What Is the Probability of Distributing Remaining Trump Cards in Bridge?

[WMDhamnekar]

North and south have ten trumps between them ( trumps being cards of specified suit).

(a) Find the probability that all three remaining trumps are in the same hand. (that is either east or west has no trumps).

(b) If it is known that king of trumps is included among the three, what is the probability that he is "unguarded" ( that is one player has the king, the other remaining two trumps)?

I didn't understand this question. Would any member of Math help Boards explain me these questions (a) and (b) in details?

Had I understood these question correctly, I would have tried to answer these questions.

How to answer these questions? What are the answers to these questions?

Any math help will be accepted.

 

[I like Serena]

This is the game of bridge.

North and south have won the bidding and one of them is playing the contract. Let's say that south is playing the contract, which is fairly typical in bridge problems.
Then west has played the first card and north has put his cards open on the table.
South must now plan how to play to maximize their score by winning as many rounds of play as possible, and also ensuring a certain minimum, which is the contract.

There are 52 cards in the game and each player starts with13 cards.
It means that south knows his own 13 cards, knows his partner's 13 cards, which are on the table, and knows the 1 card that west played, which is presumably not a trump.
One of the suits is the trump suit, which is key to the game.
Now that the cards of north are on the table, south can see that the two of them have 10 trump cards together.
That leaves the 3 missing trump cards.

Question (a) is about the worst case scenario where all missing trump cards are on 1 hand, which makes it harder for south to control the game.
So the question asks for that probability, which is:
$$P(\text{3 missing trumps are on the same hand})=\frac{\text{# distributions where either east or west has the 3 missing trumps}}{\text{total # distributions of the remaining cards}}$$

Strictly speaking we have to consider the possible distributions of all missing $12+13=25$ cards, but in a case like this we can approximate it by only considering the 3 missing cards and assume that all of their possible distributions are equally likely.
If we do that, we can simplify to:
$$P(\text{3 missing trumps are on the same hand})\approx \frac{\text{# distributions of 3 cards between 2 players with all 3 on one hand}}{\text{# distributions of 3 cards between 2 players}}$$

How many ways are there to distribute 3 cards between 2 players? How many of those have 3 cards in 1 hand?

Bonus question: how many ways to distribute the remaining 25 cards with 12 cards for west and 13 cards for east? In how many of those cases does either west or east have the 3 missing trumps?
Note that in practice it is slightly less likely that the 3 remaining trumps are in 1 hand than the approximation will show.

Question (b) is about the best case scenario if the king is one of the missing trumps, but south does have the ace.
If the king is isolated with either west or east, then south can play the ace and capture the king.
The question asks how likely it is for this plan to succeed.

So:
$$P(\text{king of the 3 missing trumps is isolated in the same hand})\approx \frac{\text{# distributions of 3 cards between 2 players with king isolated in one hand}}{\text{# distributions of 3 cards between 2 players}}$$

And the bonus question if we consider all 25 missing cards.

 

[pbuk]

That's a really good description of bridge @I like Serena, however

1. You need to ignore the card led by West because:
   1. The question does not say that West has led yet; and
   2. The card led by West does not give you any statistical information about the distribution of the three missing trumps, all it tells you is that West was dealt at least one card that was not a trump (or if W leads a trump, at least one card that was a trump).
2. The answer to Question a is easier than you suggest:
   1. Denote the player that is dealt the lowest missing trump as L and the other player as NL.
   2. The probabililty that L was also dealt the next lowest missing trump is the probability that that card is one of L's remaining 12 cards rather than the 13 cards dealt to NL, clearly $\frac {12}{25}$.
   3. If L does hold the two lowest missing trumps, the probabililty that L was also dealt the final missing trump is the probability that that card is one of L's remaining 11 cards rather than the 13 cards dealt to NL, clearly $\frac {11}{24}$.
   4. Combine the probabilities.
3. Similarly for Question b:
   1. Denote the player that is dealt the King as K and the other player as NK.
   2. The probabililty that NK was dealt the lowest missing trump is the probability that that card is one of NK's 13 cards rather than the 12 other cards dealt to K, clearly $\frac {13}{25}$.
   3. If NK does hold the lowest missing trump, the probabililty that NK was also dealt the final missing trump is the probability that that card is one of NK's remaining 12 cards rather than the 12 other cards dealt to NL, clearly $\frac {12}{24}$.
   4. Combine the probabilities.

 

[I like Serena]

That's a really good description of bridge @I like Serena, however

1. You need to ignore the card led by West because:
   1. The question does not say that West has led yet; and
   2. The card led by West does not give you any statistical information about the distribution of the three missing trumps, all it tells you is that West was dealt at least one card that was not a trump (or if W leads a trump, at least one card that was a trump).

Hey @pbuk,

I guess I went a bit overboard with details that are indeed not given as part of the problem, and that even change the problem. I know this particular poster and thought he might appreciate a link to the real world, and a way for his problem to make sense in it.

For the problem to apply to the real world, West will typically have led. And if West would have led a trump, which would be highly unusual in this case, then that would definitely have been mentioned in a problem statement, so it would be reasonable to assume that West did not lead a trump.

If we do take into account that West led and did not play a trump, then that will affect the chances of the possible distributions. It's now less likely that West has the 3 missing trumps (9.6%), since he has "less space" for them, while it is more likely that East has the 3 missing trumps (12.4%). Interestingly, the answer to the question remains the same though.

TL;DR: In the end it didn't matter that I changed the problem and made it more complicated. :)

• The answer to Question a is easier than you suggest:
  1. Denote the player that is dealt the lowest missing trump as L and the other player as NL.
  2. The probabililty that L was also dealt the next lowest missing trump is the probability that that card is one of L's remaining 12 cards rather than the 13 cards dealt to NL, clearly $\frac {12}{25}$.
  3. If L does hold the two lowest missing trumps, the probabililty that L was also dealt the final missing trump is the probability that that card is one of L's remaining 11 cards rather than the 13 cards dealt to NL, clearly $\frac {11}{24}$.
  4. Combine the probabilities.
• Similarly for Question b:
  1. Denote the player that is dealt the King as K and the other player as NK.
  2. The probabililty that NK was dealt the lowest missing trump is the probability that that card is one of NK's 13 cards rather than the 12 other cards dealt to K, clearly $\frac {13}{25}$.
  3. If NK does hold the lowest missing trump, the probabililty that NK was also dealt the final missing trump is the probability that that card is one of NK's remaining 12 cards rather than the 12 other cards dealt to NL, clearly $\frac {12}{24}$.
  4. Combine the probabilities.

Nice! I had indeed not considered this approach.

 

[WMDhamnekar]

My answer to question (a) = $2 \displaystyle\frac{\binom{23}{10}}{\binom{26}{13}}= \displaystyle\frac{11}{50} =0.22$

My answer to question (b) = $\displaystyle\frac{\binom{23}{12}} {\binom{26}{13}} = \displaystyle\frac {13} {50}=0..26$

My answers to (a) and (b) matches with authors answers.

I agree with author William Feller's answers.