whatever the lowest number drawn is.

What is the expected result of the game?

March 10th, 2020 at 8:46:39 PM
permalink

Quote:ThatDonGuyA casino opens a game that works as follows:

There are 2020 balls, numbered 1, 2, 3, ..., 2019, 2020

1000 of them are drawn, and you win whatever the lowest number drawn is.

What is the expected result of the game?

A brute force method with Excel gave me an EV of 2.01898101897439.

But I bet there's an elegant way to do the math.

But I bet there's an elegant way to do the math.

March 10th, 2020 at 8:58:16 PM
permalink

2021/1001 =~ 2.019

(N + 1) / (selections + 1)

(N + 1) / (selections + 1)

It’s all about making that GTA

March 11th, 2020 at 4:33:24 PM
permalink

Quote:Ace22021/1001 =~ 2.019

(N + 1) / (selections + 1)

Correct, but how about some details?

March 11th, 2020 at 5:00:06 PM
permalink

I calculated the values for n=5 and n=6. The pattern was clear so I didn’t have to derive the formula.Quote:ThatDonGuy

Correct, but how about some details?

It’s all about making that GTA

March 11th, 2020 at 5:52:01 PM
permalink

Quote:Ace2I calculated the values for n=5 and n=6. The pattern was clear so I didn’t have to derive the formula.

"The pattern was clear"?

So, in that case:

2

^{2}- 1 = 3

2

^{3}- 1 = 7

2

^{5}- 1 = 31

2

^{7}- 1 = 127

"The pattern is clear" that 2 raised to the power of a prime number, minus 1, is also prime

2

^{11}- 1 = 2047 = 23 x 89...

However, I'm going to be nice and provide the solution:

Before I begin, note that I use (n)C(k) to denote the number of combinations of n things taken k at a time; this is also C(n,k) or Combin(n,k).

This puzzle involves something called the Hockey Stick Principle:

(n)C(n) + (n+1)C(n) + (n+2)C(n) + ... + (n+k)C(n) = (n+k+1)C(n+1)

The name comes from the shape of the line of numbers when this is done on Pascal's Triangle.

Anyway, I will solve the generic case of subsets of K numbers out of {1, 2, 3, ..., N}

There are a total of (N)C(K) subsets

Let S be the smallest number in the subset

There are N-S numbers greater than S, and K-1 of them make up the rest of the numbers in the subset, so there are (N-S)C(K-1) subsets with the smallest number S.

There are (N-1)C(K-1) with smallest number 1, (N-2)C(K-1) with smallest number 2, and so on through (N-(N-(K-1)))C(K-1) with the smallest number N-(K-1), which is the one subset {N-K+1, N-K+2, ..., N}.

The sum of the smallest numbers = 1 x (N-1)C(K-1) + 2 x (N-2)C(K-1) + 3 x (N-3)C(K-1) + ... + N-(K-1) x 1

= ( (N-1)C(K-1) + (N-2)C(K-1) + (N-3)C(K-1) + ... + (K-1)C(K-1) ) + ( (N-2)C(K-1) + (N-3)C(K-1) + (N-4)C(K-1) + ... + (K-1)C(K-1) ) + ... + ( (K)C(K-1) + (K-1)C(K-1) ) + (K-1)C(K-1)

Apply the Hockey Stick Principle to each element in the sum

The sum = (N)C(K) + (N-1)C(K) + (N-2)C(K) + ... + (K+1)C(K) + (K)C(K)

Apply it again to the sum to get (N + 1)C(K + 1)

The expected return = (N + 1)C(K + 1) / (N)C(K)

= ( (N + 1)! / ( (K + 1)! (N - K)! ) ) / ( N! / K! (N - K)! )

= (N + 1)! / ( (K + 1)! (N - K)! ) ) x K! (N - K)! / N!

= (N + 1) N! K! (N - K)! / ( (K + 1) K! (N - K)! ) N!)

= (N + 1) / (K + 1)

March 12th, 2020 at 12:47:58 AM
permalink

We are going to need the Wizard to continue posting math problems. Because apparently we will all be quarantined or sheltered in our homes for the next month without any sports events to watch.

So, its going to be math and books and movies. And turtles all the way down.

So, its going to be math and books and movies. And turtles all the way down.

So many better men, a few of them friends, are dead. And a thousand thousand slimy things live on, and so do I.

July 3rd, 2021 at 7:32:33 PM
permalink

Total possible combinations of 1000 balls = 2020 combination 1000

In possible combinations of 1000 balls

Sets having ball of min value 1 = 2020-1 Combination 1000-1

Sets having ball of min value 2 = 2020-2 Combination 1000-1

Sets having ball of min value 3 = 2020-3 Combination 1000-1

- - - -

- - - -

Sets having ball of min value 1020 = 2020-(2020-1000) Combination 1000-1

Sets having ball of min value 1021 = 2020-1021 Combination 1000-1

Using above pattern we can sum of series or avg or prob. whatever we need.

In possible combinations of 1000 balls

Sets having ball of min value 1 = 2020-1 Combination 1000-1

Sets having ball of min value 2 = 2020-2 Combination 1000-1

Sets having ball of min value 3 = 2020-3 Combination 1000-1

- - - -

- - - -

Sets having ball of min value 1020 = 2020-(2020-1000) Combination 1000-1

Sets having ball of min value 1021 = 2020-1021 Combination 1000-1

Using above pattern we can sum of series or avg or prob. whatever we need.

July 3rd, 2021 at 9:02:14 PM
permalink

Quote:gordonm888We are going to need the Wizard to continue posting math problems. Because apparently we will all be quarantined or sheltered in our homes for the next month without any sports events to watch.

So, its going to be math and books and movies. And turtles all the way down.

Why did this old post of mine from 16 months ago randomly get re-posted in the thread in July 2021?

Is there a gremlin in the gears?

So many better men, a few of them friends, are dead. And a thousand thousand slimy things live on, and so do I.