# What are the odds of being dealt exactly the same 2 hole cards in consecutive hands?

I mean exactly the same two cards e.g. A♥ and 5♠, not just any A5 off suit.

• number of different hole card combinations times itself, i believe – t123 Jul 30 '13 at 17:53
• Likelihood of one hand, multiplied by itself, (n) times, Where n=consecutive. I believe. Too short to be an answer really. One hand is 1 in 1326 for NLHE – Toby Booth Jul 30 '13 at 21:45
• what is this question good for? – amigal Jul 31 '13 at 21:22
• Sorry, I should have explained. The online poker site I regularly use has added a relatively new feature and a quirk of this is that you're often dealt exactly the same two hole cards consecutively. Just wanted to know exactly how small the odds of this occurring were before I raised hell about it... – Robbie Dee Aug 1 '13 at 9:47

The way to figure this this type of question out is card by card. So the first card in the second hand is 2 in 52 (It can match either of the cards in the previous hand). × 1 in 51 ( it has to match the card that was not matched by the first card and there are 51 choises left). So 2 in 52 = 1 in 26 x 1 in 51 = 1 in 26 x 51 = 1 in 1326

• To get the hand once, yes. Twice in a row it's (1/1326) * (1/1326) – Toby Booth Jul 30 '13 at 21:48
• Great, thanks Toby. Post as an answer and I'll close the question. – Robbie Dee Jul 30 '13 at 23:41
• For "twice the same hand", the first hand doesn't matter as long as the second hand is the same. So the answer is 1/1326 (and not (1/1326)^2). – azimut Jul 31 '13 at 12:37
• Yes. The chances of getting a specific hand twice in a row is 1/(1326^2) but if you don't care what the first hand is it's 1/1326. – msergeant Jul 31 '13 at 14:04

To understand how to calculate these types of questions yourself, here is how you would do it:

Since this hand could be dealt two ways (A♥ and 5♠ or 5♠ and A♥) and be the "same hand" for the purposes of poker, you would need to be dealt either of those two cards on the first hole card that you receive and the other card on the second hole card that you receive.

To calculate the likelihood of being dealt a specific hand once (and in this case, the "specific hand" would be "the same hand that I was dealt last time), the math looks like this:

``` 2 cards 1 card 2 1 ---------------- × -------------------------- = ------ = ------ 52 cards in deck 51 cards remaining in deck 2652 1326 ```

Then, the odds that a particular, predetermined hand will be dealt two hands in a row would be the odds of it being dealt the first time multiplied by the odds of it being dealt on the second hand (the same):

``` 1 1 1 ------ × ------ = ---------- 1326 1326 1,758,276 ```

Also known as: 1,758,276 to 1 (which is the answer to your question).

• The probability of getting a hand on the first deal is 100%; because it's not any particular hand. Then the second hand has to match the first, and it's 1/1326. – dcaswell Aug 24 '13 at 1:30
• @user814064: Correct, but this is to be dealt a particular hand twice in a row. I'll clarify. – lnafziger Aug 24 '13 at 13:11

You only have to square the 1326 if you're going to pick a specific hand to begin with. Otherwise it's 1/1×1/1326.