# The Naked Scientists Forum

### Author Topic: How is probability calculated?  (Read 5477 times)

#### Thebox

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##### How is probability calculated?
« on: 10/12/2015 14:14:28 »
The entire internet is saying that three random variables,

N
N
N

Has a 4/52 chance of a specific variable

what, there is 3 variables not 52 am I missing something here?

at the moment my intelligence is being insulted.
« Last Edit: 13/12/2015 11:42:11 by chris »

#### chiralSPO

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« Reply #1 on: 10/12/2015 15:20:19 »
You have specified three variables picked at random from a distribution, but you have not specified the type of distribution. It seems to me that they are assuming the distribution is a deck of cards (which 4/52 would work for), but they would be incorrect if you are discussing coin tosses, a continuous uniform distribution from 0 to 1, a normal distribution or a chi squared distribution etc. etc.

Is the question regarding cards?

#### Thebox

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« Reply #2 on: 10/12/2015 16:26:45 »
You have specified three variables picked at random from a distribution, but you have not specified the type of distribution. It seems to me that they are assuming the distribution is a deck of cards (which 4/52 would work for), but they would be incorrect if you are discussing coin tosses, a continuous uniform distribution from 0 to 1, a normal distribution or a chi squared distribution etc. etc.

Is the question regarding cards?

If the three variants are unknown variants , would it matter if they were playing cards, numbers or even xmas cards?

The choice is clearly of three unknown variants?

?/3

Ok ,, lets give the unknown variants values,  ones a christmas card, ones an ace from a deck of cards, and ones a number 1.

NNN

Pick one, what is the chance of an ace?
« Last Edit: 10/12/2015 16:36:32 by Thebox »

#### chiralSPO

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« Reply #3 on: 10/12/2015 23:22:56 »
You have specified three variables picked at random from a distribution, but you have not specified the type of distribution. It seems to me that they are assuming the distribution is a deck of cards (which 4/52 would work for), but they would be incorrect if you are discussing coin tosses, a continuous uniform distribution from 0 to 1, a normal distribution or a chi squared distribution etc. etc.

Is the question regarding cards?

If the three variants are unknown variants , would it matter if they were playing cards, numbers or even xmas cards?

The choice is clearly of three unknown variants?

?/3

Ok ,, lets give the unknown variants values,  ones a christmas card, ones an ace from a deck of cards, and ones a number 1.

NNN

Pick one, what is the chance of an ace?

1/3

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#### Colin2B

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« Reply #4 on: 11/12/2015 00:07:06 »
1/3
I agree with ChiralSPO

The entire internet is saying that three random variables,

N
N
N

Has a 4/52 chance of a specific variable
The only way this could be true is if you were selecting the 3 Ns by picking the top cards from 3 decks of 52, then the 4/52 represents the probability of a particular face value eg ace.

#### Thebox

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« Reply #5 on: 11/12/2015 00:19:18 »

The only way this could be true is if you were selecting the 3 Ns by picking the top cards from 3 decks of 52, then the 4/52 represents the probability of a particular face value eg ace.

No Colin seriously, I really am telling the truth, just your 1/3 answer was needed mate , please continue to answer and I will show you.   I will add  the next part on Chirals post.

#### Thebox

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« Reply #6 on: 11/12/2015 00:21:13 »

1/3

Take 5 coins, toss each coin and place the coin on a table or counter, result side up.

hide the 5 results and ask another party to pick one of the 5 results,

What is their chance of heads from the results you know?
« Last Edit: 11/12/2015 00:23:41 by Thebox »

#### chiralSPO

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« Reply #7 on: 11/12/2015 00:41:47 »
If you know they are all tails, then they have no chance of getting heads. If three of them are heads, they have a 60% chance of getting heads.

I know this is counter-intuitive because it sounds like the results change based on what you know. This is not the case. It works out because the chances that you get a biased result in one direction is exactly equal to the chances of getting the biased result in the other direction.

When the coins are first tossed, there is a 1 in 32 chance that all coins are heads, but there is also a 1 in 32 chance that all coins are tails. There is a 5/32 chance of there being exactly 1 head and 4 tails, but also a 5/32 chance of getting 4 heads and 1 tail. Then there is a 10/32 chance that it is 3 heads and 2 tails, and a 10/32 chance that it is 3 tails and 2 heads. If you count it all up, at the end, it's still a perfect 50/50 shot.

As long as the one peeking at the results isn't betting about them against people who have not seen, this type of game should be quite fair.

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#### Thebox

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« Reply #8 on: 11/12/2015 02:44:48 »
If you know they are all tails, then they have no chance of getting heads. If three of them are heads, they have a 60% chance of getting heads.

I know this is counter-intuitive because it sounds like the results change based on what you know. This is not the case. It works out because the chances that you get a biased result in one direction is exactly equal to the chances of getting the biased result in the other direction.

When the coins are first tossed, there is a 1 in 32 chance that all coins are heads, but there is also a 1 in 32 chance that all coins are tails. There is a 5/32 chance of there being exactly 1 head and 4 tails, but also a 5/32 chance of getting 4 heads and 1 tail. Then there is a 10/32 chance that it is 3 heads and 2 tails, and a 10/32 chance that it is 3 tails and 2 heads. If you count it all up, at the end, it's still a perfect 50/50 shot.

As long as the one peeking at the results isn't betting about them against people who have not seen, this type of game should be quite fair.

Can I ask why you are counting them all up at the end? that  makes absolutely no sense when the results are random and unknown.  The choice is out of 5, all 5 could be heads like you admitted, there may be no heads, so why add them up when the chance is unknown and a variate?

#### Colin2B

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« Reply #9 on: 11/12/2015 09:36:27 »
Can I ask why you are counting them all up at the end? that  makes absolutely no sense when the results are random and unknown.  The choice is out of 5, all 5 could be heads like you admitted, there may be no heads, so why add them up when the chance is unknown and a variate?
We've been through this many times before in the other threads, I'm sure we even wrote out this heads/tails sequence in detail along with examples of card sequences.
What you are dealing with is more a sort of predestination theory, not probability.

So you are saying you only want answers that fit your theory???
Hardly seems worth answering in that case.

This is why I didn't bother to answer your post about "are they pulling my leg". That post covers things we've explained in detail in other threads and reveals more about your theory than you realise if you bothered to understand it. We've spent a lot of time trying to explain probability and I don't really want to start going over old ground again.

#### Thebox

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« Reply #10 on: 11/12/2015 14:11:39 »

So you are saying you only want answers that fit your theory???
Hardly seems worth answering in that case.

This is why I didn't bother to answer your post about "are they pulling my leg". That post covers things we've explained in detail in other threads and reveals more about your theory than you realise if you bothered to understand it. We've spent a lot of time trying to explain probability and I don't really want to start going over old ground again.

No predestination involved Colin , just maths. So you know I do have people other places who have agreed with me.   I am not insane,

''I agree with the op. The coins have been flipped so there is a result.that means the chance to pick the result of a new dice isn't 1/2 anymore. You'd have to know the result of the flips to know the chance of being right. If the choice was made before the coin flips the probability would be 1/2.

But let's say it turned out:

H T H H T. Then the probability of picking heads would be 3/2.''

« Last Edit: 11/12/2015 16:25:47 by Thebox »

#### Thebox

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« Reply #11 on: 11/12/2015 16:47:15 »
To clarify
123
123
123

If I can see the above and you can't see this, and I offer you a choice of the first values, and I tell you , your odds are 1/3,and 2 is a winner, I am committing fraud?

#### alancalverd

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« Reply #12 on: 11/12/2015 21:05:51 »
What is their chance of heads from the results you know?
If you already know there are n heads, then you know the probability of their picking a head is n/5.

Quote
If I can see the above and you can't see this, and I offer you a choice of the first values, and I tell you , your odds are 1/3,and 2 is a winner, I am committing fraud?

Yes, because you know the quoted odds are a lie. You are knowingly offering a defective product. People have been imprisoned for selling raffle tickets with no winning number, and it is even illegal to draw the third prize first: the buyer's expectation is to have an equal chance of winning the first prize.

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#### Thebox

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« Reply #13 on: 11/12/2015 22:30:19 »

If you already know there are n heads, then you know the probability of their picking a head is n/5.

Thank you interesting,

So if we take 52 random cards, one from each deck of 52 decks, and in considering what we have discussed,

what is the chance of an ace?

I can tell you there is a 1.6% that there is no aces at all in your random 52 , and a 20.3% chance in there being exactly 4 aces.

#### alancalverd

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« Reply #14 on: 11/12/2015 23:47:23 »
Each card has a 1/13 chance of being an ace, and all the picks are independent, so the probability of picking an ace is 1/13, no matter how many decks you begin with.

#### Colin2B

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« Reply #15 on: 12/12/2015 00:24:28 »
So if we take 52 random cards, one from each deck of 52 decks, and in considering what we have discussed,

what is the chance of an ace?
Your question is ambiguous, it can be interpreted in different ways.
In addition to how Alan has interpreted it, I can think of at least other 2 ways. Answers to both these 2 questions are 6.74% and 98.4%

Can you tell me what questions these answers relate to.

You need to be much clearer how your questions are worded because it can lead to confusion.

All of these answers, including the different patterns that can occur when selecting cards from multiple decks, we have answered in the different threads on this forum.

However, picking a pack from 52 packs does not answer your poker theory.

#### Thebox

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« Reply #16 on: 12/12/2015 01:44:14 »

Your question is ambiguous, it can be interpreted in different ways.
In addition to how Alan has interpreted it, I can think of at least other 2 ways. Answers to both these 2 questions are 6.74% and 98.4%

Can you tell me what questions these answers relate to.

You need to be much clearer how your questions are worded because it can lead to confusion.

All of these answers, including the different patterns that can occur when selecting cards from multiple decks, we have answered in the different threads on this forum.

However, picking a pack from 52 packs does not answer your poker theory.

I have no idea at this time what your percentages are related to. I know Alan answered 1/13 so I can only presume he misinterpreted the question like you mentioned.

Have 52 decks of cards in a row, , from each deck  slide the top card off each deck and discard the rest of the cards.

A to B
A to B
A to B
A to B

etc for 52 decks.

leaving 52 of B and no more A

BBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBB

I am not asking you what is the chance an individual card is an ace, I am asking you what is the chance of an ace out of the  52 choices. How many aces are there in these 52 random's.

#### Thebox

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« Reply #17 on: 12/12/2015 04:19:55 »
Prob(0 aces in new deck) = 0.01557
Prob(1 aces in new deck) = 0.06749
Prob(2 aces in new deck) = 0.14340
Prob(3 aces in new deck) = 0.19917
Prob(4 aces in new deck) = 0.20332
Prob(5 aces in new deck) = 0.16265

Prob(6 aces in new deck) = 0.10618
Prob(7 aces in new deck) = 0.05814
Prob(8 aces in new deck) = 0.02726
Prob(9 aces in new deck) = 0.01110
Prob(10 aces in new deck) = 0.00398

Prob(11 aces in new deck) = 1.266E-3
Prob(12 aces in new deck) = 3.605E-4
Prob(13 aces in new deck) = 9.243E-5
Prob(14 aces in new deck) = 2.146E-5
Prob(15 aces in new deck) = 4.530E-6

Prob(16 aces in new deck) = 8.729E-7
Prob(17 aces in new deck) = 1.540E-7
Prob(18 aces in new deck) = 2.496E-8
Prob(19 aces in new deck) = 3.722E-9
Prob(20 aces in new deck) = 5.118E-10

Prob(21 aces in new deck) = 6.499E-11
Prob(22 aces in new deck) = 7.631E-12
Prob(23 aces in new deck) = 8.295E-13
Prob(24 aces in new deck) = 8.353E-14
Prob(25 aces in new deck) = 7.796E-15

Prob(26 aces in new deck) = 6.746E-16
Prob(27 aces in new deck) = 5.414E-17
Prob(28 aces in new deck) = 4.028E-18
Prob(29 aces in new deck) = 2.778E-19
Prob(30 aces in new deck) = 1.775E-20

Prob(31 aces in new deck) = 1.050E-21
Prob(32 aces in new deck) = 5.740E-23
Prob(33 aces in new deck) = 2.899E-24
Prob(34 aces in new deck) = 1.350E-25
Prob(35 aces in new deck) = 5.786E-27

Prob(36 aces in new deck) = 2.277E-28
Prob(37 aces in new deck) = 8.205E-30
Prob(38 aces in new deck) = 2.699E-31
Prob(39 aces in new deck) = 8.074E-33
Prob(40 aces in new deck) = 2.187E-34

Prob(41 aces in new deck) = 5.333E-36
Prob(42 aces in new deck) = 1.164E-37
Prob(43 aces in new deck) = 2.256E-39
Prob(44 aces in new deck) = 3.845E-41
Prob(45 aces in new deck) = 5.697E-43

Prob(46 aces in new deck) = 7.224E-45
Prob(47 aces in new deck) = 7.685E-47
Prob(48 aces in new deck) = 6.671E-49
Prob(49 aces in new deck) = 4.538E-51
Prob(50 aces in new deck) = 2.269E-53

Prob(51 aces in new deck) = 7.415E-56
Prob(52 aces in new deck) = 1.188E-58

Just compare all that to a standard deck,

a standard deck

prob 0 aces = 0%
prob 1 ace =0%
prob 2 aces= 0%
prob 3 aces =0%
prob 4 aces = 100%
prob 5 aces = 0%

etc etc 0% the rest of the way

#### Colin2B

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« Reply #18 on: 12/12/2015 10:06:58 »
You need to be clearer.
I answered this question, so did Alan.
what is the chance of an ace?

I have no idea at this time what your percentages are related to.
Well, you should because those same answers are contained in that long list someone gave you - give or take rounding.
My answers are for the probability that there is one and only one ace in the constructed pack, and secondly that there is at least one ace in the pack.

I'm now convinced that someone is having a laugh, because they are not giving you full information about what this list means. We have always been straight with you and tried to explain the difference between individual probability and that of sequences (and given examples of how that list is worked out), and how it relates to your poker theory.

What he is saying is that if you take the deck constructed from the other 52 decks, and choose a card at random, then there is a 1/13 probability it will be an ace just the same as in a standard deck.
(Note, this also works for a deck constructed from a infinite number of decks shuffled together, or if you take a card from a pack, write down its value, replace the card, shuffle, and then pick another card, repeat 52 times.)

There are some other important points which the other forums are not telling you.
- the list you have been given shows the probability (not the actual outcome) is the same for every deck so constructed, so deck skipping has no long run effect.
- the online poker games do not use this method of constructing decks for the games, they use simulators.

If you think the simulator is wrong, you need to take your argument to the gaming board who will investigate as they do testing of these simulators. They are currently asking for input.
http://www.gamblingcommission.gov.uk/Technical-standards.aspx

Remember, we are not trying to have a laugh at your expense, just to help you understand probability.
« Last Edit: 12/12/2015 10:09:15 by Colin2B »

#### alancalverd

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« Reply #19 on: 12/12/2015 12:38:49 »
How many aces are there in these 52 random's.

#### Thebox

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« Reply #20 on: 12/12/2015 13:42:28 »
How many aces are there in these 52 random's.

My head feels like it is in the twilight zone....about 4?   where the hell do you get about 4 from? It is 79.7% chance that there is not 4, saying 4 is equivalent to predicting the lottery results.
« Last Edit: 12/12/2015 14:20:01 by Thebox »

#### Thebox

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« Reply #21 on: 12/12/2015 13:54:09 »
''What he is saying is that if you take the deck constructed from the other 52 decks, and choose a card at random, then there is a 1/13 probability it will be an ace just the same as in a standard deck. ''

I know this , this is not even the question I am asking.

What I am asking is this -

AA4A4

We can clearly see that for an ace in this revealed example, the chance of an ace is 3/5.

So if we have

1.................................................52 random cards from 52 decks, the new deck contains more aces than 4, less than 4, or equal to 4,

So forgetting after the pick for now of 1/13,   what is the chance of the pick, picking an ace.
« Last Edit: 12/12/2015 14:25:29 by Thebox »

#### Colin2B

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« Reply #22 on: 12/12/2015 14:21:48 »
...I know this....
If you really know it - by which I mean understand it - then you would know how he arrives at "about 4"

this is not even the question I am asking.
Well, it is in the words you used.

You have been given a long list of probabilities, did anyone explain exactly what it means.  Did they explain about expectation? Did they explain about the mode of data? Did they suggest you put the data in a histogram?
Do you understand what it means to have a probability of < 10-3 let alone 10-50?
They're having a laugh, but we tell it how it is.

#### Thebox

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« Reply #23 on: 12/12/2015 14:29:00 »
...I know this....
If you really know it - by which I mean understand it - then you would know how he arrives at "about 4"

this is not even the question I am asking.
Well, it is in the words you used.

You have been given a long list of probabilities, did anyone explain exactly what it means.  Did they explain about expectation? Did they explain about the mode of data? Did they suggest you put the data in a histogram?
Do you understand what it means to have a probability of < 10-3 let alone 10-50?
They're having a laugh, but we tell it how it is.

I only understand my Hypothesis Colin, and how to work out some probability, I do not need to understand all probability,  when I can see the problem has clear as light. I am not being arrogant Colin, there is a problem, a few people have agreed and do see the problem.

Who can I trust any more...

I take the top card from a deck and repeat for 52 times with 52 decks, how do I work out the chance of the 52 random's contain exactly 4 aces?

Can you show me please to confirm my understanding of this?

I get that the X-axis is  discrete random variables where as the Y-axis is continuous random variables.

integral Y = 0-52

P(X=a)=0

52²

f(x)=

f(y)=X²=0

« Last Edit: 12/12/2015 15:26:29 by Thebox »

#### Colin2B

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« Reply #24 on: 12/12/2015 16:40:05 »
I do not need to understand all probability,
No one expects you to understand all probability, just enough to understand your hypothesis. That is why I put those items down, you need to understand what they mean.

I take the top card from a deck and repeat for 52 times with 52 decks, how do I work out the chance of the 52 random's contain exactly 4 aces?

Can you show me please to confirm my understanding of this?
You've already got/been given this sequence 20.33%, but remember it is for all sequences in the pack not just one.
Also remember what Alan is saying, when you create a pack from 52 others, if you do it a lot of times you will on average get 4 aces in the created packs. This is why understanding mode and histograms is important, otherwise you will make incorrect assumptions about your hypethesis.

I get that the X-axis is  discrete random variables where as the Y-axis is continuous random variables.
No you are mistaken or misinformed. Both x and y are discrete not continuous.

integral Y = 0-52

P(X=a)=0

52²

f(x)=

f(y)=X²=0
No, you are misinformed

a few people have agreed and do see the problem.

Who can I trust any more...
There are also people on this site who believe that gravity is due to air pressure!!

We have explained the differences in the sequences between what you call X & Y, why they occur, and why they don't give the result you think they do.
We have never attempted to mislead you.