Define a perfect hand as a hand that...

1) ...has all four suits, and

2) ...has only the top cards in every suit (AK, AKQJ, etc)

or...

3) ...does not have all four suits,

4) ...has only the top cards in the suits you have, and

5) ...has at least seven of one suit.

Thanks!

For the first part, the possible suit breakdowns, and the number of hands for each, are:

10, 1, 1, 1 (4)

9, 2, 1, 1 (12)

8, 3, 1, 1 (12)

8, 2, 2, 1 (12)

7, 4, 1, 1 (12)

7, 3, 2, 1 (24)

7, 2, 2, 2 (4)

6, 5, 1, 1 (12)

6, 4, 2, 1 (24)

6, 3, 3, 1 (12)

6, 3, 2, 2 (12)

5, 5, 2, 1 (12)

5, 4, 3, 1 (24)

5, 4, 2, 2 (12)

5, 3, 3, 2 (12)

4, 3, 3, 3 (4)

4, 4, 3, 2 (12)

4, 4, 4, 1 (4)

For the second part:

13 (4)

12 1 (12)

11 2 (12)

11 1 1 (12)

10 3 (12)

10 2 1 (24)

9 4 (12)

9 3 1 (24)

9 2 2 (12)

8 5 (12)

8 4 1 (24)

8 3 2 (24)

7 6 (12)

7 5 1 (24)

7 4 2 (24)

7 3 3 (12)

That is 476 hands, out of a possible 6,350,013,559,600; the probability is 119 / 1,587,503,389,900, or about 1 in 13,340,364,621.

Check your grand total for an extra 0… There are “only” 635,013,559,600 different 13-card hands, so odds are 1 perfect bridge hand in every 1,334,062,100 hands.

Notice that the first type creates fully dominant perfect hands. In other words, having all 4 Aces AND a perfect hand precludes any other perfect hand from existence on the same deal.

What are the odds of 2, 3, or 4 players each having perfect hands on the same deal? (HINT: These must be of the second type.)

Example…. A spades. A Hearts. A Diamonds K Diamonds Q Diamonds

A, K, Q, J, 10, 9, 3, 2 Clubs is actually a perfect hand, in the sense that correct (easy) play makes the highest possible bid every time.

Quote:SOOPOOFWIW….. There are more ‘perfect’ bridge hands, if you would define a ‘perfect’ hand as one that will ALWAYS be able to take all 13 tricks at a bid of 7 NT. If you have all the aces, then you can still have a true ‘perfect’ hand as long as you have the top 6 in a suit, you can have any of the other cards in that suit.

Example…. A spades. A Hearts. A Diamonds K Diamonds Q Diamonds

A, K, Q, J, 10, 9, 3, 2 Clubs is actually a perfect hand, in the sense that correct (easy) play makes the highest possible bid every time.

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Yes, I see why. In your example above, you have 8 Clubs, and even if the remaining 5 clubs were all held by only one of your opponents (worst case), your hand still dominates. Still works if you drop the 3C from your hand, and give it to the same opponent.

In general, your A through 9 of Clubs exhausts the remaining six or fewer cards, so any rank for your 7th through 10th will be unchallenged.

C(7,6) = C(7,1) = 7

C(7,5) = C(7,2) = 21

C(7,4) = C(7,3) = 35

To further SOOPOO’s point, I assert that 4 Aces aren’t necessary for this to hold. Any hand where one player holds 7 or more cards AND has Ace thru 9 of the same suit will suffice, no matter how many suits in hand.

Quote:ThatDonGuyI think I counted all of them:

For the first part, the possible suit breakdowns, and the number of hands for each, are:

10, 1, 1, 1 (4) x 35 = 140

9, 2, 1, 1 (12) x 35 = 420

8, 3, 1, 1 (12) x 21 = 252

8, 2, 2, 1 (12) x 21 = 252

7, 4, 1, 1 (12) x 7 = 84

7, 3, 2, 1 (24) x 7 = 168

7, 2, 2, 2 (4) x 7 = 28

6, 5, 1, 1 (12)

6, 4, 2, 1 (24)

6, 3, 3, 1 (12)

6, 3, 2, 2 (12)

5, 5, 2, 1 (12)

5, 4, 3, 1 (24)

5, 4, 2, 2 (12)

5, 3, 3, 2 (12)

4, 3, 3, 3 (4)

4, 4, 3, 2 (12)

4, 4, 4, 1 (4)

For the second part:

13 (4)

12 1 (12) x 7 = 84

11 2 (12) x 21 = 252

11 1 1 (12) x 21 = 252

10 3 (12) x 35 = 420

10 2 1 (24) x 35 = 840

9 4 (12) x 35 = 420

9 3 1 (24) x 35 = 840

9 2 2 (12) x 35 = 420

8 5 (12) x 21 = 252

8 4 1 (24) x 21 = 504

8 3 2 (24) x 21 = 504

7 6 (12) x 7 = 84

7 5 1 (24) x 7 = 168

7 4 2 (24) x 7 = 168

7 3 3 (12) x 7 = 84

That is 6,780 hands, out of a possible 635,013,559,600; the probability of 1 or more players being dealt a perfect hand is 1 in approximately 93.7 million.

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Four perfect hands: An event never seen before (right?)

FWIW, I agree with the odds calculation in the article for 4 perfect hands.

Quote:gordonm888Without 4 aces, there is always the chance that you will be out bid (at the 7-level) and then have the other team keep leading the suit you don't have.

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There’s also this! Thanks, gordon. Experienced players could make it happen without the perfect hand as well… Never played myself. Parents were in a contract bridge club for years when I was a kid. I only know how it is similar to spades…

Quote:camaplStanding on TDG’s shoulders (thanks, Don!) See adjustments in bold. Note that C(x,y) is combinatorial notation for choosing y items from a list of x length, or the COMBIN() function.

C(7,6) = C(7,1) = 7

C(7,5) = C(7,2) = 21

C(7,4) = C(7,3) = 35

To further SOOPOO’s point, I assert that 4 Aces aren’t necessary for this to hold. Any hand where one player holds 7 or more cards AND has Ace thru 9 of the same suit will suffice, no matter how many suits in hand.Quote:ThatDonGuyI think I counted all of them:

For the first part, the possible suit breakdowns, and the number of hands for each, are:

10, 1, 1, 1 (4) x 35 = 140

9, 2, 1, 1 (12) x 35 = 420

8, 3, 1, 1 (12) x 21 = 252

8, 2, 2, 1 (12) x 21 = 252

7, 4, 1, 1 (12) x 7 = 84

7, 3, 2, 1 (24) x 7 = 168

7, 2, 2, 2 (4) x 7 = 28

6, 5, 1, 1 (12)

6, 4, 2, 1 (24)

6, 3, 3, 1 (12)

6, 3, 2, 2 (12)

5, 5, 2, 1 (12)

5, 4, 3, 1 (24)

5, 4, 2, 2 (12)

5, 3, 3, 2 (12)

4, 3, 3, 3 (4)

4, 4, 3, 2 (12)

4, 4, 4, 1 (4)

For the second part:

13 (4)

12 1 (12) x 7 = 84

11 2 (12) x 21 = 252

11 1 1 (12) x 21 = 252

10 3 (12) x 35 = 420

10 2 1 (24) x 35 = 840

9 4 (12) x 35 = 420

9 3 1 (24) x 35 = 840

9 2 2 (12) x 35 = 420

8 5 (12) x 21 = 252

8 4 1 (24) x 21 = 504

8 3 2 (24) x 21 = 504

7 6 (12) x 7 = 84

7 5 1 (24) x 7 = 168

7 4 2 (24) x 7 = 168

7 3 3 (12) x 7 = 84

That is 6,780 hands, out of a possible 635,013,559,600; the probability of 1 or more players being dealt a perfect hand is 1 in approximately 93.7 million.

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Where do you get "x7" from in the {12 1} hands?

I have a feeling you interpreted the problem differently than I did.

A 10-1-1-1 hand must have A, K, Q, J, 10, 9, 8, 7, 6, 5 of one suit, and the other three Aces. It looks like you're treating it as if, as long as you have Ace through 8 of a suit, the other cards in that suit can be anything.

EDIT: To further answer your question, the “7 choose x” comes from choosing however many ranks needed (in addition to the top 6) out of the 7 remaining (2 thru 8).

Quote:ThatDonGuyQuote:camaplStanding on TDG’s shoulders (thanks, Don!) See adjustments in bold. Note that C(x,y) is combinatorial notation for choosing y items from a list of x length, or the COMBIN() function.

C(7,6) = C(7,1) = 7

C(7,5) = C(7,2) = 21

C(7,4) = C(7,3) = 35

To further SOOPOO’s point, I assert that 4 Aces aren’t necessary for this to hold. Any hand where one player holds 7 or more cards AND has Ace thru 9 of the same suit will suffice, no matter how many suits in hand.Quote:ThatDonGuyI think I counted all of them:

For the first part, the possible suit breakdowns, and the number of hands for each, are:

10, 1, 1, 1 (4) x 35 = 140

9, 2, 1, 1 (12) x 35 = 420

8, 3, 1, 1 (12) x 21 = 252

8, 2, 2, 1 (12) x 21 = 252

7, 4, 1, 1 (12) x 7 = 84

7, 3, 2, 1 (24) x 7 = 168

7, 2, 2, 2 (4) x 7 = 28

6, 5, 1, 1 (12)

6, 4, 2, 1 (24)

6, 3, 3, 1 (12)

6, 3, 2, 2 (12)

5, 5, 2, 1 (12)

5, 4, 3, 1 (24)

5, 4, 2, 2 (12)

5, 3, 3, 2 (12)

4, 3, 3, 3 (4)

4, 4, 3, 2 (12)

4, 4, 4, 1 (4)

For the second part:

13 (4)

12 1 (12) x 7 = 84

11 2 (12) x 21 = 252

11 1 1 (12) x 21 = 252

10 3 (12) x 35 = 420

10 2 1 (24) x 35 = 840

9 4 (12) x 35 = 420

9 3 1 (24) x 35 = 840

9 2 2 (12) x 35 = 420

8 5 (12) x 21 = 252

8 4 1 (24) x 21 = 504

8 3 2 (24) x 21 = 504

7 6 (12) x 7 = 84

7 5 1 (24) x 7 = 168

7 4 2 (24) x 7 = 168

7 3 3 (12) x 7 = 84

That is 6,780 hands, out of a possible 635,013,559,600; the probability of 1 or more players being dealt a perfect hand is 1 in approximately 93.7 million.

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Where do you get "x7" from in the {12 1} hands?

I have a feeling you interpreted the problem differently than I did.

A 10-1-1-1 hand must have A, K, Q, J, 10, 9, 8, 7, 6, 5 of one suit, and the other three Aces. It looks like you're treating it as if, as long as you have Ace through 8 of a suit, the other cards in that suit can be anything.

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A perfect hand in Bridge is one that is guaranteed to win all 13 tricks, no matter how the other 39 cards are distributed amongst your opponents and partner. With a 10-1-1-1 hand you don't need the 10 highest cards in a suit, because once you have led the 3 highest cards (AKQ) your opponents will have no more cards of that suit and thus the other 7 cards of that suit will simply win on length when they are led (presumably this is your trump suit or otherwise that you are in 7NT.)

For perfect hands of a 10-1-1-1 distribution the number of combinations is c(10,7): A,K,Q of a suit plus any 7 of the 10 ranks 2-J of the suit.

For a 12-1 distribution of suits. all you need for a perfect hand is an Ace in both of your two suits, assuming your 12 card suit is declared as the trump suit. Thus for 12 -1 hands, the number of combinations is c(11,10).

Quote:camaplSo, even higher than my figures… Hadn’t even occurred to me that the more cards you have in your biggest suit, the fewer “royals” you need. Perhaps I should stick to the VP strategies I need to massage! lol

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Nah. You're a good analyst/mathematician and I make a point to read all of your posts - because they're interesting. Please, keep being curious.