wizard
Mood: |
| posted on 16/4/2006 at 12:46 AM |
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50-50???
Situation:
A five handed game.
Hand No. 2.
As dealer you are dealt 2 Aces.
It is No Trump.
The other 4 players all bid zero.
What do you bid?
In reality I bid 2 and went down 2 tricks.
Was this a 50-50 choice? Another player commented that I had taken a risky gamble but then conceded that it was a 50-50 chance. But was it really?
It would have been foolish to have bid zero although in the actual play of the hand I would have made my bid if I had bid zero. But think about the
possibilities.
When I made zero tricks I went down 20 points. This was the worst case scenario. However 2 other players each went down 10 points. and the other 2
players got 20 points each. So 2 players gained 40 points on me and the other 2 gained 10 points on me.
Now consider the worst case scenario if I had bid zero and had made 2 tricks. I would again go down 20 points but all of the other 4 players would
have gone up 20 points. This means that all 4 of them would have gained 40 points on me.
Conclusion: The better bid was for 2 tricks because if I did not make it the downside was less tragic.
But wait! What about the best case scenario?
If I bid 2 and make it I get 40 points and gain 20 on all other players.
If I bid zero and make it I stay even with 2 players and gain 30 points on the other 2 players.
Again, the better bid is 2 because although your chances of success are 50-50 the possible results are much more favorable with the 2 bid.
wizard (April 06)
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The Boss
Mood: |
| posted on 16/4/2006 at 02:47 AM |
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Interesting situation!
I think you made the best choice.
Another possibility would have been to bid 1. (If CR was ON, and you were the only leader after Round #1, you could not have bid 2.)
I'm not the best in mathematical probabilities, but you would have had 50% of chances to win the 1st card. And then if you would have escaped the 1st
card, you would have had 25% ** of chances to win with the 2nd card.
So overall, were the odds better to make 0, 1 or 2 tricks in this situation? I don't know. But for the reasons you mentionned, (points) I think it was
worth bidding 2.
** = Maybe a little more, since there's the possibility of Jesters (1, 2 or even 3) beeing played before your turn, increasing the odds of your card
becoming the lead off suit. (Obviously, there was no Wizard beeing dealt here.)
Note: -Well, this is getting complicated,  but there was also a 2-3% chances of
the game crashing at this point, increasing your odds to escape beeing set in this situation...! 
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The Boss
Mood: |
| posted on 17/4/2006 at 10:53 PM |
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Yes 2! Another proof...
Shouldn't the odds be:
2 Tricks: 50%
1 Trick: 12,5%
0 Trick: 37,5%
And if YES, then that's another proof that the best choice was to bid 2. If I'm wrong, please tell me.
Wizard reply: Yeah, you're wrong!
The odds of taking the first trick are 50-50. However if you take trick 1 you are almost certain to take trick 2 unless someone is insane or has
misclicked with a wizard.
However if you lose trick one then the odds of taking trick 2 are 25% as there are 3 other suits that could be led.(approximately as we are not
accounting for the missing cards made up of trick 1).This means the odds for taking 1 trick are 1 in 8 or 12.5 %
The odds for taking zero tricks are 50-50 for trick one and then 25% for trick 2. This means 12.5 % over all.
Why doesn't the total =100%? Because the odds for trick 2 change after completion of trick 1. For example the odds for taking trick 2 after taking
trick 1 can be considered to be 100% not 25%.
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The Boss
Mood: |
| posted on 19/4/2006 at 02:43 AM |
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Me wrong? I don't think so...
quote:
Originally posted by The Boss
Shouldn't the odds be:
2 Tricks: 50%
1 Trick: 12,5%
0 Trick: 37,5%
And if YES, then that's another proof that the best choice was to bid 2. If I'm wrong, please tell me.
Wizard reply: Yeah, you're wrong!
The odds for taking zero tricks are 50-50 for trick one and then 25% for trick 2. This means 12.5 % over all.
Why doesn't the total =100%? Because the odds for trick 2 change after completion of trick 1. For example the odds for taking trick 2 after taking
trick 1 can be considered to be 100% not 25%.
I think you're wrong there. The odds for all the possibilities must always equal 100%.
If we don't take trick #1, the odds cannot be of 25% of taking trick #2 and 25% of not taking it. (What about the other 50% ???) Since there's 4
suits, I say the odds are 25/75.
So I stand on my position. Overall, when it was your turn to bid, the odds were like this:
2 Tricks: 50%
1 Trick: 12,5%
0 Trick: 37,5%
Now, let's see what Mr. Wizard has to say about this. 
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Wizbucket
Mood: No Mood. |
| posted on 4/7/2008 at 03:11 AM |
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the way I see it
Wizard wanted a post, so here goes.
I figure it a little differently. You are figuring each trump has an equal possibility of being played, when I figure there are only 12 possible
cards it each suite that you have a ace it that can be played on the first lead, that will allow you to make your bid. 24 out of the 57 unknown cards.
Logicaly you can rule out that the 4 wizards are not in play, and the remaining 3 jesters are not an issue since a trump will be played after them.
That leaves 50 cards unaccounted for or a 48% probability of making your 2 bid and a 52% probability of missing a 2 bid. Most people use the even
possibility of any trump, I think that they forget to remove the cards in their hand from the probability. There is a slightly higher probability of
failing, but the payout is worth the chance this early in the game. I know that someone will say that no one will lead one of the other aces, or
probably not a face card, even in your suit, then you have to figure and refigure..... lol still there are only 24 out of the remaining cards that
will allow you to make your 2 bid....
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