Number of ways to shuffle a deck of cards
That's about 8.07 x 1067, or 8 with 67 zeroes after it.
If a million people each shuffled a million decks of cards every second from now until the end of the earth (best guess = 8 billion years), that would be only 2.56 x 1029 decks. With all those decks, in all that time, it's unlikely that exactly there would ever be two matching decks of cards.
Another large number is the possible number of Bitcoin addresses: 1.4 x 1048. Anyone can generate a new Bitcoin address using random data (or have a service do it for you), but how do you make sure you didn't choose an address that someone else already chose? For example, if someone created a paper wallet but has never used it, it won't appear on the blockchain, so how do you know whether someone else already chose your address? The answer is you don't need to know: If a million people each created a million new Bitcoin addresses every second for billions of years...same principle. Created randomly, no one will ever create a random Bitcoin address that someone else already created.
Yeah, a collision is possible, but not like winning the lottery is possible. Far, far longer than that. It's like winning Powerball six times in a row.
I read this and all I can think is lets do it. You create the data post it here with a deployment
date & time and we will see how many we can get to participate in the Great Bitcoin Wallet
Collision of 2022.
Not really, I used a calculator.Quote: rainmanGood stuff MB. I am fully willing to admit your brain is significantly more powerful than mine.
link to original post
maybe if you have a powerful computer with a program that runs for days?
https://web2.0calc.com/
Quote: MichaelBluejayThe number of distinct results for shuffling a deck of cards is 52!, or 52 x 51 x 50 x 49, etc.
That's about 8.07 x 1067, or 8 with 67 zeroes after it.
link to original post
Although this number sounds gigantic, it is not horrible. In a blackjack game, it’s much smaller because card suits do not matter at all. What is the number of possible deck compositions for a one-deck blackjack game?
Quote: acesideQuote: MichaelBluejayThe number of distinct results for shuffling a deck of cards is 52!, or 52 x 51 x 50 x 49, etc.
That's about 8.07 x 1067, or 8 with 67 zeroes after it.
link to original post
Although this number sounds gigantic, it is not horrible. In a blackjack game, it’s much smaller because card suits do not matter at all. What is the number of possible deck compositions for a one-deck blackjack game?
link to original post
If you ignore suits and consider the 52 card deck to be comprised of 13 ranks with four identical cards per rank then the number of possible deck compositions is (I think) = 52!/ (13*4!) = 2.6 x 1065.
Someone needs to check me on this.
Quote: gordonm888Quote: acesideQuote: MichaelBluejayThe number of distinct results for shuffling a deck of cards is 52!, or 52 x 51 x 50 x 49, etc.
That's about 8.07 x 1067, or 8 with 67 zeroes after it.
link to original post
Although this number sounds gigantic, it is not horrible. In a blackjack game, it’s much smaller because card suits do not matter at all. What is the number of possible deck compositions for a one-deck blackjack game?
link to original post
If you ignore suits and consider the 52 card deck to be comprised of 13 ranks with four identical cards per rank then the number of possible deck compositions is (I think) = 52!/ (13*4!) = 2.6 x 1065.
Someone needs to check me on this.
link to original post
Why not 52!/[(4!)^13].
Quote: odiousgambitfor the often seen 6 decks in BJ, my usual online calculator refuses to guess how many combinations there are
maybe if you have a powerful computer with a program that runs for days?
https://web2.0calc.com/
link to original post
I think it depends on what you consider a meaningful equivalence.
If you outright ignore suits, and 10 J Q K are all interchangeable, you get a surprisingly low number of "different" shuffled decks.
In some games, suit and the X ranks are significant, and you get a different answer.
If one J♣ is not equivalent to any of the others, then I think it should be 312!. This is a very large number followed by 76 zeroes.
Quote: DieterQuote: odiousgambitfor the often seen 6 decks in BJ, my usual online calculator refuses to guess how many combinations there are
maybe if you have a powerful computer with a program that runs for days?
https://web2.0calc.com/
link to original post
I think it depends on what you consider a meaningful equivalence.
If you outright ignore suits, and 10 J Q K are all interchangeable, you get a surprisingly low number of "different" shuffled decks.
link to original post
Consider a one-deck blackjack game, there are
51!/[16!x(4!)^9]
different shuffles.
Is this calculation correct?
Quote: acesideQuote: DieterQuote: odiousgambitfor the often seen 6 decks in BJ, my usual online calculator refuses to guess how many combinations there are
maybe if you have a powerful computer with a program that runs for days?
https://web2.0calc.com/
link to original post
I think it depends on what you consider a meaningful equivalence.
If you outright ignore suits, and 10 J Q K are all interchangeable, you get a surprisingly low number of "different" shuffled decks.
link to original post
Consider a one-deck blackjack game, there are
51!/[16!x(4!)^9]
different shuffles.
Is this calculation correct?
link to original post
I'm not a combinatoricist. That said, the parts all seem to make sense.
I get stuck trying to count beyond 99 on my fingers, so someone smarter should verify.
6 * 52 / e = 1.15 * 10^2 = aQuote: odiousgambitfor the often seen 6 decks in BJ, my usual online calculator refuses to guess how many combinations there are
maybe if you have a powerful computer with a program that runs for days?
https://web2.0calc.com/
link to original post
a^312 = 4.75 * 10^18 * 10^624 = 4.75 * 10^642 = b
b * (2pi312)^.5 = 2.10 * 10^644 =~ 312!
That will be even more accurate than the 52! calculation, which was within 17 relative bips of the exact answer, assuming my late-night arithmetic is correct.
Quote: odiousgambitfor the often seen 6 decks in BJ, my usual online calculator refuses to guess how many combinations there are
maybe if you have a powerful computer with a program that runs for days?
https://web2.0calc.com/
link to original post
If I am not mistaken, number of combinations the player's first two cards is: 48516. Player's first two cards + dealer's card: 5013320
pi is in that? amazingQuote: Ace26 * 52 / e = 1.15 * 10^2 = aQuote: odiousgambitfor the often seen 6 decks in BJ, my usual online calculator refuses to guess how many combinations there are
maybe if you have a powerful computer with a program that runs for days?
https://web2.0calc.com/
link to original post
a^312 = 4.75 * 10^18 * 10^624 = 4.75 * 10^642 = b
b * (2pi312)^.5 = 2.10 * 10^644 =~ 312!
That will be even more accurate than the 52! calculation, which was within 17 relative bips of the exact answer, assuming my late-night arithmetic is correct.
link to original post
a google is 10^100 and so the answer has 544 more zeroes, so to speak, than a google if your late night crunching is right
Amendment: I think you’d have to divide by 6!^52 to get the number of distinct ways to shuffle six decks (52 unique cards per deck). So divide 312! by 3.81 * 10^148. Which is still infinity for all practical purposesQuote: Ace26 * 52 / e = 1.15 * 10^2 = aQuote: odiousgambitfor the often seen 6 decks in BJ, my usual online calculator refuses to guess how many combinations there are
maybe if you have a powerful computer with a program that runs for days?
https://web2.0calc.com/
link to original post
a^312 = 4.75 * 10^18 * 10^624 = 4.75 * 10^642 = b
b * (2pi312)^.5 = 2.10 * 10^644 =~ 312!
That will be even more accurate than the 52! calculation, which was within 17 relative bips of the exact answer, assuming my late-night arithmetic is correct.
link to original post
Quote: DobrijQuote: odiousgambitfor the often seen 6 decks in BJ, my usual online calculator refuses to guess how many combinations there are
maybe if you have a powerful computer with a program that runs for days?
https://web2.0calc.com/
link to original post
If I am not mistaken, number of combinations the player's first two cards is: 48516. Player's first two cards + dealer's card: 5013320
link to original post
The player's starting two-card hand number is as small as 55. The player's first two cards + dealer's upcard combination is as small as 550.
Quote: acesideQuote: DobrijQuote: odiousgambitfor the often seen 6 decks in BJ, my usual online calculator refuses to guess how many combinations there are
maybe if you have a powerful computer with a program that runs for days?
https://web2.0calc.com/
link to original post
If I am not mistaken, number of combinations the player's first two cards is: 48516. Player's first two cards + dealer's card: 5013320
link to original post
The player's starting two-card hand number is as small as 54. The player's first two cards + dealer's upcard combination is as small as 702.
link to original post
Why 702 instead of 540?
Do the same equivalencies not apply to the dealer hand?
Quote: acesideYou are right. I revised these numbers.
link to original post
OK... why 550 instead of 540?
(I'll assume tpyo instead of logical error.)
As it's been grinding in my brain, I also don't understand the 51! from earlier. (I'll assume another tpyo. I swear, I'm not trying to bust your chops.)
Quote: MichaelBluejayThe number of distinct results for shuffling a deck of cards is 52!, or 52 x 51 x 50 x 49, etc.
That's about 8.07 x 1067, or 8 with 67 zeroes after it.
If a million people each shuffled a million decks of cards every second from now until the end of the earth (best guess = 8 billion years), that would be only 2.56 x 1029 decks. With all those decks, in all that time, it's unlikely that exactly there would ever be two matching decks of cards.
My favorite way of describing how big 52! is...
If every star in the Milky Way had a trillion planets...
And each planet had a trillion people...
And each person had a trillion decks of cards...
And they were shuffling all of them 1,000 times a second...
And they've been doing that SINCE THE BIG BANG...
...they'd JUST NOW be starting to repeat permutations.
Quote: DieterQuote: acesideYou are right. I revised these numbers.
link to original post
OK... why 550 instead of 540?
(I'll assume tpyo instead of logical error.)
As it's been grinding in my brain, I also don't understand the 51! from earlier. (I'll assume another tpyo. I swear, I'm not trying to bust your chops.)
link to original post
I looked back into a few posts and found some problems. Let me correct them all now.
For a blackjack game, the player's starting two-card hand number is 55. The player's first two cards and dealer's upcard combination number is 550.
If we consider a one-deck blackjack game, there are these many different shuffles:
52!/[16!x(4!)^9] = 8.20752E+13
