[CSGibson]

During dinner we had a 15 minute argument about this scenario:

You hold Jxxx opposite AQ98 in a suit. the opponent in front of the AQ98 is known to have 3 to the king. What is the probability that they also have the T?

[JLOGIC]

50 % lol. Either player is equally likely to have Tx of the remaining non K cards.

Edit: The more formal way of thinking about it if you want to prove it to your friend is say that the missing cards are KT432. Opponent onside can have:

KT4, KT3, KT2, vs K43, K42, K32. So there are an equal amount of combos where the guy has the T or he doesn't. You don't need to know math at all to get these problems or basically any bridge problems right, it's a myth.

[JLOGIC]

Actually I think I somehow rather embarassingly messed this or some variation of this up in some thread before and said the "long" hand was more likely to have the ten so I shouldn't say lol, but the answer is pretty obvious.

[JLOGIC]

Give this one to the person who argued about this incorrectly for 15 mins:

What are the odds of making a contract that is on 2 out of 3 finesses?

Spoiler

50 % ZOMG!!!!!!!!!

might blow their mind lol.

[CSGibson]

ok, that's what I kept on arguing, but two people tried to convince me it was wrong, and they used mathy things which sounded convincing.

[han]

Hmm, is it very bad if I like JLOGICs first argument much better than his second argument?

[Mbodell]

It might depend on how you know they have 3 and how you know they have the K (I.e., there might be a little more information that you learned with that). But yeah, this is a 50% thing if you just know magically what the OP says and nothing else.

[Vampyr]

JLOGIC, on 2013-February-23, 02:06, said:

Give this one to the person who argued about this incorrectly for 15 mins:

What are the odds of making a contract that is on 2 out of 3 finesses?

Spoiler

50 % ZOMG!!!!!!!!!

might blow their mind lol.

How about 1 out of 2 finesses?

Spoiler

75%

[jogs]

CSGibson, on 2013-February-23, 01:32, said:

During dinner we had a 15 minute argument about this scenario:

You hold Jxxx opposite AQ98 in a suit. the opponent in front of the AQ98 is known to have 3 to the king. What is the probability that they also have the T?

For your question post #2 by JLOGIC is correct(50%).

If you had asked LHO known to have the king with no stipulation of number of trumps, then RHO would have 13 slots for the ten and LHO would have 12 slots. LHO would have the ten 12/25ths of the time.

[inquiry]

I hate to disagree with justin (you other guys, not so much) , so the odds of me being right is well below 50%, but it is my understanding that the odds of West having the the heart 10 is only 40%, and the odds of EAST holding the ten is 60%.

Now how did I rationalize such a number when, as justin points out, the possible combinations of the suits are three KTx and three with Kxx? The answer is that East's heart play would not restricted.

So with the first key card (the King) know to be in West hand, West has two places that can be the Ten, to East's three places that can be the ten. So lets do a mental or physical exercise. Take the KT432 out of a deck. Give "west" the King, and randomly deal the other five cards, two to West and three to East. Do this 100 Times (or better yet, do this mathematically thousands of times). You will find that, on average, the ten will end up with east 60% of the time. The reason being if you "know" West has the king (say from the bidding) and three cards (say from count in other suits), the vacant space in his heart suit iis "two" to "three" in East's hand.

There are 10, equally likely combinations which are:
K32 T54
K42 T53
K43 T52
K52 T43
K53 T42
K54 T32
KT2 543
KT3 542
KT4 532
KT5 432

As you can see (bolded Blue) there are six cases where East has the ten, to only four carse where west has the Ten. I wait for someone to prove this "logic" wrong.

[fromageGB]

inquiry, on 2013-February-23, 08:49, said:

I wait for someone to prove this "logic" wrong.

See if this counts.
"West has two places that can be the Ten, to East's three places that can be the ten."
So W has 3 cards, E has 3 cards, S has 4 cards, N has 4 cards. 3+3+4+4=14.
Number of cards in suit = 13
(At least there are here in UK. I have never played in USA)

[FM75]

inquiry, on 2013-February-23, 08:49, said:

I hate to disagree with justin (you other guys, not so much) , so the odds of me being right is well below 50%, but it is my understanding that the odds of West having the the heart 10 is only 40%, and the odds of EAST holding the ten is 60%.

Now how did I rationalize such a number when, as justin points out, the possible combinations of the suits are three KTx and three with Kxx? The answer is that East's heart play would not restricted.

So with the first key card (the King) know to be in West hand, West has two places that can be the Ten, to East's three places that can be the ten. So lets do a mental or physical exercise. Take the KT432 out of a deck. Give "west" the King, and randomly deal the other five cards, two to West and three to East. Do this 100 Times (or better yet, do this mathematically thousands of times). You will find that, on average, the ten will end up with east 60% of the time. The reason being if you "know" West has the king (say from the bidding) and three cards (say from count in other suits), the vacant space in his heart suit iis "two" to "three" in East's hand.

There are 10, equally likely combinations which are:
K32 T54
K42 T53
K43 T52
K52 T43
K53 T42
K54 T32
KT2 543
KT3 542
KT4 532
KT5 432

As you can see (bolded Blue) there are six cases where East has the ten, to only four carse where west has the Ten. I wait for someone to prove this "logic" wrong.

Ben this would only be right if the suit had 14 cards.

[fromageGB]

I don't like jlogic's way of deducing this, though, it is impossible for me at the table to work out what all the combinations are and how many of then include the T. I just say "apart from K, there are 4 other cards. W has 2, E has 2 (using inquiry's compass), so the probability of any card in W is 2/4".

[inquiry]

FM75, on 2013-February-23, 09:19, said:

Ben this would only be right if the suit had 14 cards.

Ah, i thought it was 4-3 suit.. with 4-4 suit, of course, justin was right... I knew I shouldn't disagree with him... I missed one of the "x's" in Jxxx.

Thanks.

[aguahombre]

JLOGIC, on 2013-February-23, 02:06, said:

Give this one to the person who argued about this incorrectly for 15 mins:

What are the odds of making a contract that is on 2 out of 3 finesses?

Spoiler

50 % ZOMG!!!!!!!!!

might blow their mind lol.

Heh. Back in the 70's Hal Kandler sprung this on his students before a club game. Then, later he overheard one of them trying to stump his experienced partner by asking "What are the odds that 2 out of 3 finesses will work?"

[JLOGIC]

fromageGB, on 2013-February-23, 09:24, said:

I don't like jlogic's way of deducing this, though, it is impossible for me at the table to work out what all the combinations are and how many of then include the T. I just say "apart from K, there are 4 other cards. W has 2, E has 2 (using inquiry's compass), so the probability of any card in W is 2/4".

It should not be scary or hard to work out all the combinations at the table. How many combinations of Tx are there? Obviously 3 because there are 3 low spot cards (T4, T3, T2). How many combinations of 2 small? It's 2+1 (43, 42, 32). Alternatively, 2 low is facing KTx, and we know again pretty easily there are 3 possible combos of KTx because there are 3 cards.

In this case just saying its 2-2 non ten cards is fine, but normally in bridge hands you need to do the combinations. Here is a common example.

Kx opp AT98x. You need 4 tricks from the suit. You play the king and everyone follows. Now you lead low towards the AT98 and they follow low. Is the T the percentage play or the ace?

The ten wins on QJxx, and the ace wins on Qxxx and Jxxx. Qxxx and Jxxx are the same as Qx or Jx in the other hand, there are 8 combinations of that (how do I know so quickly? Because there are 4 low spot cards, times two). QJxx is the same as 2 low which is 6 combinations (3+2+1). 54, 53, 52, 43, 42, 32. I am pretty sure this is some mathematical concept (my dad always says CHOOSE or something), but this pattern is evident if you do any of these away from the table, 3 small cards as 2 small has 2+1 combos, 4 small cards has 3+2+1, 5 small cards has 4+3+2+1 ways of being 2 small etc.

So for 4 tricks it is right to play the ace by a margin of 8 to 6. Now, what if you could win on 3 tricks with only 1 loser in the suit also (aka, you could pick up QJxxx). Well, now the ten gains on 4 lesser likely combos as well, stiff small on our right (easy to come up with the number 4, there are 4 low cards). So now the ten would be the right play. This is all very doable at the table with little knowledge of math.

[fromageGB]

A nice example, and you are right, it doesn't need any maths. Not sure if my brain can handle the strain, but it is not every hand you need to do this. I must practice.

I've always worked out the combinations of 2 cards from 4 as 4*3/2*1 = 6, and I have never noticed the fact that this equals 3+2+1. And 2 from 5 is 5*4/2*1 = 10 = 4+3+2+1. Pretty.

I think the trick at the table must be the visualisation/realisation that the chances of QJxx is the same as xx on the other side. I would be tempted to attempt to work out the chance of QJxx being in 4 cards of 6, and get bogged down.

[lalldonn]

JLOGIC, on 2013-February-23, 13:53, said:

I am pretty sure this is some mathematical concept (my dad always says CHOOSE or something)

If you have X cards and want to know how many combinations of Y cards you can make from it where order doesn't matter, it's "X choose Y" which is X!/[Y!(X-Y)!]. So combinations of 2 cards you can make from 4 is (4*3*2*1)/[(2*1)(2*1)] = (4*3)/(2*1) = 6. Or if you want to know how many different 8 card spade suits are possible it's (13*12*11*10*9)/(5*4*3*2*1) = 1287. Yay I'm redeemed from needing Han's help earlier.