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scoutharmon
scoutharmon
Joined: Feb 22, 2018
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February 22nd, 2018 at 11:39:21 AM permalink
Hi all,

First time users/poster here. Please be merciful :-)

I hope someone could assist in calculating this probability of an unusual event happening in American Roulette.

Watching roulette the other day, I saw a "full house" of numbers show up on the board.

With 5 consecutive spins, the numbers: 13, 5, 5, 13, 5 appeared.

What is the probability and calculation of that occurring after 5 spins?

And within a 10 spin span ,what are the odds that the same type of series would appear?


Thank you for any help!

Scoutie H.
ThatDonGuy
ThatDonGuy
Joined: Jun 22, 2011
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February 22nd, 2018 at 12:22:50 PM permalink
Assuming a double-zero wheel:
There are 385 = 79,235,168 possible sets of results for five spins.

There are 38 different numbers that can be the "three of a kind", and for each one, there are 37 that can be the "pair".
There are 10 ways to arrange a set of 5 items where 3 are identical to each other, and the other 2 are identical to each other.
The total number of full houses = 38 x 37 x 10 = 14,060.

The probability in 5 spins is 14,060 / 79,235,168, or about 1 / 5635.

I am not sure what you are asking about in 10 spins - do you mean, the probability that one number will appear (at least) three times, and (at least) one other, different, number will appear (at least) twice?
scoutharmon
scoutharmon
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February 22nd, 2018 at 1:23:15 PM permalink
Thank you very much.

As for the second part of the question, I was wondering if there were ten random spins on an American roulette table, would the odds of that "full house" sequence occuring consecutively at anytime during those 10 spins would have the same odds as the 1/5635 you had calculated.

Thanks again
ChesterDog
ChesterDog
Joined: Jul 26, 2010
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February 22nd, 2018 at 3:08:38 PM permalink
Quote: scoutharmon

Hi all,

First time users/poster here. Please be merciful :-)

I hope someone could assist in calculating this probability of an unusual event happening in American Roulette.

Watching roulette the other day, I saw a "full house" of numbers show up on the board.

With 5 consecutive spins, the numbers: 13, 5, 5, 13, 5 appeared.

What is the probability and calculation of that occurring after 5 spins?

And within a 10 spin span ,what are the odds that the same type of series would appear?


Thank you for any help!

Scoutie H.



Your questions bring to mind Poker for RouletteTM, a side bet invented by a regular poster on Wizard of Vegas.

On a related note, what's the probability of eleven 18s in a row?
prozema
prozema
Joined: Oct 24, 2016
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February 22nd, 2018 at 5:59:09 PM permalink
Looks like the math guys are jumping on this one, but I see they all forgot to welcome you to the board... So, I'll cover that.

Thanks for the post, and welcome to the board.

I'm not going to take a crack at the math problem because 1) I'd do it wrong and 2) I'm not sure how the answer would be useful... I can say by looking... You are talking about longshots... But I am curious as to why the length of the Longshot matters...
DJTeddyBear
DJTeddyBear
Joined: Nov 2, 2009
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February 22nd, 2018 at 6:49:07 PM permalink
Quote: ChesterDog

Your questions bring to mind Poker for RouletteTM, a side bet invented by a regular poster on Wizard of Vegas.

That would be me. Thanks. You beat me to it. :)

I have math for three, four and five spin version. Here痴 a direct link to the five spin section of the math page:
http://www.poker-for-roulette.com/math_pay.html#five_spin

On that page, I list the Full House odds at 0.00017719 or 1 in 5,643.5. Note that this differs slightly from the prior response because I rank a FH consisting of 0 and 00 with the all green jackpot result.

But to the ten spin question: did you mean 5 CONSECUTIVE spins within a set of 10 spins? There are 6 such groups so simply multiply (or divide) the above numbers by 6.

Or did you mean ANY five results within a ten spin group? If so, I as, what痴 the point, but add that the multiply/divide factor is 720!
I invented a few casino games. Info: http://www.DaveMillerGaming.com/ 覧覧覧覧覧覧覧覧覧覧覧覧覧覧覧覧覧覧 Superstitions are silly, childish, irrational rituals, born out of fear of the unknown. But how much does it cost to knock on wood? 😁
ThatDonGuy
ThatDonGuy
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February 23rd, 2018 at 6:40:52 AM permalink
Quote: DJTeddyBear

But to the ten spin question: did you mean 5 CONSECUTIVE spins within a set of 10 spins? There are 6 such groups so simply multiply (or divide) the above numbers by 6.


I don't think it's as simple as that, as you are counting multiple full houses separately.
For example, if the first six spins are 10, 20, 10, 20, 10, 20, then you are counting the 10-10-10-20-20 full house in spins 1-5 and the 20-20-20-10-10 full house in spins 2-6 as separate results.
gordonm888
gordonm888
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February 23rd, 2018 at 9:07:32 AM permalink
Quote: prozema

You are talking about longshots... But I am curious as to why the length of the Longshot matters...



Curiosity is a virtue - and I think the OP was just being virtuous.
So many better men, a few of them friends, are dead. And a thousand thousand slimy things live on, and so do I.
scoutharmon
scoutharmon
Joined: Feb 22, 2018
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February 23rd, 2018 at 9:56:47 AM permalink
That is a great site you have! bookmarked.

In order to clarify my question may I present this example? Perhaps it makes 0 difference to the odds or maybe it does.

You step up to a Roulette Table - American 0/00.

On your first 5 bets, a full house occurs. Let's say the numbers are 10,2,2,2,10. You had made bets on 5 consecutive spins and that event occurs, right from your start.

vs.

On your first 10 bets, a full house occurs somewhere within those 10 spins. Let's say the numbers that come up are 33,4,8,10,2,2,2,10,28,9. So in this example, the event pops up in the middle area of the first 10 spins that you bet on.

Would the odds of the full house occurring in both examples each be 1 in 5,643.5?

Probably a very basic question but appreciate it.

Thanks

PS- If you haven't seen it already, George Carlin's take on Religion is a masterpiece. Easy to find his skit on YouTube.
gordonm888
gordonm888
Joined: Feb 18, 2015
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Thanks for this post from:
scoutharmon
February 23rd, 2018 at 11:44:52 AM permalink
Given that you are defining a full house as needing to occur on 5 consecutive spins, and that you want to know the probability of getting a full house in 10 spins:

It is very difficult to calculate this probability exactly because of sequences of 10 spins that have more than one full house like this:

2 2 8 8 8 2 2 8 8 8

In that 10 spin sequence there are 6 different full houses.

When calculating probabilities, it is very easy to make the error of "double counting" the full houses in these kind of sequences. Specifically, the sequence
2 2 8 8 8 2 2 8 8 8
is one sequence out of 79,235,1682 that has 6 full houses and not 6 sequences out of 79,235,1682 that have 1 full house.

So,if you want to know how many full houses you will see, on average, in 10 spins then I think the answer is 1/939.25

But if you want to know the probability of having one or more full houses in a 10 spin sequence then I don't know the correct answer. My rough guess is about 1/966. Edit: 1/955
Last edited by: gordonm888 on Feb 23, 2018
So many better men, a few of them friends, are dead. And a thousand thousand slimy things live on, and so do I.

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