Perfect Pairs Math

Most websites show that on a 6 deck blackjack game the odds of getting a perfect pair is 0.016077 or 1.6077%
That makes sense because you have 5 remaining cards that can match your card for a perfect pair out of the 311 remaining cards in the shoe.

5/311 = 0.016077 or 1.6077%

The odds of getting a blackjack in the same 6 deck game is 0.04749 or 4.4749%
That makes sense cause you take the 24 possible aces and divide them by the total 312 cards. Multiply that by the 96 possible ten value cards divided by the 311 remaining cards in the shoe. Then multiply that by 2 because you can get either the ace first or the tan value first. 2 possible outcomes.

(24/312) * (96/311) * 2 = 0.04749 or 4.4749%

Now if this is all true? Then why wouldn't the odds of getting a perfect pair be 0.0006183 or 0.06183%???
Applying the math from the odds of getting a blackjack to getting a perfect pair. You take the number of same suited cards in a six deck game. 6 divided by the total 312 cards. Multiplied by the 5 remaining matching cards divided by the remaining 311 cards in the shoe. Multiplied by 2 because of the two different outcomes.

(6/312) * (5/311) * 2 = 0.0006183 or 0.06183%

What is the correct way to calculate the odds of getting a perfect pair?

Also if I were able to make 1,000,000 perfect pair bets, how would I calculate the average amount of times it would come up?

Thanks!!!

The first way is correct (5/311).

To get the expected number of wins in 1 million bets, multiply 5/311 by 1,000,000 = 16077.

The 2nd way does work for Perfect Pairs, but it's not as straightforward. The blackjack calculation is essentially:

Ways to get an Ace * ways to get a 10 + ways to get a 10 * ways to get an Ace.

You can shorten the calculation by just doing one of them and doubling it. But for perfect pairs, any card can be paired. So it's:

Ways to get the As * ways to get another one + ways to get Ac * ways to get another one + ... 52 times in total.

this is: 1/52 * 5/311 * 52

which is 5/311

Awesome Thanks!!!

Quote: CrystalMath

To get the expected number of wins in 1 million bets, multiply 5/311 by 1,000,000 = 16077.

Thanks for the reply!

With that math can I assume that in the long run I should see a perfect pair roughly about every 62.2 dealt hands?

(1,000,000/16077) = 62.2

If this is true, say i were at a full table with seven players total, would that mean that i'm expected to see a perfect pair within 62.2 of my own dealt hands in front of me or within 62.2 dealt hands on the table?

Also say I were dealt 1,000 hands and within that 1,000 hands not one perfect pair came up. What are the odds of this situation occurring and how would i calculate it?
Am I x times more likely to receive one after the 1,000 no-shows of perfect pairs?

I ask this because I once read in a roulette strategy book that if a certain number hasn't come up in a certain amount of spins you are x times more likely to see it come up.
They gave the example of if you were to bet continuously on the number "5" on a single 0 wheel you have a 1/37 chance of it coming up. If you haven't seen the number "5" come up within say 111 spins you are 3x more likely to see it come up the next spin.

(37*3) = 111

I also read online that this is something called the gamblers fallacy. Where you expect something to happen because it hasn't yet. So I assume the book was wrong and no matter what I still have 1/37 chance on the next spin. Or with perfect pairs still a 5/311 chance on the next deal no matter what has occurred those first 1,000 hands.

Thanks!!!

" I ask this because I once read in a roulette strategy book that if a certain number hasn't come up in a certain amount of spins you are x times more likely to see it come up. "

BURN THAT BOOK !

Buzz is right, burn the book.

You would expect to see it every 62.2 hands on average. This means that you will personally get it every 62.2 hands, but at a table of 7, it will happen much more frequently, roughly every 8.8 games.

The chance that you won't see it in 1000 hands is (306/311)^1000 = 0.00000009142, or about 1 in 11 million. After witnessing this, you are no more likely to see it.

Quote: CrystalMath

The chance that you won't see it in 1000 hands is (306/311)^1000 = 0.00000009142, or about 1 in 11 million. After witnessing this, you are no more likely to see it.

Awesome!!!

Where did you get the number 306 from? And can you explain how you got (306/311)^1000. Sorry I'm not that good with math.

Thanks!

The chance you get a perfect pair on any single hand is 5/311, so the chance you don't is 1 - 5/311 = 306/311.

Perfect! thanks!

Ok so I understand that with every individual dealt hand I will still have a 5/311 chance of getting a perfect pair no matter what previously happened.

The smaller the sample size observed, the more variation there is? Correct? I expect to average 1 every 62.2 hands. If i made 63 bets I could easily not see one within that 63. Or easily get multiple within the 63 hands. The closer I approach infinite bets, the closer I get to an actual 5/311 outcome? Correct?

Lets say in a sample size of 10,000 hands I received 135 perfect pairs. 10,000 * (5/311) = 160.7717. I'm short 25 perfect pairs of my expected wins. But if I were able to approach infinite hands, at some point I would have to get an excess of 25 wins to make the (5/311) average work. So couldn't you say that individually each time I receive my first card, I will have exactly a 5/311 chance of getting a perfect pair with the second card. Each dealt hand is independent of the last. But in the infinite deals I'm expected to win, at some points in the future, more than whats expected in order to be at 5/311.

I'm also curious that as cards are being dealt that 5/311 changes up or down very slightly within each shoe. For instance you notice that a 4 of clubs came out earlier in the shoe and you now have a four of clubs in front of you and you are waiting to be given the next card as the dealer is dealing. 49 of the total cards have been dealt thus far. You now have a 4/262 chance of getting another 4 of clubs.

5/311 = 1.6077% chance is now 4/262 = 1.5267% chance. I now have a slightly smaller chance of getting a perfect pair at that current point in the shoe. With every card pulled the odds change slightly.

I assume that it will equal out at the end back to 1.6077%

Thanks.

theoretical and actual results are rarely the same.
just because you have a 'deficit' from the expected average,
does not mean you will have a 'surplus' in the future.
as you pointed out, you should consider the probability on each hand as an individual number.
the cards have no memory that you are due an increased probability.
you are correct that the probability changes as the composition of the shoe changes.

Quote: danepeterson

Ok so I understand that with every individual dealt hand I will still have a 5/311 chance of getting a perfect pair no matter what previously happened.

The smaller the sample size observed, the more variation there is? Correct? I expect to average 1 every 62.2 hands. If i made 63 bets I could easily not see one within that 63. Or easily get multiple within the 63 hands. The closer I approach infinite bets, the closer I get to an actual 5/311 outcome? Correct?

Yes

Quote: danepeterson

Lets say in a sample size of 10,000 hands I received 135 perfect pairs. 10,000 * (5/311) = 160.7717. I'm short 25 perfect pairs of my expected wins. But if I were able to approach infinite hands, at some point I would have to get an excess of 25 wins to make the (5/311) average work. So couldn't you say that individually each time I receive my first card, I will have exactly a 5/311 chance of getting a perfect pair with the second card. Each dealt hand is independent of the last. But in the infinite deals I'm expected to win, at some points in the future, more than whats expected in order to be at 5/311.

Not exactly. There is no need for you to ever make up the excess loss of 25 wins. Using your example, you hit 135/10,000. Now, lets ad 311 billion plays and say that you got exactly 5 billion wins out of that. Now, your win ratio is 5,000,000,135/311,000,010,000. So, although you neve made up those 35 losses, in the long run, you can still approach the expected outcome.

Quote: danepeterson

I'm also curious that as cards are being dealt that 5/311 changes up or down very slightly within each shoe. For instance you notice that a 4 of clubs came out earlier in the shoe and you now have a four of clubs in front of you and you are waiting to be given the next card as the dealer is dealing. 49 of the total cards have been dealt thus far. You now have a 4/262 chance of getting another 4 of clubs.

Yes, it would change based on the number of pairs already removed from the shoe and the number of cards remaining. You could theoretically deal 1/6 of the shoe with no card repeating. At this point, you would have 5 decks remaining, and your chance of a perfect pair would be 4/259, which is a little worse than 5/311.

Quote: CrystalMath

Not exactly. There is no need for you to ever make up the excess loss of 25 wins. Using your example, you hit 135/10,000. Now, lets ad 311 billion plays and say that you got exactly 5 billion wins out of that. Now, your win ratio is 5,000,000,135/311,000,010,000. So, although you neve made up those 35 losses, in the long run, you can still approach the expected outcome.

AWESOME THANKS!!!!!!!! You really cleared things up for me!