Post by: guest39538 on 10/12/2015 14:14:28
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.
Post by: chiralSPO on 10/12/2015 15:20:19
Is the question regarding cards?
Post by: guest39538 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?
Post by: chiralSPO 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
Post by: Colin2B on 11/12/2015 00:07:06
1/3I agree with ChiralSPO
The entire internet is saying that three random variables,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.
N
N
N
Has a 4/52 chance of a specific variable
Post by: guest39538 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.
Please answer
Post by: guest39538 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?
Post by: chiralSPO on 11/12/2015 00:41:47
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.
Post by: guest39538 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?
Post by: Colin2B 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.
.. 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.So you are saying you only want answers that fit your theory???
Please answer
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.
Post by: guest39538 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.''
added -'' I understand what OP is trying to say. Your chances of picking heads depends on how many heads came in 5 flips. If head comes all 5 times, then your chances of picking heads is 100% , if it came all tails, your chances are 0%. If it came H T T T T, your chances are 20% and so on... But that is completely irrelevant to any kind of math problem. You can`t...[SNIP]... going to result in heads. If you flip the coin 5 times there will be 2.5 heads on average in that sample ( it doesn`t matter mathematically that OP chose the odd number of flips to confuse things ). So, in any sample of 5 flips , there will be 2.5 heads on average. Therefore 2.5/5, still 50% ,I agree that the one who has to pick can't calculate the odds. But the one who did the flips can.''
Post by: guest39538 on 11/12/2015 16:47:15
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?
Post by: alancalverd 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.
Post by: guest39538 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.
Post by: alancalverd on 11/12/2015 23:47:23
Post by: Colin2B 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,Your question is ambiguous, it can be interpreted in different ways.
what is the chance of an ace?
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.
Post by: guest39538 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.
Post by: guest39538 on 12/12/2015 04:19:55
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
Post by: Colin2B on 12/12/2015 10:06:58
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.
Alan's reply is very important for understanding 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.
Post by: alancalverd on 12/12/2015 12:38:49
How many aces are there in these 52 random's.About 4
Post by: guest39538 on 12/12/2015 13:42:28
How many aces are there in these 52 random's.About 4
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.
Post by: guest39538 on 12/12/2015 13:54:09
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.
Post by: Colin2B 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.
OK let's check your understanding.
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.
Post by: guest39538 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.
OK let's check your understanding.
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=0Post by: Colin2B 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?You've already got/been given this sequence 20.33%, but remember it is for all sequences in the pack not just one.
Can you show me please to confirm my understanding of this?
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 = 0-52No, you are misinformed
P(X=a)=0
52²
f(x)=
f(y)=
a few people have agreed and do see the problem.There are also people on this site who believe that gravity is due to air pressure!!
Who can I trust any more...
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.
Post by: guest39538 on 12/12/2015 17:28:26
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.
I don't consider you have discussed and explained any comparison of X and Y.
[ Invalid Attachment ]
Would you say this was a discrete and continuous diagram when the x axis is in constant shuffle ? ,
If I asked you to pick any square from any x axis you have a 1/2 chance of white and a 1/2 chance of black. But If I offered you a choice from any Y axis, you can observe the physics is ?/? and 1/2 and that is the paradox I am trying to explain.
Post by: guest39538 on 12/12/2015 19:50:30
I thank the other members also, but for my own sanity and to stop my self coming on forums, will you please just ban me now for real.
Post by: Colin2B on 12/12/2015 21:39:42
Words like expectation, discrete, continuous have very specific meanings in probability.
Y like X is discrete because it cannot take any value just integers 1 to 52. A measurement like length is continuous because it can take any value. It is important you call things by their correct names otherwise you will always be misunderstood.
I wish you luck in your quest, but you might have better luck if you were looking for the Holy Grail.
Let us know how you get on with the gaming board.
Who should you trust?
Trust in God, but keep your powder dry.
Post by: alancalverd on 13/12/2015 00:22:15
How many aces are there in these 52 random's.About 4
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.
You have picked 52 cards at random. Each one came from a full deck and thus had a 1 in 13 probability of being an ace. 52/13 = 4. This is the "expected" value, and if you repeated the experiment several times, you would expect the average number of aces to converge towards it. However there is obviously a finite probability that you could get any number from 0 to 52 aces in your selection, and that distribution is clearly skewed: the likelihood of 3 or 5 is larger than 2 or 6, but you can't get less than 0 (the probability of zero is (12/13)52= 1.5%), whilst the likelihood of 52 is (1/13)52 = 1.2 x 10-58 - nonzero but very small indeed.
Stating 4 would indeed be a bit like predicting the result of a lottery, which is why I said "about 4" - this is mathematically sensible and if I had to bet on any number, it would obviously be 4.
Post by: guest39538 on 13/12/2015 10:13:56
have 52 decks of cards lined up, slide off the top card of each deck, like in the earlier scenario, then slide the cards back onto the top of the original decks, you are having the first card,
you are offered a choice of top card, would you agree that this changes things compared to using a single deck?
About 4 that changes every turn, is not the same has a constant 4.
Would you agree that X 4/52 is not equal to Y <,>,=4/52 less than, more than or equal to 4.
P(a)/x=4
p(a)/y=<,>,4=0
where (a) is any one of 4 different suits of a single value such has an ace, and X and Y =52
and added, the time problem. Courtesy of Khan.
[ Invalid Attachment ]
Post by: guest39538 on 13/12/2015 11:09:03
[ Invalid Attachment ]
Post by: alancalverd on 13/12/2015 12:18:37
you are offered a choice of top card, would you agree that this changes things compared to using a single deck?No. The probability of any top card being an ace remains at 1/13 because you haven't changed anything.
Quote
P(a)/x=4is obvious nonsense if P is a probability. No probability can exceed 1.
p(a)/y=<,>,4=0
Post by: guest39538 on 14/12/2015 11:29:59
you are offered a choice of top card, would you agree that this changes things compared to using a single deck?No. The probability of any top card being an ace remains at 1/13 because you haven't changed anything.QuoteP(a)/x=4is obvious nonsense if P is a probability. No probability can exceed 1.
p(a)/y=<,>,4=0
Hi Alan, 4 was 400%, each individual ace being 100% itself.
You say 1/13, I am confused by this ,
a dice 1/6
a coin 1/2
a roulette wheel 1/37
a deck of cards 1/52
aces in a deck of cards 4/52
52 random cards 1/13? should your answer not end with 52?
Where are you getting 1/13 from that would suggest a known odds/chance?
Post by: chiralSPO on 14/12/2015 15:23:52
Post by: guest39538 on 14/12/2015 15:38:43
4/52 = 1/13
4/52 =4/52
1/13=1/13
4/52 is not equal to 1/13
I have not offered a choice of 13, I offered a choice of 52,
below 4 choices of X axis 1/13 and 4 choices of Y axis 4/4
[ Invalid Attachment ]
Clearly there is something different in the x and y axis to output P's in this diagram
[ Invalid Attachment ]
Post by: Colin2B on 14/12/2015 15:48:28
4/52 =4/52This is an example of what is frustrating about you. You claim to know maths and probability, but fail to understand the most basic of the maths. This is why there are so many misunderstandings.
1/13=1/13
4/52 is not equal to 1/13
Probability, to answer your question, is measured by dividing the number of items (or cases) favourable to what you are trying to find, divided by the total number of possibilities (cases).This is a fraction, but is usually shown as a decimal number between 0 and 1.
So for a coin, if we want the probability of a head then there is one head (favourable case) but 2 possibilities head and tail. So probability is 1/2 as a fraction, or 0.5 as a decimal fraction, or 50% as a percentage. But fractions can be shown in different way so 1/2=2/4=4/8=8/16 so if you divide 8/16 you still get 0.5 as with 1/2, so they are all saying the same thing.
OK, I know you will say you knew all that, but your questions indicate you don't.
Post by: guest39538 on 14/12/2015 15:55:38
4/52 =4/52This is an example of what is frustrating about you. You claim to know maths and probability, but fail to understand the most basic of the maths. This is why there are so many misunderstandings.
1/13=1/13
4/52 is not equal to 1/13
Probability, to answer your question, is measured by dividing the number of items (or cases) favourable to what you are trying to find, divided by the total number of possibilities (cases).This is a fraction, but is usually shown as a decimal number between 0 and 1.
So for a coin, if we want the probability of a head then there is one head (favourable case) but 2 possibilities head and tail. So probability is 1/2 as a fraction, or 0.5 as a decimal fraction, or 50% as a percentage. But fractions can be shown in different way so 1/2=2/4=4/8=8/16 so if you divide 8/16 you still get 0.5 as with 1/2, so they are all saying the same thing.
OK, I know you will say you knew all that, but your questions indicate you don't.
No Colin I understand, the point is why are you removing something away, narrowing it down when the scenario of 52 has not contracted in any way.
example - AAAA......1-13........52 there is no 1/13 in this example, the aces are aligned in the top 13.
I offered you 1 of 52 random cards, I did not offer you to pick 13 cards then pick one of them.
Post by: Colin2B on 14/12/2015 16:02:27
4/52=0.0769=7.69%.
1/13=0.0769=7.69%.
They are exactly the same.
If you don't get this you really don't understand maths and probability.
Post by: alancalverd on 14/12/2015 16:06:48
52 random cards 1/13? should your answer not end with 52?
Where are you getting 1/13 from that would suggest a known odds/chance?
4/52 = 1/13 in my universe.
Post by: guest39538 on 14/12/2015 16:11:18
No one is removing anything. Do the sums
4/52=0.0769=7.69%.
1/13=0.0769=7.69%.
They are exactly the same.
If you don't get this you really don't understand maths and probability.
I understand Colin, everybody seem's to be missing the point,
Take 52 top cards from 52 shuffled decks, forming new deck Y. Shuffle the Y deck, take the top card, look at it, the chance of that card being an ace is 4/52 or 1/13 or 7.69%, yes I agree with this but only if we know that there is exactly 4 aces in new deck Y. There is only a 20.3% chance that Y contains exactly 4 aces, and an even less chance to be 4 individual aces. There is a 1.6% chance there is no aces at all. There is a variable of percentage to how many aces Y contains.
Lets say you your top card is an ace from Y , and its the ace of hearts. draw the second card , what is the chance that card is also an ace, and also the ace of hearts?
It should be zero percent in a standard deck, premise for argument and discussion.
Draw the 20th card and the 34 th card, what is the chance they are also the ace of hearts?
Post by: Colin2B on 14/12/2015 16:39:43
everybody seem's to be missing the point,No one is missing the point. You asked Alan a very specific question about expectation and he answered correctly.
if you look at the probability distribution of the number of aces in your constructed deck you will see that 4 aces has the highest probability, so over a large number of games you would expect to get around 4, same as for a standard deck.
If you look back at previous threads we have been over this before. We never said the sequences are the same, just that the long run probabilities are the same.
Post by: alancalverd on 14/12/2015 21:57:25
Take 52 top cards from 52 shuffled decks, forming new deck Y. Shuffle the Y deck, take the top card, look at it, the chance of that card being an ace is 4/52 or 1/13 or 7.69%,
No. The expectation value is 1/13, but unlike a proper deck, it is conceivable that the new deck contains anything from 0 to 52 aces, as you say.
Quote
Lets say you your top card is an ace from Y , and its the ace of hearts. draw the second card , what is the chance that card is also an ace, and also the ace of hearts?1/52