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So the situation was the following yesterday: 3 players in game:

I have two kings in my hand, and with the flop I receive a quad of kings, player 2 folds, player 3 goes all in - I call. He gets the royal flush hand only with the river.

Since we're playing with friends and not money, we give a few chips to the player who was eliminated so he can try again.

The next round the same player had another royal flush with the flop already.

What probability is there to make such a thing (considering only that the first round, there was a royal flush with the river, and all four quads with; and that the second round had a royal flush with the flop).

Here's the log (without player 2 who folded both rounds):

Round 1

Player 1: A♠Q♠

Player ME: 9♡K♣

Flop till River: J♠K♡T♠K♢K♠

Round 2

Player 1: A♣Q♣

Player ME: T♡8♡

Flop till River: J♣K♣T♣K♠Q♠

I know this may seem unbelievable and hoaxed, since the player 1 had the same card values, and the flop and the turn were the same valued-card.

If looking at ALL the coincidences (and just to make sure, it was his first royal flush and his second in his entire life; we're playing still very often at school) how big would be the probability of such a rare happening?

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You can look up how to work out the odds in detail for a royal flush, I'll just give you a really quick breakdown, but they basically work out as 1 : 649,739 or 0.000154%.

Four suits, but only one combination works and total possible hands in poker is 2,598,960 (52 C 5). So 4 in 2,598,960 or 1 in 649,739.

When wanting to work out the probability of two events happening one after the other it's just multiple the two results together. So 0.000154% x 0.000154% = 0.000000023716%. Or 1 in 422,160,768,121

If you want to use all 7 cards it's easy enough to do as well. Similar idea as above, take all possible hands, which for 7 cards is 133,784,560 (52 C 7).

When using 7 cards to calculate we must figure out how the 6th and 7th card end up changing the probability. To do this we need to simple factor in the next two cards or (47 C 2) which gives you 1081. Given again we know there are 4 suits that means we multiple the above by 4 to give 4,324 total possible ways the last two cards can give a royal flush. Once we plug in these numbers we can get the percentage similarly like above.

4,324 in 133,784,560 or 1 in 30,940 which gives 0.00323%

For this event to occur twice in a row, like above would be 0.00323% x 0.00323% = 0.0000104329% or 1 in 957,283,600. So better than above where 5 cards are used.

So basically in both cases of 5 cards and 7 cards a really damn long shot.

  • Sure let me edit to include to use all 7. I didn't as typically speaking people for hold'em only use 5 cards to do probability for their hand. – Grinch91 Sep 27 at 10:43
  • Updated to include the maths for 7 cards used. – Grinch91 Sep 27 at 10:56
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well, im not an expert in probability or statistics but if the odds of getting a royal flush in a 7-cards game like Texas holdem (where you make the best 5-card hand out of 7 total cards) is 1:30,000 (roughly). then i would say its 30,000 X 30,000 to get 2 consecutive royal flushes. so 1:1,000,000 hands (roughly).

so yeah, extremely unlikely. i would say it is way more probable that one of your friends can fix cards...

  • Your odds are too good. – Grinch91 Sep 23 at 13:09

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