[fred]

---

The contract is 7D and the lead is a trump.

Rightly or wrongly, you play the first 11 tricks like this:

1. Win trump in dummy (RHO follows)
2. Play trump to hand (RHO discards a club)
3-4. Cash top spades (both follow)
5. Spade. LHO ruffs and you overruff.
6. Trump to hand (RHO discards a club)
7. Cash King of clubs (both follow small)
8. Heart to dummy (both follow small)
9. Cash Ace of clubs, discarding a spade (both follow small)
10. Heart to hand (RHO follows with an honor and LHO follows small)
11. Trump. LHO discards a small club, you discard the Jack of clubs from dummy, and RHO also discards a small club.

To summarize, you know that RHO started with 4 spades and 1 diamond. He is either 2-6 or 3-5 in hearts and clubs. The only outstanding club is the Queen.

In the 2-card ending, you lead a heart and LHO follows small.

Three questions:

1) What are the odds that finessing the 9 of hearts will win?
2) What would the odds be if dummy had the 10 of hearts instead of the 9?
3) What would the odds be if dummy had a small club instead of the Jack?

Fred Gitelman
Bridge Base Inc.
www.bridgebase.com

[bluecalm]

I always hoped someone teaches me how to solve such problems
Awaiting replies...

[hanp]

There are 3:2 times as many 5-4 splits as 6-3 splits. (9 choose 4 compared with 9 choose 3).

The last question is easiest. If we don't have the club jack then RHO having 10x or Jx + 6 clubs is 4:3 times as likely as J10x plus 5 clubs. So in that case we should hook.

If there was no club jack but the heart 9 was the 10 then we should compare xx + 6 clubs with Jxx + 5 clubs. In that case the odds give 3:2 in favor of the drop.

If we put the club jack back in, we notice that we should take the hands where LHO started with the club queen and 4 hearts out of the equation, in that case he would have been squeezed to part with a heart. On 3/9 of the hands where LHO started with 3 clubs, he has the club queen. So we should take out 1/3 of those hands, in effect multiplying the odds for the finesse with 2/3.

That means that when dummy has the 9, hooking is 2/3 of 4:3, or 8:9. So play for the drop.

When dummy has the 10 the odds are 9:4 in favor of the drop, which seems intuitive.

Now I'm going to edit all the mistakes I surely made, but I'm going to post anyway in case cherdano is reading.

Thanks Roger for the corrections.

[hanp]

I don't see a mistake. That means I can get some food and will read my mistakes later.

[hrothgar]

I hesitate to suggest this - especially since I should be working on something real - however, it seems as if this sort of problem is well suited for a Bayesian formulation. (Updating your priors as more information becomes available about the hand).

I might take a look at this tonight...

[rogerclee]

hanp, on Aug 25 2010, 11:10 AM, said:

There are 3:2 times as many 5-4 splits as 6-3 splits. (9 choose 4 compared with 9 choose 3).

The last question is easiest. If we don't have the club jack then RHO having 10x or Jx + 6 clubs is 4:3 times as likely as J10x plus 5 clubs. So in that case we should hook.

If there was no club jack but the heart 9 was the 10 then we should compare xx + 6 clubs with Jxx + 5 clubs. In that case the odds give 3:2 in favor of the drop.

If we put the club jack back in, we notice that we should take the hands where LHO started with the club queen and 4 hearts out of the equation, in that case he would have been squeezed to part with a heart. On 3/9 of the hands where LHO started with 3 clubs, he has the club queen. So we should take out 1/3 of those hands, in effect multiplying the odds for the finesse with 2/3.

That means that when dummy has the 9, hooking is 2/3 of 4:3, or 8:9. So play for the drop.

When dummy has the 10 the odds are 9:4 in favor of the drop, which seems intuitive.

Now I'm going to edit all the mistakes I surely made, but I'm going to post anyway in case cherdano is reading.

I think this is what you meant.

[fred]

Han - thanks for your thoughts - I had been hoping you would weigh in on this

If you and I disagree on the answer to a math problem, I would guess the odds would be about 99% that you are right, but please consider this reasoning to question 1 (which yields a different answer than what you got):

We can answer question 1 by comparing these numbers:

1) The number of ways that RHO could be dealt honor-doubleton of hearts and Qxxxxx of clubs

2) The number of ways that RHO could be dealt J10x of hearts and 5 small clubs

Do you agree with this?

The number of cases for both Qxxxxx and xxxxx are the same so this can be simplified to comparing the number of honor-doubleton cases to the number of J10x cases.

I am sure you agree with this.

That is 8:4 or 2:1.

I am sure you agree with this.

But apparently you don't agree with my conclusion that the answer to question 1 is that the odds are 2 to 1 in favor of the finesse (you got 9 to 8).

Using the same sort of reasoning leads to a pure guess for problem 2 (versus your 9:4 in favor of the drop).

FWIW my bridge instincts suggest that you are right, but I can't quite pinpoint the flaw in the above. I do have an idea, but in a effort not to further embarass myself, I won't post it just yet

Fred Gitelman
Bridge Base Inc.
www.bridgebase.com

[Apollo81]

If dummy had the ♥T, my instincts are that the drop is 2:1 ish, given that we can apply restricted choice only on trick 11 as opposed to tricks 10-11.

[rogerclee]

Under the conditions

1) Someone has 8 cards in hearts and clubs
2) There are 9 possible clubs he could have and 6 possible hearts he could have.
3) He has either 5 clubs and 3 hearts or 6 clubs and 2 hearts.

I get that he is 2:1 to be 5-3, as opposed to 6-2.

(This is binomial(9,5) * binomial(6,3) vs. binomial(9,6) * binomial(6,2))

I am not sure if this is applicable in this situation, someone should feel free to let me know.

[Apollo81]

Apollo81, on Aug 25 2010, 03:06 PM, said:

If dummy had the ♥T, my instincts are that the drop is 2:1 ish, given that we can apply restricted choice only on trick 11 as opposed to tricks 10-11.

I can't see a flaw in han's reasoning in the case where an opponent holds the ♣J, and I can't see a flaw in fred's reasoning in the case where the ♥9 and ♣J are in dummy. Combining fred's reasoning if the ♥10 is in dummy with the restricted choice on trick 11 (the first time RHO had the option to play the ♣Q) says to me that the drop is a 2:1 favorite in this situation.

han, I don't understand why it is necessary to remove hands where LHO had ♣Qxx in the cases were the ♣J is in dummy: don't we know that this wasn't happening anyway by looking at his play to trick 11?

for this sort of problem, i would think that in general my opinions are the least likely to be right out of all respondents so far -- please show me where i'm going wrong

[hanp]

fred, on Aug 25 2010, 01:44 PM, said:

We can answer question 1 by comparing these numbers:

1) The number of ways that RHO could be dealt honor-doubleton of hearts and Qxxxxx of clubs

2) The number of ways that RHO could be dealt J10x of hearts and 5 small clubs

Do you agree with this?

I think this part of what you say is not correct. The fact that RHO played small on the 11th trick gives no information because at that time (when we had already played the club jack from the dummy) he could play the queen or small from Qxxxxx. To get the correct answer, we should also consider Qxxxx of clubs on our right.

This kind of argument can get a bit messy. Best is to ignore the card RHO plays. If RHO "falsecards" exactly the right amount of time from Qxxxxx there is no information to get from the play of the low club on our right.

Let me try to phrase it in terms of strategies, I think it is easier. If we play for the drop no matter whether RHO plays low or the queen, we win if RHO started with J10x of hearts and Qxxxx or xxxxx of clubs, but lose if RHO started with Jx or 10x of hearts and Qxxxxx of clubs.

♥J10x and ♣xxxxx is exactly equally likely as ♥Hx and ♣Qxxxxx, but ♥J10x and ♣Qxxxx is a bit more likely than ♥J10x and ♣xxxxx, and thus also more likely than ♥Hx and ♣Qxxxxx.

Therefore we should play for the drop.

[fred]

hanp, on Aug 25 2010, 08:09 PM, said:

fred, on Aug 25 2010, 01:44 PM, said:

We can answer question 1 by comparing these numbers:

1) The number of ways that RHO could be dealt honor-doubleton of hearts and Qxxxxx of clubs

2) The number of ways that RHO could be dealt J10x of hearts and 5 small clubs

Do you agree with this?

I think this part of what you say is not correct. The fact that RHO played small on the 11th trick gives no information because at that time (when we had already played the club jack from the dummy) he could play the queen or small from Qxxxxx. To get the correct answer, we should also consider Qxxxx of clubs on our right.

This kind of argument can get a bit messy. Best is to ignore the card RHO plays. If RHO "falsecards" exactly the right amount of time from Qxxxxx there is no information to get from the play of the low club on our right.

Let me try to phrase it in terms of strategies, I think it is easier. If we play for the drop no matter whether RHO plays low or the queen, we win if RHO started with J10x of hearts and Qxxxx or xxxxx of clubs, but lose if RHO started with Jx or 10x of hearts and Qxxxxx of clubs.

♥J10x and ♣xxxxx is exactly equally likely as ♥Hx and ♣Qxxxxx, but ♥J10x and ♣Qxxxx is a bit more likely than ♥J10x and ♣xxxxx, and thus also more likely than ♥Hx and ♣Qxxxxx.

Therefore we should play for the drop.

Thanks very much, Han.

Yes, this makes sense and it is actually pretty similar to the theory I was afraid to post out of fear of further embarassing myself (though you definitely did a better job of explaining it to me than I did explaining it to myself!).

Indidentally, the reason I found this problem interesting was because I could not really decide if the Queen of clubs should be treated of as a pure x (as it is in question 3), a pure non-x (like all the known cards in spades and diamonds), or as something in between.

Fred Gitelman
Bridge Base Inc.
www.bridgebase.com

[MFA]

fred, on Aug 25 2010, 07:31 PM, said:

1) What are the odds that finessing the 9 of hearts will win?

Finesse: 47,0%, drop: 53,0%

Quote

2) What would the odds be if dummy had the 10 of hearts instead of the 9?

Finesse: 30,8%, drop: 69,2%

Quote

3) What would the odds be if dummy had a small club instead of the Jack?

Finesse: 57,1%, drop: 42,9%

Seems to be the same results as the other posters.
I calculated by looking at the relevant distributions:

1)
♥Hxxx, ♣xxx - ♥Hx, ♣Qxxxxx
compared to
♥xxx, ♣xxxx - ♥HHx, ♣xxxxx [♣Q irrelevant as a specific card]

2)
♥Jxxx, ♣xxx - ♥xx, ♣Qxxxxx
compared to
♥xxx, ♣xxxx - ♥Jxx, ♣xxxxx [♣Q irrelevant as a specific card]

3)
♥Hxxx, ♣xxx - ♥Hx, ♣xxxxxx
compared to
♥xxx, ♣xxxx - ♥HHx, ♣xxxxx

I counted the number of distributions and then calculated odds.

[nigel_k]

I can't do maths any more but I ran a computer simulation (100,000 random E/W hands consistent with the conditions) and it was 2:1 in favour of the finesse.

Isn't hanp's reasoning missing the restricted choice element, i.e. only half of the J10x probability counts because they could have played the other card?

[JLOGIC]

nigel_k, on Aug 25 2010, 04:35 PM, said:

Isn't hanp's reasoning missing the restricted choice element, i.e. only half of the J10x probability counts because they could have played the other card?

No. Counting both Hx and JTx is the same as counting half of JTx.

[pooltuna]

JLOGIC, on Aug 25 2010, 04:54 PM, said:

nigel_k, on Aug 25 2010, 04:35 PM, said:

Isn't hanp's reasoning missing the restricted choice element, i.e. only half of the J10x probability counts because they could have played the other card?

No. Counting both Hx and JTx is the same as counting half of JTx.

Nevertheless the simulation shows some anomaly exists either in the calculations or the simulation assumptions

[nigel_k]

Ok I understand why we might want to include the Qxxxx cases, though we know the current hand is not one of those cases. But don't you then need to also include the cases where RHO has xxxxxx (and LHO Qxx)?

Anyway after redoing the simulation completely disregarding the ♣Q I get 57% (4:3) in favour of the finesse.


The layouts break down as follows:

JTx, Qxxxx 70 cases ( 8C4 )
JTx, xxxxx 56 cases ( 8C5 )
Total 126

Jx, xxxxxx 28 cases ( 8C6 )
Jx, Qxxxxx 56 cases ( 8C5 )
Tx, xxxxxx 28 cases ( 8C6 )
Tx, Qxxxxx 56 cases ( 8C5 )
Total 168

Which is a 4:3 ratio. Without the xxxxxx cases it is 126:112 which is the 53% that MFA got.

[bluecalm]

After running my simulation on large sample assuming all clubs are equals the results are:

JTx - 42.8%
Hx - 57.2%

which seems to be in line with MFA's result.

[bluecalm]

As to the first question my simulation gives:

Hx - 47.2%
JTx - 52.8%

which again is in line (almost) with MFA's result.

I'am jealous of you guys who can figure it out without simulations. I am always afraid I will make some stupid mistake counting combinations or just misrepresent the problem. I lost trust in my ability to solve such problems on paper long time ago

[dburn]

Deleted - that was nonsense. I think I'd better think it out again.