# The 538 Riddler:New Casino Game

Welcome to my weekly look at 538’s The Riddler. Happily my answer to last week’s riddle proved to be correct once again, and I was happy to see some commenters who came to the same conclusions. Thanks for reading and commenting.

Due to a lack of time with Easter and all, I’m not very confident of my answer to this week’s Riddler, but I’ll let you decide. Here’s the riddle:

Suppose a casino invents a new game that you must pay \$250 to play. The game works like this: The casino draws random numbers between 0 and 1, from a uniform distribution. It adds them together until their sum is greater than 1, at which time it stops drawing new numbers. You get a payout of \$100 each time a new number is drawn.

For example, suppose the casino draws 0.4 and then 0.7. Since the sum is greater than 1, it will stop after these two draws, and you receive \$200. If instead it draws 0.2, 0.3, 0.3, and then 0.6, it will stop after the fourth draw and you will receive \$400. Given the \$250 entrance fee, should you play the game?

Specifically, what is the expected value of your winnings?

Okay, when I read that part it seemed like this was going to be a no-brainer, but then Ollie added:

[Editor’s note: You know how sometimes your boss/teacher asks you not to bring your laptop to a meeting/class, and you’re all like, “Jeez, how am I gonna tweet during the meeting/class, then?” Well, I’m going to pull rank and ask that you don’t bring laptops to this Riddler. Next week I will highlight — nay, celebrate! — the elegant, pencil-and-paper solutions over the brutal, cold, silicon ones. We’re all on the honor system with one another, right? RIGHT?]

No computers! Pencil and paper? Egads. Okay, so I did a little bit of working this out and I came up with an answer I thought might be right. First of all, I assumed that you are guaranteed to see at least 2 numbers, because with all the real numbers between 0 and 1, the chances of getting 1 exactly are one over infinity, or zero. So the minimum amount you are going to lose in this game is \$50. And it seems to me that the chances of seeing a third number are 50%, since the average value of the first number will be 0.5 and the chances that the second number will be greater than 0.5 is… 0.5, or 50%.

Okay, so right away we see that there’s a 50% chance we lose \$50 (-\$250 + \$200), and a 50% chance that we win \$50 (-\$250 + \$300), with the possibility of more money to come. This is a wonderful casino game; I would very much like to find the casino that offers it. A game like this might explain how somebody could bankrupt a casino. (Sad!)

So we have a 50% chance of seeing only 2 numbers and losing \$50. What are the chances of seeing only 3 numbers? Well, I figure if 2 random numbers are below 1, then there’s a 2/3 chance that the 3rd number will break the ceiling. So 2/3 of what is remaining (1/2) is 1/3. So 1/2 the time we see 2 and only 2 numbers and 1/3 of the time we see 3 and only 3 numbers, and now we have 1/6 (1/2 – 1/3) remaining.

Chance of only 2 numbers: 1/2. 1 – 1/2 = 1/2 remaining.
* 2/3  =
Chance of only 3 numbers: 1/3. 1/2 – 1/3 = 1/6 remaining.
* 3/4 =
Chance of only 4 numbers: 1/6 * 3/4 = 1/8. 1/6 – 1/8 = 1/24 remaining.
* 4/5 =
Chance of only 5 numbers: 1/24 * 4/5 = 1/30. 1/24 – 1/30 = 1/120 remaining.
* 5/6 =
Chance of only 6 numbers: 1/120 * 5/6 = 1/144.

And so on.

So to get the Expected Value (EV) we can take the odds of each payout * its value and get:

\$-50/2 + \$50/3 + \$150/8 + \$250/30 + \$350/144 = \$21.18

Plus some increasingly smaller numbers for seeing 7, 8, 9, 10? numbers.

Ok, now that we got our EV answer, let’s cheat! (If you don’t want to see whether a computer confirms or disproves my EV answer, stop reading here.)

I wrote some code in Go (are you sick of Go yet, because I am not.)

```package main

import (
"fmt"
"math/rand"
)

func doFloatTrial() int {
total := 0.0
numbers := 0

for {
numbers++
total += rand.Float64()
if total >= 1.0 {
break;
}
}
return numbers
}

func main() {

trials := 1000000

var results []int

cash := 0.0

for i := 0; i < trials; i++ {
cash -= 250.0
numbers := doFloatTrial()
cash += 100.0 * float64(numbers)
for a := len(results); a <= numbers; a++ {
results = append(results, 0)
}
results[numbers]++
}

fmt.Printf("EV = %.3f\n", cash / float64(trials));
for i := 0; i < len(results); i++ {
fmt.Printf("Results[%d] = %d (%.2f) \n", i, results[i], float64(trials)/float64(results[i]))
}

}
```

And here’s the output:

```EV = 21.826
Results[0] = 0 (+Inf)
Results[1] = 0 (+Inf)
Results[2] = 499604 (2.00)
Results[3] = 334041 (2.99)
Results[4] = 124717 (8.02)
Results[5] = 33341 (29.99)
Results[6] = 6939 (144.11)
Results[7] = 1157 (864.30)
Results[8] = 184 (5434.78)
Results[9] = 15 (66666.67)
Results[10] = 2 (500000.00)```

Hurray! Our EV from the code is \$21.83 and we can see that the odds of drawing each count of numbers is (with variance) what we had calculated by hand. Math! Follow this link to run the code yourself in your browser.

Thanks for reading. I’m not 100% sure of this one so I look forward to better approaches to this problem in the comments.

## Author: Wiesman

Husband, father, video game developer, liberal, and perpetual Underdog.

## 5 thoughts on “The 538 Riddler:New Casino Game”

1. Pretty much what I got. My answer was 100e. The odds of busting after n rolls is simply 1/n!. The EV for any number of rolls, then, is 100 * n/n! = 100/(n – 1)!. The EV of the whole thing is the infinite series of 1001/(n – 1)! from n = 1 to infinity, which is the same as 100/n! from n=0 to infinity, which is e. The infinite sum of 1/n! from n=0 to infinity is e, so our answer is 100e = \$271.828183..

1. greggg230 says:

Sorry, those are obviously the odds of NOT busting. E.g. you have a 1/3! = 1/6 chance of having a sum under 1 after 3 rolls.

2. e! I never even thought of checking whether the value might resolve to e. That’s pretty cool. Math, man, amirite? Thanks for posting.

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