June 10th, 2020 at 4:02:24 AM
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Hi Wizard,

I wanted to ask you a question related to Banker decisions in Baccarat. I know that it is a well established fact that Banker decisions comes often than Player decisions statistically because the Banker hands are completed last and most of the third card drawing rules for Banker depends on the Player's total. Therefore, over the long term Banker's odds are 50.68% and Players odds are 49.32% (omitting Tie decisions)

With that being said over 1000 pure Banker/Player decisions, Banker is supposed to lead Player by approximately 13 decisions more. However, this is not true for each and every sample of 1000 hands.

So, my question is what is the maximum possible number of decisions that Banker decisions will Trail Player decisions in those 1000 hands? Is there any formula to calculate this over say 1800 hands?

Thank you.

Sinister

I wanted to ask you a question related to Banker decisions in Baccarat. I know that it is a well established fact that Banker decisions comes often than Player decisions statistically because the Banker hands are completed last and most of the third card drawing rules for Banker depends on the Player's total. Therefore, over the long term Banker's odds are 50.68% and Players odds are 49.32% (omitting Tie decisions)

With that being said over 1000 pure Banker/Player decisions, Banker is supposed to lead Player by approximately 13 decisions more. However, this is not true for each and every sample of 1000 hands.

So, my question is what is the maximum possible number of decisions that Banker decisions will Trail Player decisions in those 1000 hands? Is there any formula to calculate this over say 1800 hands?

Thank you.

Sinister

June 10th, 2020 at 4:53:05 AM
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I had 2 shoes out of 10 that went 50 to 25 one way or the other, and it was the opposite of what I was betting on, so yay counterfeit computer casino games.

June 10th, 2020 at 5:18:32 AM
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The formula you can use runs as follows.Quote:sinister1...Is there any formula to calculate this...

Let N = number of hands - in your case 1000 or 1800.

Let p = Probability of the event occuring - in this case a win happens about 50.68% (but it's easier to use 50% for these kinds of things to get an estimate).

Let q = (1-p) - in this case a loss.

Average (#wins) = Np

Std Deviation is SQRT (Npq).

Then you use the normal curve to determine how likely the difference from the mean your result is (or would be).

N=1000

Average ~= 507 wins (here you need to use the accurate percentage)

Std Dev = SQRT (1000*.5*.5) ~= 16 (here you can use the estimate).

So it is fairly likely you might be having more players after 1000 hands.

N=100000.

Avg=50680 (using your figs)

StdD = (SQRT(100000*.5*.5) ~= 158.

So it's fairly unlikely (680/158 ~= 4 Sds) that you would see more players than bankers.

(See https://en.wikipedia.org/wiki/Normal_distribution )

June 10th, 2020 at 6:32:06 AM
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Thanks charliepatrick. May I ask what does the value of Standard Deviation (in your examples 16 and 158) signify?

When Std Dev is 16, how does it help to conclude that it is likely that the Players decisions will be more after 1000 hands?

Does the lower or larger value of Std Dev tells something about the difference (swing) between decisions?

My original question was in a set of 1000 decisions, how many max possible decisions Banker will lag by compared to Player's? - you calculated 16, does that mean the max. decisions Banker lags by will be 507 -16 = 491 Banker with 493 +16 = 509 Player. So, Banker will lag by 491 - 509 = -18 decisions at the max?

Cheers

When Std Dev is 16, how does it help to conclude that it is likely that the Players decisions will be more after 1000 hands?

Does the lower or larger value of Std Dev tells something about the difference (swing) between decisions?

My original question was in a set of 1000 decisions, how many max possible decisions Banker will lag by compared to Player's? - you calculated 16, does that mean the max. decisions Banker lags by will be 507 -16 = 491 Banker with 493 +16 = 509 Player. So, Banker will lag by 491 - 509 = -18 decisions at the max?

Cheers

June 10th, 2020 at 7:27:23 AM
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Let's look at coin tosses, they're very similar to your situation but I'll give an example where the maths is easier.

If 100 coins are tossed, then the average number of heads will be 50. The question is what range of heads might we expect.

The Standard Deviation = SQRT(100*.5*.5) = SQRT(25) = 5. That means the result could be 5 either way etc.

1 SD - 68% of the time the result will be 45-55.

2 SDs - 95% of the time the result will be 40-60

3 SDs - 99.7% of the time the result will be 35-65.

If your case the SD is 16 for 1000 hands, and 158 for 100k hands.

Using similar logic for 1000 hands

1 SD - 68% of the time the result will be within 16 either way of the mean = 491 to 523.

Using similar logic for 100k hands

1 SD - 68% of the time the result will be within 158 either way of the mean = 50522 to 50838.

4 SDs - 99.99% of the time the result will be within 158*4 either way = 50048 to 51312.

So with 1000 hands is is quite possible for the player to be ahead. After 100k hands it is almost impossible.

If 100 coins are tossed, then the average number of heads will be 50. The question is what range of heads might we expect.

The Standard Deviation = SQRT(100*.5*.5) = SQRT(25) = 5. That means the result could be 5 either way etc.

1 SD - 68% of the time the result will be 45-55.

2 SDs - 95% of the time the result will be 40-60

3 SDs - 99.7% of the time the result will be 35-65.

If your case the SD is 16 for 1000 hands, and 158 for 100k hands.

Using similar logic for 1000 hands

1 SD - 68% of the time the result will be within 16 either way of the mean = 491 to 523.

Using similar logic for 100k hands

1 SD - 68% of the time the result will be within 158 either way of the mean = 50522 to 50838.

4 SDs - 99.99% of the time the result will be within 158*4 either way = 50048 to 51312.

So with 1000 hands is is quite possible for the player to be ahead. After 100k hands it is almost impossible.