Mitch Hedberg math (dice) problem

Quote:

My lucky number is four billion. That doesn't come in real handy when you're gambling. "Come on four billion!... F---! Seven. Not even close. I need some more dice. Four billion divided by six... at least. -- Mitch Hedberg

This got me thinking on calculating the probability of rolling 4 billion with the minimum number of dice.

Here's how far I got.

You'd need 6666666667 dice to roll 4 billion.

If you rolled a 6 on 6666666666 dice and a 4 on the other die, the probability of that is 1 in 87791495225240054872976680384307270233204389574760109739369.

Where I am getting lost is the other possibilities... for example, you could roll a 6 on 6666666665 dice and two 5s. And so on.

I have no idea how to calculate the permutations -- does it matter which die you roll below 6 on -- and how to integrate them into the total odds.

Can anyone give me some guidance?

Best post ever.

RIP Mitch

With 666,666,667 dice there are two situations that yield exactly 4 billion:

A) 666,666,665 sixes and 2 fives. Since any 2 of the 666,666,667 dice can be fives, there are combin(666666667,2) = 222,222,222,111,111,111 combinations for this case.

B) 666,666,666 sixes and a four. Since any one die can be the four, there are combin(666666667,1) = 666,666,667 combinations for this case.

Added together, there are a total of 222,222,222,777,777,778 ways to roll 4 billion out of 6^666666667 possible outcomes. The fraction would therefore look like:



Since the denominator is too large to work with on a typical computer (over 518 million digits), a different approach is required to reduce the fraction. The base-6 log of 222,222,222,777,777,778 is approximately 22.292308182317007125968919384765. So, dividing both the numerator and denominator by 6^{22.292308182317007125968919384765} gives the following approximation:



Suffice it to say, the probability of 666,666,667 dice being rolled to a sum of exactly 4,000,000,000 is (an) infinitesimal.

Thank you very much, JB. It was the combinations part that threw me. Very informative.

Another good craps on TV problem.

In a Dave Chappelle skit, "The World Series of Dice" A character "Grits n Gravy" is known for rolling 77 7s in a row at a casino in Vegas.

What's the probability of that? (1/6)^77 = my calculator broke...

Use Excel instead.

(1/6)^7 = 1.2088E-60

(FYI: E-60 means you slide that decimal sixty places to the left.)

Quote: TheArchitect

What's the probability of that? (1/6)^77 = my calculator broke...

1 in 827,268,102,990,819,696,904,779,987,451,100,917,723,545,245,377,785,847,873,536.

Quote: JB

Quote: TheArchitect

What's the probability of that? (1/6)^77 = my calculator broke...

1 in 827,268,102,990,819,696,904,779,987,451,100,917,723,545,245,377,785,847,873,536.

So that's roughly 40% less than 1 in a googol?

Quote: TheArchitect

So that's roughly 40% less than 1 in a googol?

No, for a couple of reasons:

1) 1 in a googol is a smaller number than the above figure.

2) The difference in the number of decimal digits between two numbers is not the same thing as their arithmetic difference. Your question is similar to the following one: "So 2 is roughly 75% less than 2000?". A single-digit number is much lower than 25% of a 4-digit number; it is roughly 1 thousandth. Likewise, a 60-digit number is much lower than a 100-digit number; it is roughly one-10,000,000,000,000,000,000,000,000,000,000,000,000,000th (one ten-duodecillionth).

What would be correct is to say that there are 40% fewer digits in the decimal representation of the denominator of the figure above than there are in the denominator of 1 in a googol.

Quote: JB

Quote: TheArchitect

So that's roughly 40% less than 1 in a googol?

No, for a couple of reasons:

1) 1 in a googol is a smaller number than the above figure.

2) The difference in the number of decimal digits between two numbers is not the same thing as their arithmetic difference. Your question is similar to the following one: "So 2 is roughly 75% less than 2000?". A single-digit number is much lower than 25% of a 4-digit number; it is roughly 1 thousandth. Likewise, a 60-digit number is much lower than a 100-digit number; it is roughly one-10,000,000,000,000,000,000,000,000,000,000,000,000,000th (one ten-duodecillionth).

What would be correct is to say that there are 40% fewer digits in the decimal representation of the denominator of the figure above than there are in the denominator of 1 in a googol.

I really didn't know that. Thank you JB!