This is to work out the probability of an event occurring in a bridge hand.
Aces are worth 4 points, kings 3, queens 2, and jacks 1. I can establish that my partner and I hold 33 points between us, and therefore the opponents have 7. I want to calculate the probability that their 7 points are made up of the ace and king of the same suit. It can be any of the suits.
In case clarification is needed, we hold 26 cards and the opponents hold 26.
Can some of you math whizzes help please?
Are you assuming that you’re not able to look at your own cards? For example, if you can see that you’re holding all four kings, then the probability that the opponents hold AK is zero.
–Mark
Good point. I would know my own cards but not my partner’s. I would not need to know the probability in any situation where I held all four aces or all four kings, or even either an ace or a king in all four suits.
As a first pass, we could ignore those conditions and just calculate the raw probability of AK irrespective of anything else.
I’m getting:
920,662,591,680 possible 26-card hands that have seven points
Of which 5,006,710,800 are AK suited
So 5,006,710,800 / 920,662,591,680 = about a 0.054 chance, or a little over one in 200.
Wouldn’t you have asked for any aces you did not hold, and then if you got the right answer, asked for kings enroute to bidding your slam?
I think you made a typo – 5,006,710,800 / 920,662,591,680 is 0.0054, not 0.054, but it is indeed about 1 in 200.
However I used a simpler approach and I get a different number. I’m ignoring all non-face cards, and I find that there are 284 ways to make a 7 with the face cards. Of those, 4 are AK of the same suit, so I get 4/284 = 0.0141 = 1.41%, which is a little less than 3 in 200.
–Mark
4/284 is the answer I first came up with, too. Then I thought, are all 284 equally likely? I don’t think they are.
For example, if the seven points come from AK, then we must choose 24 cards from the 36 non-scoring cards, to make up the rest of the 26 cards. That’s (36 choose 24).
If, on the other hand, the seven points come from KQJJ, it’s (36 choose 22) for the non-scoring cards, which is more than (36 choose 24).
Ideally, yes. However, the most recent hand in question is one where I would have raised partner’s 15-17 1NT direct to 6NT, but our opponent only bid 4NT on the grounds that they could be missing a cashing AK. His hand was AK xx KQxx AQT9x. His partner passed with 15 points (QJxx AJT Axxx Kx) and they made all 13 tricks. My view is that because of a small chance of losing the first 2 tricks, he gave up a large chance of making slam. So I was interested in being able to quantify “small chance”.
And asking for aces and kings would not tell you if the missing ones were in the same suit.
Well first off, if they open 15-17 NT, then the minimum they have is 33 points. They might have 34 or 35. So bidding should not be based on assuming only 33 points. 15 point hands are 44% of all 15-17 point hands. This conditional probability will be higher, I think, when you hold 18 points, but still I’d think you could conclude there’s at least a 50% chance you have more than 33 points.
Secondly, in this case a Gerber 4C would have determined that no aces were missing if you wished to be careful. Error completely to 18 point hand holder. 15 point hand holder exonerated. His pass was proper given his minimum and no 5 card suit.
But if they do, opener would bid 6NT - definitely with 17, likely with 16.
Responder’s hand evaluates at more like 20 points because of the good club suit, but I did not really want to turn this into a bridge thread, I was interested in how likely/unlikely it was that there are two top losers. I used a hand dealer and zero AK came up out of 64 deals, but I was also interested in the math.
True, but even if there is an ace missing, I still want to be in 6NT.
I completely agree.
Let’s start with a simpler problem. Select 26 cards at random. If the number of HCPs in the 26 cards is exactly 7, what is the chance they include Ace and King of the same suit?
Start by enumerating all the ways to get seven HCP. There aren’t that many. Let’s break them down according to whether there are 2, 3, 4 or 5 honor cards. (The indented lines are the lines I input to the bc calculater.)
A,K
p = c(36,24) * 16
A,Q,J or K,K,J or K,Q,Q
q = c(36,23) * (64+24+24)
A,J,J,J or K,Q,J,J or Q,Q,Q,J
r = c(36,22) * (16 + 96 + 16)
K,J,J,J,J or Q,Q,J,J,J
s = c(36,21) * (4 + 24)
p counts the number of ways to have 24 non-honors (there are 36 non-honors altogether) multiplied by the number of ways to have an Ace and a King (4 Aces, 4 Kings yields 4x4). Similarly, q is the number of ways to have 23 non-honors and either A,Q,J (444 ways) or K,K,J (6 ways to select a pair of Kings times 4 Jacks) or K,Q,Q (also 6*4). After computing these numbers, calculate
p / (p + q + r + s) / 4
(We divide by 4 to change “probability of A and K” to “probability of A and K in same suit.”)
This gives .00543816 if I’ve made no mistake, or about 0.5%.
This assumes no other information. (Just the fact that your hands are balanced changes the availability of non-honors.)