What are the chances?

[Brucce]

In consecutive hands of receiving exactly the same 8s5c ? What is the probability?

 

[puzzlefish]

The chance of getting the exact same cards twice in a row is 1 in 1326.

 

[Luvepoker]

The chance of getting the exact same cards twice in a row is 1 in 1326.

Just curious, where did you find that answer? I have always wanted to know the odds of that happening.

 

[rhoudini]

The chance of getting the exact same cards twice in a row is 1 in 1326.

Just curious, where did you find that answer? I have always wanted to know the odds of that happening.

Hey Brucce, puzzlefish and Luvepoker, how are you guys doing? Please allow me to join the discussion.
I think that this number is actually bigger, and let me explain why.

1 in 1326 actually is the probability of getting any specific hand (including the suit of the cards), here's why: let P be the chance of getting some hand.
Not considering the order, the chance of getting the first card is 2/52, because two cards are available.
The chance of getting the second card is now 1/51.
Because both events need to occur to satisfy our condition, P is the multiplication of both odds.
P = 2/52 * 1/51 = 1/1326.

Now, to repeat the process for another hand, we need to again multiply the probability of getting a specific hand 2 times in a row.
In this case,
P(of 2 in a row) = 1/1326 * 1/1326 = 1 in 1,758,276

A quite unlikely scenario hehe.
Again: this considers the specific suits. To not consider the suits, we would have to multiply P by the number of possible combinations for a hand (for example, for AA, there are 6 combinations, so the chance of having AA is 6/1326 = 1 in 221; to receive AA in two consecutive hands, multiply 1/221 * 1/221 = 1/48,841).

Hope it makes sense.

 

[Luvepoker]

Hey Brucce, puzzlefish and Luvepoker, how are you guys doing? Please allow me to join the discussion.
I think that this number is actually bigger, and let me explain why.

1 in 1326 actually is the probability of getting any specific hand (including the suit of the cards), here's why: let P be the chance of getting some hand.
Not considering the order, the chance of getting the first card is 2/52, because two cards are available.
The chance of getting the second card is now 1/51.
Because both events need to occur to satisfy our condition, P is the multiplication of both odds.
P = 2/52 * 1/51 = 1/1326.

Now, to repeat the process for another hand, we need to again multiply the probability of getting a specific hand 2 times in a row.
In this case,
P(of 2 in a row) = 1/1326 * 1/1326 = 1 in 1,758,276

A quite unlikely scenario hehe.
Again: this considers the specific suits. To not consider the suits, we would have to multiply P by the number of possible combinations for a hand (for example, for AA, there are 6 combinations, so the chance of having AA is 6/1326 = 1 in 221; to receive AA in two consecutive hands, multiply 1/221 * 1/221 = 1/48,841).

Hope it makes sense.

Glad you join in the discussion. I should have remembered 1326 is the number of possible combinations that could be dealt. I think you answer is correct and it makes sense.

Those are some long odds indeed. Oddly enough you answered something perfectly I had wondered about for years by accident. I was dealt pocket aces twice in a row at the casino years ago and always wondered what the true odds were. Thanks mate.

 

[puzzlefish]

Yes, I made an error and forgot it's twice in a row.

 

[eberetta1]

Online, I would not be surprised. Hard to bluff with such cards.

 

[MK_]

....got 83 of hearts twice in a row today, fun times👍

 

[rastapapolos]

Yes, I made an error and forgot it's twice in a row.

it's 1/1326 * 1/1326 = 1 in 1 758 276

 

[DiegoRamos]

In consecutive hands of receiving exactly the same 8s5c ? What is the probability?

enough to happen lol

 

[Brucce]

Thanks for the information. It really is a difficult and unlikely combination to occur. It happened to me in a tournament on the poker stars App in consecutive hands and I asked about the history of the tournament rounds and noticed that other card combinations were also repeated many times but in different suits. If you want to check it out, can I post that sequence here?

 

[christovam]

Hey Brucce, puzzlefish and Luvepoker, how are you guys doing? Please allow me to join the discussion.
I think that this number is actually bigger, and let me explain why.

1 in 1326 actually is the probability of getting any specific hand (including the suit of the cards), here's why: let P be the chance of getting some hand.
Not considering the order, the chance of getting the first card is 2/52, because two cards are available.
The chance of getting the second card is now 1/51.
Because both events need to occur to satisfy our condition, P is the multiplication of both odds.
P = 2/52 * 1/51 = 1/1326.

Now, to repeat the process for another hand, we need to again multiply the probability of getting a specific hand 2 times in a row.
In this case,
P(of 2 in a row) = 1/1326 * 1/1326 = 1 in 1,758,276

A quite unlikely scenario hehe.
Again: this considers the specific suits. To not consider the suits, we would have to multiply P by the number of possible combinations for a hand (for example, for AA, there are 6 combinations, so the chance of having AA is 6/1326 = 1 in 221; to receive AA in two consecutive hands, multiply 1/221 * 1/221 = 1/48,841).

Hope it makes sense.

I confess that I didn't understand much.

 

[christovam]

This has happened to me with a 4 of hearts and 7 of clubs.