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Author Topic: Maths question ?  (Read 7637 times)
bookiebasher
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« on: August 23, 2016, 01:54:32 PM »

From the following information can someone tell me the probability of the following happening on a roulette wheel ?

1. The probability of 5 being the highest % of plays and 17 being the joint 2nd highest ?

2. The probability of 5 , 17 , 35 being 1st , joint 2nd  and joint 6th highest in % of plays ?


      number drawn     % of plays
1         5                     4.5
2         17                    4.1
3         33                    4.1
4         25                    3.9
5         13                    3.7
6         35                    3.5
7         11                    3.5

Just curious Wink

Should add this is a UK roulette with a single zero.
« Last Edit: August 23, 2016, 01:56:55 PM by bookiebasher » Logged
atdc21
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« Reply #1 on: August 23, 2016, 02:02:14 PM »

dont know but the 'number 'is irrelevant, the % or odds will be the same for any given number and or combo, but you prob know that , so my post is pretty unhelpful  Cheesy
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Cf
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« Reply #2 on: August 23, 2016, 02:16:14 PM »

By number of plays you mean landed on?
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bookiebasher
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« Reply #3 on: August 23, 2016, 02:22:49 PM »

dont know but the 'number 'is irrelevant, the % or odds will be the same for any given number and or combo, but you prob know that , so my post is pretty unhelpful  Cheesy

Yeah  , the actual numbers drawn are irrelevant in terms of odds/probabilities.

Obviously any number can have the highest % of plays so for any one single number it is 37-1 including zero.

Trying to work out the double with a joint 2nd has got me a tad confused

The second question I haven't got the foggiest.

Doobs is the man for the job I think Wink
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bookiebasher
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« Reply #4 on: August 23, 2016, 02:23:18 PM »

By number of plays you mean landed on?


Yep
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Longines
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« Reply #5 on: August 23, 2016, 04:29:35 PM »

From the following information can someone tell me the probability of the following happening on a roulette wheel ?

1. The probability of 5 being the highest % of plays and 17 being the joint 2nd highest ?


2. The probability of 5 , 17 , 35 being 1st , joint 2nd  and joint 6th highest in % of plays ?


      number drawn     % of plays
1         5                     4.5
2         17                    4.1
3         33                    4.1
4         25                    3.9
5         13                    3.7
6         35                    3.5
7         11                    3.5

Just curious Wink

Should add this is a UK roulette with a single zero.

1. Probability of 5 being most popular = 37/1. Probability of 17 being 2nd most popular of the remaining numbers = 36/1. Probability of both = 37/1 x 36/1 = 1332/1.

2. Depends on how specific you want to be regarding "joint 2nd" and "joint 6th" i.e. exactly the same number of rolls or just the same number to one decimal place as a percentage - I think the answer differs depending on which one you choose. Not 100% sure what the maths is irrespective of which one you want  Grin
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bookiebasher
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« Reply #6 on: August 23, 2016, 04:55:44 PM »

From the following information can someone tell me the probability of the following happening on a roulette wheel ?

1. The probability of 5 being the highest % of plays and 17 being the joint 2nd highest ?


2. The probability of 5 , 17 , 35 being 1st , joint 2nd  and joint 6th highest in % of plays ?


      number drawn     % of plays
1         5                     4.5
2         17                    4.1
3         33                    4.1
4         25                    3.9
5         13                    3.7
6         35                    3.5
7         11                    3.5

Just curious Wink

Should add this is a UK roulette with a single zero.

1. Probability of 5 being most popular = 37/1. Probability of 17 being 2nd most popular of the remaining numbers = 36/1. Probability of both = 37/1 x 36/1 = 1332/1.

2. Depends on how specific you want to be regarding "joint 2nd" and "joint 6th" i.e. exactly the same number of rolls or just the same number to one decimal place as a percentage - I think the answer differs depending on which one you choose. Not 100% sure what the maths is irrespective of which one you want  Grin

That's how I worked out a pretty basic probability , so if the numbers 5 17 35 came out the most it would be 37 x 36 x 35 which equates to 46,620-1.

The argument I was having is that the guy from the gaming company supplying the machines said that it was within normal parameters of play but he was only looking at one specific number each time. My argument was that the 3 numbers in question , 5 17 35 , were 1st , joint 2nd , joint 6th in the % of times they came out. The sample size was not over 1 session either , it was over many sessions and many plays.

He argues that its "variance" and nothing unusual for a guy to bet mainly 3 specific numbers and for them to have those percentages. My argument is that I can understand 1 possibly 2 numbers in the top percentages but to get 3 in the top 6 the probability must be extremely high.

What that probability is , is the question.
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Doobs
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« Reply #7 on: August 23, 2016, 05:18:50 PM »

Don't understand.  Why are you picking these 3 numbers?  One number is always going to be most drawn.  Are you suspicious because these are the 3 most popular numbers for instance.
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Most of the bets placed so far seem more like hopeful punts rather than value spots
bookiebasher
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« Reply #8 on: August 23, 2016, 06:23:28 PM »

Don't understand.  Why are you picking these 3 numbers?  One number is always going to be most drawn.  Are you suspicious because these are the 3 most popular numbers for instance.

A guy kept backing those 3 numbers. He had other numbers but those 3 were his "bankers" , usually putting the max of £13.80 on one of them and around £5-£10 on the other two.

Over the course of around 10 days he won every time he played except once when he lost a nominal amount.

The actual numbers themselves are obviously irrelevant but I am trying to get a handle on the probability of those numbers
continually coming out like they did day after day. The machine guy says it's nothing unusual and the percentages are what you can expect.

My point to him was that the 3 numbers were ranked 1st , joint 2nd and joint 6th in there "hit" rate.

Trying to express that in a probability was pretty hard to do but it must be a big bloody number !!

The losses I incurred during that period will take around 6 months to recoup IF the machines run at "normal " variance.

So I am just trying to get some perspective on how the machines under performed shall we say.

« Last Edit: August 23, 2016, 06:26:49 PM by bookiebasher » Logged
doubleup
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« Reply #9 on: August 23, 2016, 06:33:37 PM »


You need to provide a complete breakdown of the number of spins and the times each number came up. 
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bookiebasher
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« Reply #10 on: August 23, 2016, 06:49:48 PM »


You need to provide a complete breakdown of the number of spins and the times each number came up. 

Ok , I am on the case and will report for duty in the morning hopefully  thumbs up
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DMorgan
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« Reply #11 on: August 23, 2016, 09:27:41 PM »

50p? Tongue
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typhoon13
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« Reply #12 on: August 24, 2016, 09:15:49 AM »


Thinks me might get round his shop today and make him squirm a tad more
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bookiebasher
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« Reply #13 on: August 24, 2016, 09:37:51 AM »


Thinks me might get round his shop today and make him squirm a tad more

One miserable old git in the shop is enough thank you very much.
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MintTrav
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« Reply #14 on: August 24, 2016, 10:17:53 AM »

From the following information can someone tell me the probability of the following happening on a roulette wheel ?

1. The probability of 5 being the highest % of plays and 17 being the joint 2nd highest ?


2. The probability of 5 , 17 , 35 being 1st , joint 2nd  and joint 6th highest in % of plays ?


      number drawn     % of plays
1         5                     4.5
2         17                    4.1
3         33                    4.1
4         25                    3.9
5         13                    3.7
6         35                    3.5
7         11                    3.5

Just curious Wink

Should add this is a UK roulette with a single zero.

1. Probability of 5 being most popular = 37/1. Probability of 17 being 2nd most popular of the remaining numbers = 36/1. Probability of both = 37/1 x 36/1 = 1332/1.

2. Depends on how specific you want to be regarding "joint 2nd" and "joint 6th" i.e. exactly the same number of rolls or just the same number to one decimal place as a percentage - I think the answer differs depending on which one you choose. Not 100% sure what the maths is irrespective of which one you want  Grin

That's how I worked out a pretty basic probability , so if the numbers 5 17 35 came out the most it would be 37 x 36 x 35 which equates to 46,620-1.

The argument I was having is that the guy from the gaming company supplying the machines said that it was within normal parameters of play but he was only looking at one specific number each time. My argument was that the 3 numbers in question , 5 17 35 , were 1st , joint 2nd , joint 6th in the % of times they came out. The sample size was not over 1 session either , it was over many sessions and many plays.

He argues that its "variance" and nothing unusual for a guy to bet mainly 3 specific numbers and for them to have those percentages. My argument is that I can understand 1 possibly 2 numbers in the top percentages but to get 3 in the top 6 the probability must be extremely high.

What that probability is , is the question.

It's only 46,620 if they have to come out in that order. But surely it doesn't matter what order they come out - you would still feel the same way if the top three were 17 35 5 or 35 17 5, etc. So the probability of them being the three most occurring numbers (in any sequence) would be 3/37 × 2/36 × 1/35, which is 6/46,620 or 1/7,770. I think the probability of them occurring in any three nominated positions would be the same. The probability of the three numbers occurring anywhere in the top seven would be considerably lower.
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