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 I love math! 21 votes (46.66%) Math is great. 14 votes (31.11%) My religion is mathology. 6 votes (13.33%) Women didn't speak to me until I was 30. 2 votes (4.44%) Total eclipse reminder -- 04/08/2024 12 votes (26.66%) I steal cutlery from restaurants. 3 votes (6.66%) I should just say what's on my mind. 6 votes (13.33%) Who makes up these awful names for pandas? 5 votes (11.11%) I like to touch my face. 12 votes (26.66%) Pork chops and apple sauce. 10 votes (22.22%)

45 members have voted

Ace2
Joined: Oct 2, 2017
• Posts: 2393
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May 14th, 2023 at 7:55:02 AM permalink

[[{(38/18)^(5+1) - 1} / {(38/18) - 1}] - 1] / 2 =~ 38.9 spins
It’s all about making that GTA
Wizard
Joined: Oct 14, 2009
• Posts: 25471
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May 15th, 2023 at 5:31:17 AM permalink
Quote: ThatDonGuy

Quote: Wizard

What is the expected number of spins in double-zero roulette to see five reds or five blacks in a row? The previous spin landed on 0.

Let E(n) be the expected number after having n consecutive of a color; E(5) = 0
Let p be the probability of getting a particular color, and r = 1 - 2p = the probability that it comes up green

E(4) = 1 + p E(1) + r E(0)

E(3) = 1 + p E(1) + r E(0) + p E(4)
= 1 + p E(1) + r E(0) + p (1 + p E(1) + r E(0))
= 1 + p + (p + p^2) E(1) + (r + pr) E(0)

E(2) = 1 + p E(1) + r E(0) + p E(3)
= 1 + p E(1) + r E(0) + p (1 + p + (p + p^2) E(1) + (r + pr) E(0))
= (1 + p + p^2) + p (1 + p + p^2) E(1) + r (1 + p + p^2) E(0)

E(1) = 1 + p E(1) + r E(0) + p E(3)
= (1 + p + p^2 + p^3) + p (1 + p + p^2 + p^3) E(1) + r (1 + p + p^2 + p^3) E(0)
(1 - p - p^2 - p^3 - p^4) E(1) = (1 + p + p^2 + p^3) + (1 - 2p) (1 + p + p^2 + p^3) E(0)
E(1) = (1 + p + p^2 + p^3) / (1 - p - p^2 - p^3 - p^4) + ((1 - 2p) (1 + p + p^2 + p^3) / (1 - p - p^2 - p^3 - p^4)) E(0)

E(0) = 1 + (1 - 2p) E(0) + 2p E(1)
E(0) = 1 / 2p + E(1)
E(0) = 1 / 2p + (1 + p + p^2 + p^3) / (1 - p - p^2 - p^3 - p^4) + ((1 - 2p) (1 + p + p^2 + p^3) / (1 - p - p^2 - p^3 - p^4)) E(0)
(1 - (1 - 2p) (1 + p + p^2 + p^3) / (1 - p - p^2 - p^3 - p^4)) E(0) = 1 / 2p + (1 + p + p^2 + p^3) / (1 - p - p^2 - p^3 - p^4)
E(0) = (1 / 2p + (1 + p + p^2 + p^3) / (1 - p - p^2 - p^3 - p^4)) x (1 - p - p^2 - p^3 - p^4) / p^4
E(0) = (1 - p - p^2 - p^3 - p^4) / (2 p^5) + (1 + p + p^2 + p^3) / p^4
E(0) = (1 - p - p^2 - p^3 - p^4) / (2 p^5) + (2p + 2p^2 + 2p^3 + 2p^4) / (2p^5)
= (1 + p + p^2 + p^3 + p^4) / (2p^5)
= (1 - p^5) / (2 p^5 (1 - p))

For p = 9/19, this is 4,592,395 / 118,098, or about 38.8863

I agree! That one is beer worthy.

Eliot's approximation was correct to the four significant digits given. Of course, an exact expression is required for the beer.

Ace is also correct and extra credit for the formula for n spins.
“Extraordinary claims require extraordinary evidence.” -- Carl Sagan
teliot
Joined: Oct 19, 2009
• Posts: 2587
May 15th, 2023 at 3:05:33 PM permalink
What I finds odd about this problem is this related problem: what are the chances that starting with the present spin you get a sequence of exactly 5 reds or 5 blacks?

The answer is, of course, (36/38)*(18/38)^4*(20/38) = 0.025103 or 1-in-39.836. I find it odd that this is even close to the other question Mike asked (the expected number of spins to get a sequence Red or Black of length 5). I was almost going to go with 39.836 as my answer.
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Ace2
Joined: Oct 2, 2017
• Posts: 2393
May 15th, 2023 at 3:50:35 PM permalink
Quote: teliot

What I finds odd about this problem is this related problem: what are the chances that starting with the present spin you get a sequence of exactly 5 reds or 5 blacks?

The answer is, of course, (36/38)*(18/38)^4*(20/38) = 0.025103 or 1-in-39.836. I find it odd that this is even close to the other question Mike asked (the expected number of spins to get a sequence Red or Black of length 5). I was almost going to go with 39.836 as my answer.

With the last spin being zero, the chance of getting five black or five red in the next five spins is (18/38)^5 * 2 or about 1 in 21. Not close to the answer

Look at it this way. The chance of you getting two consecutive heads in the next two rolls is 1 in 4, but the expected waiting time to see two consecutive heads is six. Red/black is not much different from heads/tails…would be identical without those dang two zeroes on the wheel
It’s all about making that GTA
teliot
Joined: Oct 19, 2009
• Posts: 2587
May 15th, 2023 at 4:11:10 PM permalink
Of course, but I was giving exactly 5 in a row, not 5 or more.

I'm just saying that two unrelated questions about five in a row have surprisingly close answers.
Last edited by: teliot on May 15, 2023
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Wizard
Joined: Oct 14, 2009
• Posts: 25471
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May 25th, 2023 at 12:17:58 PM permalink
A cake is divided among 100 mathematicians, numbered 1 to 100. It is divided up as follows an in order:

Mathematician 1 gets 1% of the cake.
Mathematician 2 gets 2% of the remaining cake.
Mathematician 3 gets 3% of the remaining cake.
Mathematician 4 gets 4% of the remaining cake.

This continues until the last mathematician gets 100% of what is left.

Which mathematician gets the most cake?

No calculators allowed!

For full credit, I need an explanation why your answer is right.

I also hate to keep saying this, but please don't just throw out a link to a video that solves the problem *ahem*. Let's have some fun and try to learn something.
“Extraordinary claims require extraordinary evidence.” -- Carl Sagan
charliepatrick
Joined: Jun 17, 2011
• Posts: 2829
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May 25th, 2023 at 12:43:46 PM permalink
Person N gets N/100 of what's left. In which case person (N+1) gets (N+1)/100*(100-N)/100. This means 100N=100N+100-N2-N which means N2+N-100=0. Using N=-b2+/-SQRT(etc) gives -1+-SQRT(1+400)/2 which is about 10.
When N=10 the 10th person gets 10% and the 11th person get 11% of 90% = 9.9%. Similarly the 9th person gets 9% and the 10th person gets 10% of 91% = 9.1%. Confirming the 10th person gets the most.
Wizard
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• Posts: 25471
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May 25th, 2023 at 1:24:42 PM permalink
Quote: charliepatrick

Person N gets N/100 of what's left. In which case person (N+1) gets (N+1)/100*(100-N)/100. This means 100N=100N+100-N2-N which means N2+N-100=0. Using N=-b2+/-SQRT(etc) gives -1+-SQRT(1+400)/2 which is about 10.
When N=10 the 10th person gets 10% and the 11th person get 11% of 90% = 9.9%. Similarly the 9th person gets 9% and the 10th person gets 10% of 91% = 9.1%. Confirming the 10th person gets the most.

Agreed. Full credit for a similar solution as mine.
“Extraordinary claims require extraordinary evidence.” -- Carl Sagan
lilredrooster
Joined: May 8, 2015
• Posts: 5641
May 26th, 2023 at 2:20:00 AM permalink
.

𝗻𝘂𝗺𝗯𝗲𝗿 𝟵, 𝗻𝘂𝗺𝗯𝗲𝗿 𝟵, 𝗻𝘂𝗺𝗯𝗲𝗿 𝟵, 𝗻𝘂𝗺𝗯𝗲𝗿 𝟵, 𝗻𝘂𝗺𝗯𝗲𝗿 𝟵___________________that is part of the lyrics from the Beatles song "Revolution"

the 9 times mystery:

at least it's a mystery to me..............I'm sure some here can explain it - an easy puzzle I'm sure for some - but not for me
I haven't got a clue
nothing similar happens with any other digits that I know about

here we go:

9*5=45.......................add up the digits - 4+5=9

9*8=72.......................add up the digits - 7+2=9

9*7=63.......................add up the digits - 6+3=9

9*17=153...................add up the digits - 1+5+3 =9

9*77=693...................add up the digits - 6+9+3 and get 18............then add up those digits..............1+8=9

9*4,796=43,164..........add up the digits 4+3+1+6+4 and get 18......then add up those digits...............1+8=9

9*104,675 = 942,075.........add up the digits..........9+4+2+0+7+5 and get 27...........then add up those digits..............2+7=9

9*5,327,894 = 47,951,046...........add up the digits .........4+7+9+5+1+0+4+6 and get 36............then add up those digits.................3+6=9

𝗻𝘂𝗺𝗯𝗲𝗿 𝟵, 𝗻𝘂𝗺𝗯𝗲𝗿 𝟵, 𝗻𝘂𝗺𝗯𝗲𝗿 𝟵, 𝗻𝘂𝗺𝗯𝗲𝗿 𝟵, 𝗻𝘂𝗺𝗯𝗲𝗿 𝟵

.
"believe half of what you see and none of what you hear" - Edgar Allan Poe
charliepatrick
Joined: Jun 17, 2011
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May 26th, 2023 at 2:51:09 AM permalink
You are correct that the sum of the digits of a number divisible by 9 will, assuming you keep going, always land up as 9. The logic can be thought as follows.

For two digits numbers the number can be expressed as 10a+b, where as the total of the digits is a+b. This if a+b=9, then 9a+a+b is also a multiple of 9, and this is 10a+b.

For three digits use 100a+10b+c and observe that 99a+9b+(a+b+c) is also divisible by 9 if (a+b+c) is. If a+b+c is greater than 9 then repeat the process, and note the same logic applies for the number formed by a+b+c.

For larger numbers you can see that each power of 10 can be considered as (99..99+1) so the number breaks down to 9*something+(a+b+.....+z). So you can see the pattern applies for all numbers.

btw the same applies to numbers divisible by 3, the total wil always reduce to 3 6 or 9.