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I had a situation during a live tournament last night where I made a straight on the turn twice in a row. My starting hands were 7,4 and then 7,3. Both times the flop was 3,4,5 with a 6 on the turn followed by an ace on the river (twice). The table I was on had six players. I doubled up on both hands against the same player. The odds of this occurring must have been similar to that of winning the lottery.

What are the odds of being dealt the same hand twice in a row with the same flop, river and turn, similar to the above situation?

Grinch91
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Geetee211
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  • To give a good answer, you need to be much more specific about **exactly** what you want the odds of. – 3N1GM4 Jan 05 '17 at 15:04
  • I would offer that the odds might be better than you calculate due to the way shuffling in a tournament sometimes occurs which is less than totally random... I have played in a tournament where the exact hand (hole cards, rank and suit) another player showed down with wound up in my hand (hole cards) on the next deal, and I also received the same exact hand twice in a row at least twice. –  Jan 06 '17 at 17:27
  • First off why are you playing 7-4 7-3? You won't win much with those starting hands you got some good look – Sean Mar 23 '21 at 09:59

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first of all, you didn't get the same hole cards, you got 7,4 and 7,3

Given two cards

I think chance that neither card is the same rank would be

44/52 * 43/51 //in both situations we don't want a 7 of aces, hearts, clubs, or spades

The chance that you get one of the cards the same or two cards the same is

1 - (44/52 *43/51) //the same thing as saying prob of not getting none of the same cards

The chance that you get the same two ranked cards is

8/52*4/51*2 //2 because order doesn't matter, unless you know which card the 7 or other you got first, but give me a break if you did

So the chance that you get exactly one of the cards the same is

1 - (44/52 *43/51) - 8/52*4/51*2 This is roughly 1/4, that you were dealt these cards is pretty unexceptional

In both cases one of those flop cards was also a hole card (this means there are only three 3's or three 4's that can give us what we want, it doesn't matter that's it's three one time, and four the next because multiplication is commutative ...)

So the chances of getting this flop are

(3/50*4/49*4*/48) //if order didn't matter and it was just this set of cards you would multiply by 3choose3

The chances of getting the six are

4/47

The chances of getting an ace on the river are

4/46

We multiply all these odds together to get the odds of this specific board by the river, and we can simplify the expression to

(3*4^4)/(50*49*48*47*46)

This equals roughly 3/1,000,000, so if you square that (because the board ended up this way twice) that's 9/1,000,000,000,000, roughly one in a billion, seems too low a probability

Then again, if someone were to calculate the probability that you'd see something like this before you ever played a single poker game that you'd see something like this, it would be higher.

The difference between this and winning the lottery is that there are only a few lotteries, and it seems like there probably would have been a number of different board hole card combinations that if replicated, would have probably seemed amazing.

Here's a generalization of what you saw: the chance that you get the same different ranks (not the same cards) in the same order on the board is 4^5/(52*51*50*49*48, or 3/1,000,000

If you've seen 100,000 hands let's say, the chance you were gonna see the condition up above at least once is

1 - (1 - 3/1,000,000)^100000

Or 1/4, crazy right

Basically I think how exceptional this is depends on what about it was exceptional for you.

user3163829
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    I think it's inappropriate to square the result. If you were wondering what the odds were for this specific hand to happen twice in a row, then you might do this. But the OP didn't care what the first hand was. He just cared that the second "matched" the first (even though it really didn't). – Chris Farmer Jun 22 '15 at 20:40
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    I'm also 100% certain that a reasonable definition of "same" in this case is many times more likely than what is calculated here. This is one of those many "what were the odds of that?" questions that really doesn't have a meaningful answer. – Lee Daniel Crocker Jun 22 '15 at 21:44
  • This still seems a pretty exceptional coincidence even if you account for all the situations that are "equally rare" or "rarer". It's one in a million at least. – Yang Jun 22 '15 at 23:07