## Last Hand Bid Puzzle 2

dustin22
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### Re: Last Hand Bid Puzzle 2

While the "95%" figure is obviously made up (not sure why people feel the need to make up statistics), I too agree that an 8 bid is clearly the best here.

Can a game end in a tie on Silver Creek? If so, that's pretty stupid.

The 8 option allows us to tie (if we only take 8 tricks) and gives us another chance to win.

Bidding anything less than 8 is essentially bidding 9, as Galt points out.

More likely to take 9 tricks if you bid less than 8? Yeah probably. But I like my chances of taking 9 tricks with 2 nils on the table, I also think 8 tricks is a lock here except in a very poor layout of the cards.

That said, one intriguing reason to bid less than 8 is that you are much more likely to take 9 tricks by making a very low bid (presuming your LHO nils). These types of decisions shouldn't be made purely on the math. Psychology/reading the opponents is just as important, and if you have a read that an underbid will work, make the call.

Galt
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### Re: Last Hand Bid Puzzle 2

As I mentioned earlier, the probability that 3 of the 4 cover dimes will be on your left side is around 10%.

You ask why West would bid NIl. Why the heck would he not bid Nil?

This isn't rocket science.

What actually happened in this game is completely irrelevant.

For years I have seen people who have not been trained in probabilty theory post hands that have crazy distributions and talk after the fact about how some unusual bid or play worked or could work and how the obvious bid or play would have or could have been wrong.

The objective of any bid or play at any point in any Spades game is the following:

To win the highest percentage of games that one would play over time from the exact same position.

I could bid first seat first hand and bid Nil with the 9, 10, and Jack of Spades and make my Nil. That would not make it a good bid.

I will say it again. The only reason to not bid 8 in this situation is out of fear that one will not win 8 tricks. There is no other possible explanation. Playing scared, over time, loses lots of winnable games.

babar
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### Re: Last Hand Bid Puzzle 2

babar
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### Re: Last Hand Bid Puzzle 2

babar
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### Re: Last Hand Bid Puzzle 2

babar
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### Re: Last Hand Bid Puzzle 2

babar
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### Re: Last Hand Bid Puzzle 2

Openshut
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### Re: Last Hand Bid Puzzle 2

Here is an example you are familiar with Galt
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North/South: 315/5

East/West : 326/6

Bids:
West, 4
North, 3
East, nil
South, ?

South:
S: K, J, 8, 5, 4
H: A, 10, 5
D: K, 7, 6
C: A, Q

Posted Jun 10th, 07:40 am
Hi Galt, every time someone asks me a question will try my best to answer it.
I will make an example of how to make a calculation like this to estimate which bid is better.
Definitions:
Probability of A: P(A), this is the probability that situation A will occur. P#(A) is the probability that situation A will Not occur. If P(A) is 0.7 then P#(A) is 0.3 as either A will occur or it wont. The total probability will always be 1. P(A)+P#(A)=1.
Situations.
Situation A. We bid 6 and make our bid and not setting the cover.
Situation B. We bid 6 and set the cover and take 1 bag.
Situation C, We bid 7 and make it and no bag.
Situation D. Score is 405-466 and you win. (This is if situation A occur)
Situation E. Score is 406-386 and you win. (This is if situation B occur)
Situation F. Score is 415-386 and you win. (This is if situation C occur)
Now I will try to estimate the probability of situation A-F will succeed if they occur. As a comment this is highly some estimations and they may not be accurate.
P(A) = 0.50 thus P#(A) = 0.50
P(B) = 0.50 thus P#(B) = 0.50
To make the calculation more easy I assumed that you will either make 9 or 10 tricks always with this hand. Of course this is not right but good enough for this simple calculation. P(A) + P(B) will in fact be less than 1 because you can take 2 bags, go set etc. In this example I ASSUME P(A)=P#(B) or in other words P(A) + P(B) = 1
P(C) = 0.50 thus P#(C) = 0.50. B and C got same probability as its same situation, only thing that’s different is bid but you still play for same purpose, to take 10 tricks total.
P(D) = 0.05
P(E) = 0.55
P(F) = 0.65
To calculate the probability of a 6 bid to succeed it will be the following adding of the probability.
P(A)*P(D) + P(B)*P(E) =0.5*0.05+0.5*0.55=0.30=30%
‘’Comment to Galt, here is where you make a mistake in your calculation, you only mention the P(A)*P(D) which indeed in your example is 5% but you also have the combination P(B)*P(E) which is much higher. You can bid 6 but make 7 and then you still have more than 50% chance to win. Hope I explained it good enough Galt. ‘’
To calculate the probability of a 7 bid to succeed will be:
P(C)*P(F) =0.5*0.65=0.33=33%
More interesting will be what we could call a sort of ‘’break-even’’ point here. If we assume P(D), P(E) and P(F) are accurate we can calculate at what P(A) and what P(B) we need for the 6 bid and 7 bid to be even good. This ‘’break-even’’ point will be at P(A)= 0.67 which means if you think you got at least 33% chance to make the 7 bid you should take it.
I am very sure this calculation is very dependent on how we assume P(D), P(E) and P(F). If we change them to:
P(D) = 0.10
P(E) = 0.55
P(F) = 0.60
Then we got a different ‘’break-even’’ point. Now this point is at P(A) =0.33 which means if you think you have more than 67% chance to make the 7 bid you should take it.
I am really not sure if the 6 or 7 bid is a outstanding bid here. As I said I like them both and in my eyes there is not a mistake if you bid 6 or 7 here. Feel free to comment on what I wrote here.
****
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Perhaps you would care to show those interested how the math works, less hand waving is much appreciated.

babar
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### Re: Last Hand Bid Puzzle 2

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pls start a new topic...thx

TrashCanCharlie
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