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ChesterDog
ChesterDog
Joined: Jul 26, 2010
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October 20th, 2010 at 12:12:23 PM permalink
(Wizard, I'm sure you meant to write: (6/24)*(1/3)+(8/24)*(2/5)+(10/24)*(5/11).)
7winner
7winner
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October 20th, 2010 at 12:13:37 PM permalink
Quote: guido111


http://wizardofodds.com/askthewizard/craps-probability.html

" So the probability of making a point, given that a point was established is, (5/12)*(5/11)+(4/12)*(4/10)+(3/12)*(3/9) = 40.61%"



can one use The Wizard's above formula to calculate exactly the probability of a shooter getting only 1 point before a 7out? How about 2 points or more?
7 winner chicken dinner!
7craps
7craps
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October 20th, 2010 at 12:28:26 PM permalink
Quote: MathExtremist

I believe that's rolls per shooter. Since rolls per passline decision is 3.376, that makes the average decisions per shooter about 2.5. Of those decisions, 27.1% are point winners, so the average number of points made by the shooter is about 0.68.


0.68 is a small average for point winners

Quote: MathExtremist

In other words, if you make even one point, you're having a better-than-average roll. Not very intuitive, is it?

Yes, you really do only have a bit better than 1 in 4 chance of establishing and making a point. A big chunk of winning passline bets happens from natural 7/11 rolls: 22.2% to be precise. When you add them in you get 49.3%, the overall chance of winning a line bet.



So, when the pass line wins it wins about 45% of the time on the come out roll.
22.2%/49.3%
winsome johnny (not Win some johnny)
JimmyMac
JimmyMac
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October 20th, 2010 at 12:41:17 PM permalink
So, if I'm reading this right.. if I was to walk up to a full crap's table, I could use that typically the shooter will not hit his point? If he hits one of his points he is doing average. For him to even hit two points before crapping out and passing the dice would be above average.

Would it be fair to say then that if I go to a table and wait for a roller to hit 1 number and then play the don't pass that I would have a little better shot of winning? (not counting the come out roll where I am at a disadvantage because of the 7)

I know if you run scenario's with a million rolls, the numbers probably tell you that will not have any advantage because the dice have no memory.

But, in real life when I go to a table, I may play long enough for the dice to go around the table about 3 times.

I'm just looking for an edge for the 2 hours I'm at the table.

This site is a great forum. I really appreciate all your advice.
DJTeddyBear
DJTeddyBear
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October 20th, 2010 at 1:41:46 PM permalink
Quote: JimmyMac

Would it be fair to say then that if I go to a table and wait for a roller to hit 1 number and then play the don't pass that I would have a little better shot of winning? (not counting the come out roll where I am at a disadvantage because of the 7)

I know if you run scenario's with a million rolls, the numbers probably tell you that will not have any advantage because the dice have no memory.

You just answered your own question.

On the other hand, there are plenty of gamblers that do just that - wait for a shooter to make a point - to 'qualify' him as a good shooter.
Superstitions are silly, childish, irrational rituals, born out of fear of the unknown. But how much does it cost to knock on wood? 😁 Note that the same could be said for Religion. I.E. Religion is nothing more than organized superstition. 🤗
guido111
guido111
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October 20th, 2010 at 2:18:56 PM permalink
Quote: JimmyMac

So, if I'm reading this right.. if I was to walk up to a full crap's table, I could use that typically the shooter will not hit his point?


The shooter has a 40.61% on average of hitting any point.
Specifically,
4/10 33.33% (3/9)
5/9 40% (4/10)
6/8 45.5% (5/11)
So that makes on average a 59.39% chance of not hitting any point.
Quote: JimmyMac

If he hits one of his points he is doing average.


As stated above, the average is .68 so he is "above" average
Quote: JimmyMac

For him to even hit two points before crapping out and passing the dice would be above average.


A shooter "7outs" when he does NOT hit his point.
The term is NOT "crapping out". That ONLY applies on a comeout roll of a 2,3 and 12. The same shooter still keeps the dice after "crapping out".

Once a shooter has hit a point 1 or more times he IS already above average.
40.6% of all shooters hit 1 or more points.
16.1% of all shooters hit 2 or more points
6.66% of shooters hit 3 or more points
and so on by the table on page 1.
Wizard
Administrator
Wizard
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October 20th, 2010 at 2:32:30 PM permalink
Quote: ChesterDog


(Wizard, I'm sure you meant to write: (6/24)*(1/3)+(8/24)*(2/5)+(10/24)*(5/11).)



Thanks, you're absolutely right. I just corrected my original post. I'm going through the flu, so my mind is not thinking straight.
It's not whether you win or lose; it's whether or not you had a good bet.
JimmyMac
JimmyMac
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October 20th, 2010 at 3:10:41 PM permalink
Okay, so then what you have described then tells me that if I am playing the don't pass then my odds of winning the next established point increase every time a player hits his point.

Based on the above:

59.39% shooter will not his his point.
40.6% shooter will hit 1 or more points
16.1% shooter will hit 2 or more points
6.66% shooter will hit 3 or more points

If I take the opposite:
40.61% shooter will his his point
59.4% shooter will not hit 1 or more points
83.9% shooter will not hit 2 or more points
93.34% shooter will not hit 3 or more points

Therefore, if I choose to play the Don't Side of the table and I sit out until a shooter establishes a point and hits it and then join the game by placing my money on the Don't Pass (get past the come out roll) and the shooter establishes a 2nd point then my odds of winning go from 59.4% to 83.9%.

If I could hold out for a shooter to hit 2 points and then join the game, get past the come out roll and establish a point, I'd be at a 93.34% advantage of winning.

Do I have this correct?
MathExtremist
MathExtremist
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October 20th, 2010 at 3:16:29 PM permalink
Quote: JimmyMac

Okay, so then what you have described then tells me that if I am playing the don't pass then my odds of winning the next established point increase every time a player hits his point.
...
Do I have this correct?



No -- that's the Gambler's Fallacy. In truth, the odds don't change over time. If the shooter has just made 10 passes in a row, his chance of making the next one are still 49.3%.

The only time aggregate probabilities matter are at the beginning of your group of events. Once you're in the middle of your set of trials, what happened in the past doesn't matter at all.
"In my own case, when it seemed to me after a long illness that death was close at hand, I found no little solace in playing constantly at dice." -- Girolamo Cardano, 1563
Ayecarumba
Ayecarumba
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October 20th, 2010 at 3:22:56 PM permalink
Hehe, no. The dice don't know that they just hit two points. You have the same chance of a seven out, if you stepped up to a table that had just lost 10 points in a row (probably lots of room), or won 10 in a row (probably no room).

Edit, sorry, got interrupted before posting. Interesting how the examples are so similar...
Simplicity is the ultimate sophistication - Leonardo da Vinci

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