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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?

Quote from: chiralSPO on 10/12/2015 15:20:19You 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??/3Ok ,, lets give the unknown variants values, ones a christmas card, ones an ace from a deck of cards, and ones a number 1.NNNPick one, what is the chance of an ace?

1/3

The entire internet is saying that three random variables, NNNHas 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.

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?

.. 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

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.

What is their chance of heads from the results you know?

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?

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

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.

what is the chance of an ace?

I have no idea at this time what your percentages are related to.

How many aces are there in these 52 random's.

Quote from: Thebox on 12/12/2015 01:44:14How many aces are there in these 52 random's.About 4

...I know this....

this is not even the question I am asking.

Quote from: Thebox on 12/12/2015 13:54:09...I know this....If you really know it - by which I mean understand it - then you would know how he arrives at "about 4"Quote from: Thebox on 12/12/2015 13:54:09this 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 do not need to understand all probability,

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 = 0-52P(X=a)=052²f(x)=f(y)=X²=0

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.

Quote from: alancalverd on 12/12/2015 12:38:49Quote from: Thebox on 12/12/2015 01:44:14How many aces are there in these 52 random's.About 4My 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 are offered a choice of top card, would you agree that this changes things compared to using a single deck?

P(a)/x=4p(a)/y=<,>,4=0

Quote from: Thebox on 13/12/2015 10:13:56you 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=4p(a)/y=<,>,4=0 is obvious nonsense if P is a probability. No probability can exceed 1.

4/52 = 1/13

4/52 =4/521/13=1/134/52 is not equal to 1/13

Quote from: Thebox on 14/12/2015 15:38:434/52 =4/521/13=1/134/52 is not equal to 1/13This 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.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.

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?

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.

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%,

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?