danepeterson Joined: Apr 23, 2012
• Posts: 41
April 23rd, 2012 at 11:57:21 AM permalink
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!!!
Dane Peterson
CrystalMath Joined: May 10, 2011
• Posts: 1907
April 23rd, 2012 at 12:23:13 PM permalink
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.
I heart Crystal Math.
dwheatley Joined: Nov 16, 2009
• Posts: 1246
April 23rd, 2012 at 12:47:59 PM permalink
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
Wisdom is the quality that keeps you out of situations where you would otherwise need it
danepeterson Joined: Apr 23, 2012
• Posts: 41
April 24th, 2012 at 9:02:52 AM permalink
Awesome Thanks!!!
Dane Peterson
danepeterson Joined: Apr 23, 2012
• Posts: 41
April 28th, 2012 at 7:42:27 PM permalink
Quote: CrystalMath

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

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!!!
Dane Peterson
buzzpaff Joined: Mar 8, 2011
• Posts: 5328
April 28th, 2012 at 7:49:04 PM permalink
" 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 !
CrystalMath Joined: May 10, 2011
• Posts: 1907
April 28th, 2012 at 8:14:02 PM permalink
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.
I heart Crystal Math.
danepeterson Joined: Apr 23, 2012
• Posts: 41
April 29th, 2012 at 1:54:17 PM permalink
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!
Dane Peterson
CrystalMath Joined: May 10, 2011
• Posts: 1907
April 29th, 2012 at 8:25:06 PM permalink
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.
I heart Crystal Math.
danepeterson Joined: Apr 23, 2012