I got the same hand again with the cards in the same position!

Most people think it's the same hand, but are probably mistaken. Well, unlike in a real casino, I can check my history on 5Dimes, and I can tell you that I couldn't believe my eyes. So much so that I took a screenshot.

What are the chances of getting back to back identical hands, with all the cards in the same position? Now, the suits weren't the same (every single card changed suit), but still. I have an idea of the calculation, but the hardest part is finding the probability that all 5 cards land in the same spot too. Let's assume that suit doesn't matter for this calculation.

Perhaps this is a case of the RNG not really being random.

The probabilities of getting one exact five-card hand (or the same hand on the next next deal) is 1 in 2,598,960 with no joker.
For it to be dealt out in the same order, the odds after that fly off the roof, but perhaps the display algorithm sorts the cards on display, I dunno.
It might be a rare bug in the computer code, RNG or otherwise. A transient glitch. WTF.

And this is ignoring suit? So 23456 of diamonds and 23456 of clubs = exact same hand?

So you think I'm onto something here? Do I actually have an example of how the RNG really isn't random? Or something worse? The problem is, until you get to the billions, it's "possible", thus not proof.

So does anybody know how to calculate this exactly? Ignoring suit, what are the chances you get the same value cards in the same slots in back to back hands? And just like he put it above, it's the same as asking, what are my chances of getting 69385 in that exact order?

4/52 * 4/51 * 4/50 * 4/49 * 4/48

Assuming all five cards were different.

=> 1024/311,875,200 or about 1 in 305,000

One result cannot be the sign of a non-random RNG.

I agree with 1 in 304566. To build on this, if you play 100,000 hands of video poker you have about a 28% chance of witnessing this at least once. At up to 1000 hands per hour (a fast but not unattainable speed online), you only have to play 100 hours to have this chance.

200 hours (200,000 hands) gives you a 48% chance of witnessing this.

Wow. The first guy had me thinking I was really onto something. However, you guys didn't calculate the chance of each card being in the same position. How would you calculate that? 1/5 ^ 5, then multiply that by 1 in 305,000? That would be .00000000104918, or about 1 in 10,000,000,000? Ya...

Quote: SilentBob420BMFJ

And this is ignoring suit? So 23456 of diamonds and 23456 of clubs = exact same hand?

So you think I'm onto something here? Do I actually have an example of how the RNG really isn't random? Or something worse? The problem is, until you get to the billions, it's "possible", thus not proof.

So does anybody know how to calculate this exactly? Ignoring suit, what are the chances you get the same value cards in the same slots in back to back hands? And just like he put it above, it's the same as asking, what are my chances of getting 69385 in that exact order?

It's the ignoring of the suit that drastically reduces the probability too.

For 2, 3, 4, 5 & 6 in order (ignoring suit) then, to calculate the probability, you just multiply each fraction obtained by dividing the number of ways an event can occur by the total possible number of outcomes.

So, probability of a 2 as the first card is 4/52.

The probability of a 3 as the next card is 4/51 (as you are assuming a 2 has been drawn as the 1st card)

4 as 3rd card = 4/50, p(5) = 4/49 and p(6) = 4/48

To get the probability of them all occurring then you just multiply the fractions together.

If you had to get the suits correct as well then this would equate to 1/52 X 1/51 X 1/50 X 1/49 X 1/48 or, to put another way, the answer to the 'ignoring suit' multiplied by 4x4x4x4x4 or 4^5.

Quote: SilentBob420BMFJ

Wow. The first guy had me thinking I was really onto something. However, you guys didn't calculate the chance of each card being in the same position. How would you calculate that? 1/5 ^ 5, then multiply that by 1 in 305,000? That would be .00000000104918, or about 1 in 10,000,000,000? Ya...

Nope, I did calculate same rank in each position... assuming five different ranks.

It's just not crazy rare.

So what would the math be if you excluded position? 20/52 * 19/52 ... ?

20/52 * 16/51 * 12 /50 * 8/49 * 4/48

(any of the 5 ranks in first position, any of the 4 remaining ranks in the next position... etc)

= 122,880 / 311,875,200

=> 1 in 2,538 (to the nearest whole number).

Again assuming the cards are all different.

Which means you are more likely to see a repeating hand than a royal flush.

I've done this many times..
gotten the exact same five cards in the exact same position, BACK TO BACK.
I've also gotten a Royal Flush, and on the next hand, Quad Aces...