System for the casino??? (baccarat)

treeman · 9 July 2009

Say a table has $1000 limit.

You go in with $50

1 50
2 100
3 200
4 400
5 800
6 - hit the limit

For not a very big win "profit of 50" you only get 5 turns and risk a relatively large sum compared to your profit.

Now that you are 800 down you either start from 50 again or you keep betting table max 1000. With each 1000 loss the gap between your loses and maximum stakes keeps increasing rapidly, your profits are now in negative even if you win at table max.

If you chose to start from scratch you already are down 800 and keep increasing the odds of hitting a unlucky streak of 5 or 6 loses in a row.

drsmith · 9 July 2009

shiftyphil said:
You can not win at the casino.

If you do find a way to win, they'll call you a cheater and walk you out the door.

Hopefully it's the door out the front of the Casino and not the one out the back.

drsmith · 9 July 2009

The odds of hitting an unlicky streak of 6 losses in a row are low but when you do you lose the lot. With the Casino taking it's margin the house will ultimately win.

It's simple math.

Mr J · 9 July 2009

With the Casino taking it's margin the house will ultimately win.

Likely, but not certain.

It's simple math.

It may be simple to some, but most don't seem to be able to understand probability and expectation.

beamstas · 9 July 2009

julius said:
There is no tendency for mean reversion with a coin flip. No amount of previous heads effects the conditional expectation for future outcomes - the 50% expectation only applies to future outcomes from that point forward.

I said nothing about past results changing future expectations.

Experiment: We have just tossed 90 heads out of 100 coin tosses. 90% heads.

We are doing a test of 200 coin tosses

Do you expect the 90% heads to try and revert back towards 50% or stay at 90% for the remaining 100 tosses?

It will revert back to the mean of 50%, therefore a coin-flip is mean reverting. The more times you toss a coin, the closer it's going to get to the mean of 50%, it will never move away from 50% for a sustained period.

Mr J · 9 July 2009

It will revert back to the mean of 50%, therefore a coin-flip is mean reverting. The more times you toss a coin, the closer it's going to get to the mean of 50%, it will never move away from 50% for a sustained period.

You're a little unclear, at least to me. The coinflip itself is not mean-reverting, it's just that the probabilities over an ever-increasing sample will make it appear that way. It certainly can move from 50% over a sustained period.

beamstas · 9 July 2009

Mr J said:
You're a little unclear, at least to me. The coinflip itself is not mean-reverting, it's just that the probabilities over an ever-increasing sample will make it appear that way. It certainly can move from 50% over a sustained period.

It will tend to revert to 50%

500flips
% of heads recorded
51
48.2
49.4
51.8
51.2
48.4
49
50.8
50
53.4

Avg for 10 runs: 50.32% heads

3000 flips
% of heads recorded
50.53
51.33
49.73
49.77
49.87
51
49.37
50.23
49.6
48.63

Avg for 10 runs = 50.006% heads

This is a tiny sample but as you can see it will over time, revert to 50/50 with a larger sample.

This doesn't help with the casino at all

I wouldn't advise betting on a tail if there has just been 9 heads, though.

Mr J · 9 July 2009

This is a tiny sample but as you can see it will over time, revert to 50/50 with a larger sample

It won't necessarily revert to 50% due to variance. The sample may actually get further and further from 50%, although this is obviously not likely. I'm sure I know what you mean, I just don't agree with the way of how you are saying it. The distribution as a percentage is very likely to revert to 50% as the sample grows, but the coinflips themselves don't revert to the mean, and I assume that is why you said you don't recommend betting tails just because there have been 9 heads.

I specify that it is the distribution, not the sample, that has a high probability of reverting to the mean.

beerwm · 9 July 2009

skyQuake said:
As in trade 2 has a higher expectancy than trade 1 due to being a different system ? Otherwise why does the next trade have a higher probability of winning?

trade 2 is only available is trade 1 fails/losses.

I made the probabilities up - just making the point that future trades may have dependancy on previous trade outcomes.

beamstas · 9 July 2009

Mr J said:
It won't necessarily revert to 50% due to variance. The sample may actually get further and further from 50%, although this is obviously not likely. I'm sure I know what you mean, I just don't agree with the way of how you are saying it. The distribution as a percentage is very likely to revert to 50% as the sample grows, but the coinflips themselves don't revert to the mean, and I assume that is why you said you don't recommend betting tails just because there have been 9 heads.

I specify that it is the distribution, not the sample, that has a high probability of reverting to the mean.

Exactly what i mean

Cartman · 9 July 2009

beamstas said:
500flips

3000 flips

Beamers boss walks in ---

"Hey brad, hows work going?" ---

"Really good boss --- I've turned over some serious coin today!"

"Good lad. Keep that up and I might give u a raise!"

(Beamer smiles as he places the 20 cent piece back in his desk drawer and continues pretending to look busy ! ---- lol -----)

We wont tell him what ya been doing all day Brad !!

BurntToast · 9 July 2009

drsmith said:
The odds of hitting an unlicky streak of 6 losses in a row are low but when you do you lose the lot. With the Casino taking it's margin the house will ultimately win. It's simple math.

Odds of hitting 6 losses in a row isn't that low, 1 in 64... you'll hit that during most sessions at the table.

Runs of 10+ are not uncommon, spend an hour in a casino looking at the roulette displays and I'm sure you'll see at least one.

drsmith · 9 July 2009

Mr J said:
Likely, but not certain.

A finite number of bets may yield a winning result for the punter (if their lucky) but certainty is mathematically assured for the casino as the number of bets increases to infinity.

Mr J said:
It may be simple to some, but most don't seem to be able to understand probability and expectation.

Very true.

drsmith · 9 July 2009

BurntToast said:
Odds of hitting 6 losses in a row isn't that low, 1 in 64... you'll hit that during most sessions at the table.

Runs of 10+ are not uncommon, spend an hour in a casino looking at the roulette displays and I'm sure you'll see at least one.

I once saw the same number come up 6 times in a row in roulette. The odds of that are 37 to the power of 5 or about one in 69 million.

skc · 9 July 2009

There is only one game in the casino where you can win money consistently and over sustained period... anyone care to guess?

beamstas · 9 July 2009

Odds of x number of wins/losses in a row at 50% probability

1 - 50.00000%
2 - 25.00000%
3 - 12.50000%
4 - 6.25000%
5 - 3.12500%
6 - 1.56250%
7 - 0.78125%
8 - 0.39063%
9 - 0.19531%
10 - 0.09766%
11 - 0.04883%
12 - 0.02441%
13 - 0.01221%
14 - 0.00610%
15 - 0.00305%
16 - 0.00153%
17 - 0.00076%
18 - 0.00038%
19 - 0.00019%
20 - 0.00010%

beamstas · 9 July 2009

skc said:
There is only one game in the casino where you can win money consistently and over sustained period... anyone care to guess?

poker

skc · 9 July 2009

beamstas said:
Odds of x number of wins/losses in a row at 50% probability

1 - 50.00000%
2 - 25.00000%
3 - 12.50000%
4 - 6.25000%
5 - 3.12500%
6 - 1.56250%
7 - 0.78125%
8 - 0.39063%
9 - 0.19531%
10 - 0.09766%
11 - 0.04883%
12 - 0.02441%
13 - 0.01221%
14 - 0.00610%
15 - 0.00305%
16 - 0.00153%
17 - 0.00076%
18 - 0.00038%
19 - 0.00019%
20 - 0.00010%

The most difficult concept for people to grasp is that these odds apply to the string of results.

If I start tossing a coin now, the chance of getting 20 heads in a row is 0.0001%. If I have tossed a coin 19 times and got 19 heads, the chance of getting a 20th head is 50%.

The 20th toss has no memory of what happened in the first 19. Only the tosser does (and hence they become losers).

The thing about martingale system, even without any table limit, is that the payoff isn't actually that good. What you win on say the 20th head is only 1 unit more than what you've bet in the first 19 rounds. And that's only if the odds are truely 50-50 (which they aren't).

skc · 9 July 2009

beamstas said:
poker

Well done young brad. The only game that doesn't require you betting against the house.

Although I would accept free member parking as an answer.

beamstas · 9 July 2009

skc said:
The most difficult concept for people to grasp is that these odds apply to the string of results.

If I start tossing a coin now, the chance of getting 20 heads in a row is 0.0001%. If I have tossed a coin 19 times and got 19 heads, the chance of getting a 20th head is 50%.

The 20th toss has no memory of what happened in the first 19. Only the tosser does (and hence they become losers).

The thing about martingale system, even without any table limit, is that the payoff isn't actually that good. What you win on say the 20th head is only 1 unit more than what you've bet in the first 19 rounds. And that's only if the odds are truely 50-50 (which they aren't).

House always wins on poker because of rake.
But poker isn't about cards, its all about betting,
So you can actually be skilled at a game that at first glance looks like it is luck based.

You are right, i should have labelled it
"The chance of the next x throws being heads or tails"

Brad

System for the casino??? (baccarat)

treeman

drsmith

drsmith

Mr J

beamstas

Mr J

beamstas

Mr J

beerwm

beamstas

Cartman

BurntToast

drsmith

drsmith

skc

Goldmember

beamstas

beamstas

skc

Goldmember

skc

Goldmember

beamstas

Similar threads

System for the casino??? (baccarat)

Goldmember

Goldmember

Goldmember

Similar threads

Privacy & Transparency

Privacy & Transparency