Any Mathematicians or Statisticians Here?
Hello
I have never been very good at maths and trying to figure out this answer has begun to boggle my mind, so any sensible answer would be much appreciated. I am trying to work this out for an article that I am writing. It is not a trick question and I honestly do not know the answer. I've tried looking up a solution on-line but ended up more confused!
What are the odds (or probability) that the following could happen?
- A randomly mixed deck of playing cards (52) is placed face down. You guess what the top card is, turn it over and you are correct. You place it to one side.
The odds of this happening must be 52-1.
- Now you guess the next card, turn it over and you are again correct. You place it to one side.
The odds of guessing this correctly must now be 51-1, as there are now just 51 cards left.
- This continues right down to the remaining 2 cards, when you now have a 50/50 or 1 in 2 change of being correct.
In my mind, as you go down the deck then for each individual guess the odds decrease. But, the cumulative odds of guessing each card in sequence must increase.
Are you still with me?
So my question is: What are the odds of correctly guessing the sequence of all 52 cards in a randomly mixed deck?
I looked at a similar 'problem' with rolling a die and worked out that the odds against correctly predicting the outcome of rolling the die 6 times in succession was 6 x 6 x 6 x 6 x 6 x 6 = 46,656 : 1
However I'm stumped with the card question.
Thanks
KH
Comments
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I’m hopeless at maths and my brain doesn’t even want to contemplate the answer, however, how do the odds decrease as you go down the pack. If each time you flip a coin the odds of it being heads are always 50/50 however many times you do it so in the same way how do the odds of the cards decrease. I’ve probably got this wrong, never good at maths questions.
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Flipping a coin and rolling the die are different because the number of options remains the same each time. With the coin there are always only two sides, with the die there are always only 6 sides.
With the cards as the 'guessed' card is removed the number of cards remaining in the deck gets fewer each time, so therefor for each individual guess the odds decrease. As in the example above when you have just two cards left the chances of guessing correctly have gone down to 50/50 or 1 in 2.
It is this that make the answer more difficult, well for me anyway!
KH
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The odds of guessing all 52 cards correctly are
52 x 51 x 50 x 49 x ................and so on >>>.........................................3 x 2 to 1
A big number
80658175170943878571660636856403766975289505440883277824000000000000 to 1
I believe from my schoolboy maths that is called "Factorial" 52
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Is this any good?
David
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Apologies for coming in late but it's my birthday so the following may be incorrect due to alcoholic influences - I'll check when I'm sober.
Firstly there is difference probability and odds in the way they are worked out. They are not the same, although for very large numbers they approach the same value - see below.
A simple example: tossing a coin the probability of getting a head is 0.5 but the odds are 1:1. or one to one.
If the probability of winning something 0.25 then the odds are 1 : 3
Odd are calculated by:
(probability of something happening) divided by (1 - that probability)
Odd are usually expressed as a ratio
Probabilites must be expressed as a fraction or its equivalent.
As I said for very large numbers then the two begin to approach each other, in the example given about the die the probability is
1/6 x 1/6 x 1/6 x 1/6 1/6 x 1/6 = 1/46656 but the odds are 1 : 46655
So the probability is 1/52! and because the number is so huge that the odds will be almost be 1:52!
And the pack is not new or unshuffled.
Glass of wine please!
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Ok my turn:
if I give you the whole numbers from 1 to 15, can you rearrange them into a new order where every two adjacent numbers add to give a square number?
When I'm sober and if I remember posting I'll give you the answer. NO using google!
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Two plus two equals five - for large values of two.
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Thank you!
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I am sure many of you will have spent a happy few hours working this out.
8 1 15 10 6 3 13 12 4 5 11 14 2 7 9
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I notice a couple of sequences in the positioning of the numbers. Can you extend this beyond 15 then?
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Yes, 1 to 15 is actually the first set that makes it possible, 1 to 16 and 17 is also good but then it cannot be done till 1 to 23, 1 to 24 isn't possible but then from 1 to 25 onwards everything works.
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Thanks. The "Law of Large Numbers" again.
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My pleasure Nav.
Now a club site has an infinite number of pitches and they are all taken, then another outfit turns up...
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Ask everyone to move down by one pitch, then the new arrival can use pitch one.
OR
Call it a CL site then at least seven caravans can fit on to five pitches
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