## Poll

20 votes (46.51%) | |||

14 votes (32.55%) | |||

6 votes (13.95%) | |||

2 votes (4.65%) | |||

12 votes (27.9%) | |||

3 votes (6.97%) | |||

6 votes (13.95%) | |||

5 votes (11.62%) | |||

12 votes (27.9%) | |||

9 votes (20.93%) |

**43 members have voted**

February 14th, 2022 at 2:32:57 PM
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Take the integral from zero to infinity of:

[{(x/6)^5/120 + (x/6)^4/24 + (x/6)^3/6 + (x/6)^2/2 + (x/6) + 1} * e^(-x/6)]^6

= 2597868106693535971 / 131621703842267136

=~ 19.74 rolls

Using the very useful property that the average time for x to occur equals the sum of the probabilities that x has not occurred over all time. The above formula sets all six numbers to 0-5 occurrences.

I think this is analogous to the geometric series which tells us, for example, that the expected rolls for any single die number to appear is: 1 / (1 - 5/6) = 6 = (5/6)^0 + (5/6)^1 + (5/6)^2 + (5/6)^3...which is the sum of probabilities it has not occurred over all time

[{(x/6)^5/120 + (x/6)^4/24 + (x/6)^3/6 + (x/6)^2/2 + (x/6) + 1} * e^(-x/6)]^6

= 2597868106693535971 / 131621703842267136

=~ 19.74 rolls

Using the very useful property that the average time for x to occur equals the sum of the probabilities that x has not occurred over all time. The above formula sets all six numbers to 0-5 occurrences.

I think this is analogous to the geometric series which tells us, for example, that the expected rolls for any single die number to appear is: 1 / (1 - 5/6) = 6 = (5/6)^0 + (5/6)^1 + (5/6)^2 + (5/6)^3...which is the sum of probabilities it has not occurred over all time

It’s all about making that GTA

February 14th, 2022 at 3:29:34 PM
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Quote:Ace2So there are 6^6 = 46,656 states, correct?

link to original post

Actually, there are twice as many, but the other 46,656 states are the ones where one of the numbers has been rolled 6 times, so the expected number of rolls needed for each of those is zero.

For example:

P(5, 5, 5, 5, 5, 5) = 1 + 1/6 P(6, 5, 5, 5, 5, 5) + 1/6 P(5, 6, 5, 5, 5, 5) + 1/6 P(5, 5, 6, 5, 5, 5) + 1/6 P(5, 5, 5, 6, 5, 5) + 1/6 P(5, 5, 5, 5, 6, 5) + + 1/6 P(5, 5, 5, 5, 5, 6) = 1 + 0 + 0 + 0 + 0 + 0 + 0 = 1.

February 14th, 2022 at 7:08:34 PM
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Quote:Ace2If you roll a single die until one of the six numbers has appeared six times, what is the expected number of rolls?

Extra credit: What is the probability you can accomplish this in 18 rolls or less?

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2597868106693535971 / 131621703842267136 = Approximation: 19.73738396371749

The expression is the integral from 0 to infinity of (exp(-x/6)*(1+x/6+x^2/72+x^3/1296+x^4/31104+x^5/933120))^6 dx.

p.s. Damn, I see Ace2 already posted the answer. I know I'm late, but can I at least get half credit?

“Extraordinary claims require extraordinary evidence.” -- Carl Sagan

February 15th, 2022 at 8:42:09 AM
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Actually, you deserve at least three times that! So I'm giving you:Quote:WizardQuote:Ace2If you roll a single die until one of the six numbers has appeared six times, what is the expected number of rolls?

Extra credit: What is the probability you can accomplish this in 18 rolls or less?

link to original post

2597868106693535971 / 131621703842267136 = Approximation: 19.73738396371749

The expression is the integral from 0 to infinity of (exp(-x/6)*(1+x/6+x^2/72+x^3/1296+x^4/31104+x^5/933120))^6 dx.

p.s. Damn, I see Ace2 already posted the answer. I know I'm late, but can I at least get half credit?

link to original post

(-1/2)! credits

Last edited by: Ace2 on Feb 15, 2022

It’s all about making that GTA

February 23rd, 2022 at 4:57:30 PM
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I don't plan to return to the forum but I did want to share this one puzzle. The Austin Chronicle has a "Math Happens" column every week in which they post a puzzle, which I usually do. This week's was more interesting than usual to me, hence why I'm sharing. The object is to derive all the integers from 1-20 using only four 4's for each integer, and only the operators + - x ÷ √ ! and exponent, and not using decimals or 44. For example, 13 could be 4! ÷ √4 + 4/4. I got all of them except 19. Here's the solution for 1-32.

(Updated with the proper link to the solutions that don't use decimals.)

(Updated with the proper link to the solutions that don't use decimals.)

Last edited by: MichaelBluejay on Feb 24, 2022

February 23rd, 2022 at 5:19:33 PM
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Quote:MichaelBluejay...19...

19 = 4! - 4 - 4/4 = 24-4-1. I think including .4 and .4 (recurring) you can do all the numbers 1 thru 100. 4/9 is a very useful number to have!

February 23rd, 2022 at 8:04:33 PM
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4*4 + 4 - 4/4 = 19

It’s all about making that GTA

February 23rd, 2022 at 8:15:49 PM
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Nice try but, I think you're only supposed to use Four 4's!Quote:Ace24*4 + 4 - 4/4 = 19

February 28th, 2022 at 2:58:39 PM
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A forum member recently posted the following (regarding the ALL craps side bet):

"You only need ten rolls to hit the ALL. Its [sic] been hit in 15-17 rolls typically."

The easy math puzzles are:

1) Assuming you win the ALL bet, what's the probability you win with less than 18 rolls?

2) Considering winning bets only, how many rolls does it take, on average, to win the ALL bet

"You only need ten rolls to hit the ALL. Its [sic] been hit in 15-17 rolls typically."

The easy math puzzles are:

1) Assuming you win the ALL bet, what's the probability you win with less than 18 rolls?

2) Considering winning bets only, how many rolls does it take, on average, to win the ALL bet

It’s all about making that GTA

February 28th, 2022 at 4:07:17 PM
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Since it is limited to winning bets, ignore all rolls of 7.

Also, since it is for 17 rolls or less, assume there are 17 rolls, including any rolls that would come after a winning set.

Find all increasing 7-tuples of numbers in {2, 3, 4, 5, 6, 8, 9, 10, 11, 12}. ("Increasing" means no number is greater than the one to its right.)

Let P(2) be the number of times 2 appears in the 17 rolls, P(3) the number of times 3 appears, and so on.

For example, if the 7-tuple is (2, 3, 4, 5, 6, 8, 9), then P(2) = P(3) = ... = P(9) = 2, and P(10) = P(11) = P(12) = 1; if it is (2, 2, 2, 2, 2, 2, 2),, then P(2) = 8 and P(3) = ... = P(12) = 1.

There are 17! / (P(2)! P(3)! ... P(12)!) permutations of these 17 numbers.

For each permutation, the probability of actually rolling those numbers is (1/30)^P(2) * (2/30)^P(3) * ... * (2/30)^P(11) * (1/30)^P(12).

Calculate this value for each permutation, and add them up to get:

235,436,099,899 / 8,649,755,859,375, or about 1 / 36.7393.

The solution is the integral over x from 0 to positive infinity of:

1-((1 - e^(-x/30))(1 - e^(-x/15))(1 - e^(-x/10))(1 - e^(-2/15*x))(1 - e^(-x/6))(1 - e^(-x/5)))^2 dx

which is 5,401,372,918,634,611 / 105,826,178,618,160, or about 51.04

Last edited by: ThatDonGuy on Feb 28, 2022