March 7th, 2014 at 5:19:14 PM
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Isn't that chance 1/infinity and that limit is 0?Quote:TerribleTomI guess one could theoretically flip a coin and get tails every time from here until eternity, but in reality you are eventually going to win and when you do you're going to be up $1.

Interesting concept

eventually you will win with unlimited casino credit and no max table limits

why?

because the chance of winning (or probability) is 1/the largest number you can make

still higher than 0

is it?

http://www.mathsisfun.com/calculus/limits-infinity.html

or is that still 1/infinity = 0 or is it?

so the probability of losing every bet over infinity = 0

the probability of winning just 1 time in infinity is also 0 (1/infinity)

coins are so dirty

TITO

Sally

I Heart Vi Hart

March 7th, 2014 at 7:09:34 PM
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Maybe it's -1/12Quote:mustangsallyor is that still 1/infinity = 0 or is it?

Someday, joor goin' to see the name of Googie Gomez in lights and joor goin' to say to joorself, "Was that her?" and then joor goin' to answer to joorself, "That was her!" But you know somethin' mister? I was always her yuss nobody knows it! - Googie Gomez

March 7th, 2014 at 7:42:27 PM
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Sally,

I was reviewing this info again and came up with another questions.

You stated in this run:

"It can double that about 44% of the time (44 out of 100 or 56 times losing all 255 units)

the other times it lost all 255 units"

Are you stating the in 100 rounds you would double up 44% of the time and 56% of the time lose all 255 units?

That means based on what I was questioning that 56% of the time we will hit 14 points without rolling a 7 out? Or are you stating in 5000 rounds there were 255 times you lost it all?

I was reviewing this info again and came up with another questions.

You stated in this run:

"It can double that about 44% of the time (44 out of 100 or 56 times losing all 255 units)

the other times it lost all 255 units"

Are you stating the in 100 rounds you would double up 44% of the time and 56% of the time lose all 255 units?

That means based on what I was questioning that 56% of the time we will hit 14 points without rolling a 7 out? Or are you stating in 5000 rounds there were 255 times you lost it all?

March 7th, 2014 at 7:54:35 PM
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Quote:mustangsallyIsn't that chance 1/infinity and that limit is 0?

Interesting concept

eventually you will win with unlimited casino credit and no max table limits

why?

because the chance of winning (or probability) is 1/the largest number you can make

still higher than 0

is it?

http://www.mathsisfun.com/calculus/limits-infinity.html

or is that still 1/infinity = 0 or is it?

so the probability of losing every bet over infinity = 0

the probability of winning just 1 time in infinity is also 0 (1/infinity)

coins are so dirty

TITO

Sally

I don't understand. 1/infinity is never zero. It's undefined so it's never anything.

March 7th, 2014 at 9:55:58 PM
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yes, per 100 sessions of playQuote:vegasrvp"It can double that about 44% of the time (44 out of 100 or 56 times losing all 255 units)

the other times it lost all 255 units"

Are you stating the in 100 rounds you would double up 44% of the time and 56% of the time lose all 255 units?

double up or bust out trying

Quote:vegasrvpThat means based on what I was questioning that 56% of the time we will hit 14 points without rolling a 7 out?

where do you get 14 points from?

the don't pass loses 45% of the time on average, sometimes higher sometimes lower,

on the come out roll on a 7 or an 11

the photo shows that in 5000 roundsQuote:vegasrvpOr are you stating in 5000 rounds there were 255 times you lost it all?

2226 sessions ended by turning a 255 unit bankroll into 510 units

2774 sessions lost all 255 units - busted.

I did not track how many losses came on the come out roll and during the point round

I am sure that ratio on average is about 45/55%

when one bets the don't pass on the come out roll, a 7 or an 11 rolled right there causes the don't pass to lose and your progression to move up one step.

That can happen 3-4 times in a row before a point is ever established.

And if you do not win your parlay bet, another loss and a move up to the next step

so if you go a bunch of bets without ever winning 2 in a row, you end up losing

unless I did not understand what you wrote about playing this method.

Sally

I Heart Vi Hart

March 8th, 2014 at 1:02:49 AM
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Quote:geoffI don't understand. 1/infinity is never zero. It's undefined so it's never anything.

Exactly. The results of Martingale with an infinite bankroll are also undefined and therefore also never anything.

Here is a good rule of thumb: if you add up a bunch of numbers that are all negative, and you get a positive number (or vice versa) your math is wrong.

March 8th, 2014 at 11:38:01 AM
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To me the most important part is what you are considering a loss vs what I am considering a loss.

In order to bust out and lose 255 units you will have to lose on either a 7 / 11 on the come out roll of the player will have to make their point.

in order to get 255 units lost you will have to lose the wager based on the following sequence:

1, 1, 1, 2, 2, 4, 4, 8, 8, 16, 16, 32, 32, 64, 64.

That is 15 losses with hitting back to back 7 once point is established.

Roll example:

point come out roll is 6

Rolls of 2,5,9,8,8,5,4,11,12,6 winner means lose bet 1 move to next level

point come out roll is 7

Lose bet 1

move to next level

point come out roll is 11

lose bet 1

move to next level

point come out roll is 5

rolls of 4,8,9,3,8,6,6,5 pint winner

lose bet 2

move to next level

point come out roll is 3

winner win bet 2

move to next level double 2 (4)

point come out roll is 9

rolls of 4,6,5,10,3,5,2,7 out

winner of 8 units (less bets of 1,1,1,2 = 8-7=+1 unit for session)

Session complete move back to beginning

Based on the sequence of 1, 1, 1, 2, 2, 4, 4, 8, 8, 16, 16, 32, 32, 64, 64.

If you win you double down and if you win again you start over.

If you lose you move forward in the sequence

on the come out roll you will win with 2 / 3 lose on 7 / 11 and push on 12.

Based on win or loss move forward of double accordingly.

Based on this methodology you will have to either hit 7 / 11 on the come out roll or win 15 point (or some combination).

BEFORE

You hit 2/3 on the come out roll back to back, 7 out on the point (or some combination).

Is this what you thought? Also is this more clear?

In order to bust out and lose 255 units you will have to lose on either a 7 / 11 on the come out roll of the player will have to make their point.

in order to get 255 units lost you will have to lose the wager based on the following sequence:

1, 1, 1, 2, 2, 4, 4, 8, 8, 16, 16, 32, 32, 64, 64.

That is 15 losses with hitting back to back 7 once point is established.

Roll example:

point come out roll is 6

Rolls of 2,5,9,8,8,5,4,11,12,6 winner means lose bet 1 move to next level

point come out roll is 7

Lose bet 1

move to next level

point come out roll is 11

lose bet 1

move to next level

point come out roll is 5

rolls of 4,8,9,3,8,6,6,5 pint winner

lose bet 2

move to next level

point come out roll is 3

winner win bet 2

move to next level double 2 (4)

point come out roll is 9

rolls of 4,6,5,10,3,5,2,7 out

winner of 8 units (less bets of 1,1,1,2 = 8-7=+1 unit for session)

Session complete move back to beginning

Based on the sequence of 1, 1, 1, 2, 2, 4, 4, 8, 8, 16, 16, 32, 32, 64, 64.

If you win you double down and if you win again you start over.

If you lose you move forward in the sequence

on the come out roll you will win with 2 / 3 lose on 7 / 11 and push on 12.

Based on win or loss move forward of double accordingly.

Based on this methodology you will have to either hit 7 / 11 on the come out roll or win 15 point (or some combination).

BEFORE

You hit 2/3 on the come out roll back to back, 7 out on the point (or some combination).

Is this what you thought? Also is this more clear?

March 8th, 2014 at 11:55:34 AM
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Here is some info I gathered on this site from other locations. Does this fit into what you are seeing?

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

What is the average number of points hit by a craps shooter before he sevens out?

— JimmyMac

Given that a point is established, the probability that the shooter makes the point is pr(point is 4 or 10) ×pr(making 4 or 10) + pr(point is 5 or 9) × pr(making 5 or 9) + pr(point is 6 or 8) ×pr(making 6 or 8) = (6/24) × (3/9) + (8/24) × (4/10) + (10/24) × (5/11) = 201/495 = 0.406061.

If the probability of an event is p, then the expected number of times it will happen before failure is p/(1-p). So, the expected number of points per shooter is 0.406061/(1-0.406061) = 0.683673.

This question was raised and discussed in the forum of my companion site Wizard of Vegas .

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

I had run a sim (not me this is on the site) of 8,522,945 million dice rolls. Here are those results:

1,000,000/shooters

682796/total points wins

count points wins % or more or less

594110 0 0.5941100000

241267 1 0.2412670000 0.4058900000 0.8353770000

97983 2 0.0979830000 0.1646230000 0.9333600000

39547 3 0.0395470000 0.0666400000 0.9729070000

16136 4 0.0161360000 0.0270930000 0.9890430000

6535 5 0.0065350000 0.0109570000 0.9955780000

2596 6 0.0025960000 0.0044220000 0.9981740000

1067 7 0.0010670000 0.0018260000 0.9992410000

425 8 0.0004250000 0.0007590000 0.9996660000

188 9 0.0001880000 0.0003340000 0.9998540000

91 10 0.0000910000 0.0001460000 0.9999450000

55 11+ 0.0000550000 0.0000550000 1.0000000000

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

If I am reading this right based on 1 million shooters 55 times a shooter hit 11 or more points?

This seems like a much more realistic number.

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

What is the average number of points hit by a craps shooter before he sevens out?

— JimmyMac

Given that a point is established, the probability that the shooter makes the point is pr(point is 4 or 10) ×pr(making 4 or 10) + pr(point is 5 or 9) × pr(making 5 or 9) + pr(point is 6 or 8) ×pr(making 6 or 8) = (6/24) × (3/9) + (8/24) × (4/10) + (10/24) × (5/11) = 201/495 = 0.406061.

If the probability of an event is p, then the expected number of times it will happen before failure is p/(1-p). So, the expected number of points per shooter is 0.406061/(1-0.406061) = 0.683673.

This question was raised and discussed in the forum of my companion site Wizard of Vegas .

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

I had run a sim (not me this is on the site) of 8,522,945 million dice rolls. Here are those results:

1,000,000/shooters

682796/total points wins

count points wins % or more or less

594110 0 0.5941100000

241267 1 0.2412670000 0.4058900000 0.8353770000

97983 2 0.0979830000 0.1646230000 0.9333600000

39547 3 0.0395470000 0.0666400000 0.9729070000

16136 4 0.0161360000 0.0270930000 0.9890430000

6535 5 0.0065350000 0.0109570000 0.9955780000

2596 6 0.0025960000 0.0044220000 0.9981740000

1067 7 0.0010670000 0.0018260000 0.9992410000

425 8 0.0004250000 0.0007590000 0.9996660000

188 9 0.0001880000 0.0003340000 0.9998540000

91 10 0.0000910000 0.0001460000 0.9999450000

55 11+ 0.0000550000 0.0000550000 1.0000000000

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

If I am reading this right based on 1 million shooters 55 times a shooter hit 11 or more points?

This seems like a much more realistic number.

March 8th, 2014 at 11:56:27 AM
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The above chart can be found here:

http://wizardofvegas.com/forum/questions-and-answers/gambling/3146-average-number-of-points-hit-per-shooter-in-craps/#post33406

http://wizardofvegas.com/forum/questions-and-answers/gambling/3146-average-number-of-points-hit-per-shooter-in-craps/#post33406

March 8th, 2014 at 12:01:51 PM
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59.39% shooter will not hit 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

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