### Author Topic: BASIC DISTRIBUTION QUESTION  (Read 1358 times)

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#### Dane

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##### BASIC DISTRIBUTION QUESTION
« on: March 02, 2017, 07:57:39 AM »
Six well known DS.  Each and every one of them may come.
Or at least three adjacent  may come more than once.
What appears first?

#### Trilobite

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##### Re: BASIC DISTRIBUTION QUESTION
« Reply #1 on: March 02, 2017, 08:49:16 AM »
What do you mean by, "or at least three adjacent?

#### Trilobite

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##### Re: BASIC DISTRIBUTION QUESTION
« Reply #2 on: March 02, 2017, 08:50:25 AM »
What came first, the egg or the chicken?

The answer is sometimes the egg and sometimes the chicken.

That's roulette for ya!

#### Dane

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##### Re: BASIC DISTRIBUTION QUESTION
« Reply #3 on: March 02, 2017, 09:40:01 AM »
What do you mean by, "or at least three adjacent?

I mean three or more repeats side by side in accordance with the well known order at the table layout, i.e.
7-12, 13-18, and 19-24.
What came first, the egg or the chicken?

The answer is sometimes the egg and sometimes the chicken.

That's roulette for ya!

But some egg head     sometimes may be able to answer in accordance with probabilities.

#### Reyth

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##### Re: BASIC DISTRIBUTION QUESTION
« Reply #4 on: March 02, 2017, 10:16:59 AM »
I guess the real question is what is the difference between the odds of a 3rd unique DS hitting in a row?

I thought there were 11 DS?

#### Bayes

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##### Re: BASIC DISTRIBUTION QUESTION
« Reply #5 on: March 02, 2017, 10:25:34 AM »
Hi Dane, if I understand your question correctly, here is my solution (note that I ignore the zero because it complicates things):

The probability of each DS hitting in 6 spins is 1.54% (see example 4 on this page for the formula):

In the case of 3 adjacent, if we label the DS's 1-6 we could either have:

1,2,3 or 2,3,4 or 3,4,5 or 4,5,6. Each of these has probability (1/6)3 and since they are mutually exclusive we just add them to get the total probability of getting 3 DS's adjacent in 3 spins:

4 x (1/6)3 = 1.85%

So it's more likely that you will get 3 adjacent in 3 spins than all 6 hitting in 6 spins, but not by much.

Edit : I just noticed you said "at least" 3 adjacent, so the answer isn't quite correct, but calculating the exact probability will just increase the probability of 3 or more adjacent, so it still beats the "all hitting in 6 spins" scenario.
« Last Edit: March 02, 2017, 10:31:09 AM by Bayes »

#### Dane

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##### Re: BASIC DISTRIBUTION QUESTION
« Reply #6 on: March 02, 2017, 11:05:14 AM »
Hi Bayes, thanks for answering. Unfortunately I did not form my question good enough. Please notice that I want to compare the appearance of all six (with less than three repeats side by side) to
"at least three  adjacent" REPEATS! Which comes  probably first? No matter how many spins it takes?
With pen and paper I tried it twenty times. (Not much, I know). Ten here and ten there. A close run, I must say.

#### Bayes

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##### Re: BASIC DISTRIBUTION QUESTION
« Reply #7 on: March 02, 2017, 11:44:54 AM »
Ok Dane, but I must admit the question seems a bit ambiguous to me, maybe a simulation would be easier than a calculation.

But for two events, the one with the higher probability will occur first.

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#### Dane

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##### Re: BASIC DISTRIBUTION QUESTION
« Reply #8 on: March 02, 2017, 12:44:36 PM »
True!  It is good to know that you too find it difficult to calculate.
The answer may be used in other contexts where exactly SIX parts are involved, i.e. six numbers.

#### kav

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##### Re: BASIC DISTRIBUTION QUESTION
« Reply #9 on: March 02, 2017, 01:26:41 PM »
How you define adjacent? Is 1 adjacent to 3 (?) since they can be considered to belong to the same group of 1,2,3?

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#### Dane

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##### Re: BASIC DISTRIBUTION QUESTION
« Reply #10 on: March 02, 2017, 05:37:51 PM »
How you define adjacent? Is 1 adjacent to 3 (?) since they can be considered to belong to the same group of 1,2,3?

No, in my understanding of the word 1 is NOT adjacent to 3, since the two numbers are not side by side. I really must have explained it insufficient. Sorry.