Is this consistant with a fair game?


Dormant account
Sep 22, 2007
Copenhagen Denmark
Hello everyone. I just joined this VERY nice site, and have BlackJack question.

I've read the rules, and i will not post the name of the internetcasino yet, as i wan't your input of the possibility of this happening.

On Saturday I played a BlackJack session on an internet-casino. I doubt that the result i got, consist with a fair game.

I played 236 hands, with an average stake of 280 Euro.

I played a perfect game, which means I draw, double and stand as I should to obtain the best possible result.

The expected loss with the rules at the casino is 0.50168%. This means I am expected to lose 280 *0.0050168 = 1.404704 euros per hand

Playing 236 hands I am expected to lose 331,51 euros. The actual loss was 8.500,00 euro.

First card distribution:

Dealer: Expected

A: 27 18
T: 77 72
9: 11 18
8: 15 18
7: 18 18
6: 12 18
5: 17 18
4: 21 18
3: 22 18
2: 16 18

The interesting part here is that the dealer got the best card, an ACE, 50% more often than expected.

On the same note the dealer was expected to get the worst card, a SIX, 50% more often than he actually did.

Out of the 236 hands played, the player busted 52 times, leaving a sample of 184 hands, where the dealer had to draw cards.

The dealer is expected to bust 31% of all hands.

Expected busts: 184*0.31= 57 times

Actual busts = 49

The dealer was expected to bust 14% more often than he actually did.

The probability of a dealer bust with first card dealt:

Total hands = 236-52(busted hands by player)=184 hands sample total.

Hands busted by dealer with (3)
Times dealt = 22
Expected busts = 22*0.3756 = 8.26
Actual busts = 8

Hands busted by dealer with (4)
Times dealt = 21
Expected busts = 21*0.4028 = 8.46
Actual busts = 7

Hands busted by dealer with (5)
Times dealt = 17
Expected busts = 17*0.4289 = 7,2
Actual busts = 8

Hands busted by dealer with (6)
Times dealt = 11 (12 including blackjack for player)
Expected busts = 11*0.4208 = 4.6
Actual busts =3

Hands busted by dealer with (7)
Times dealt = 18
Expected busts = 18*0.2599 = 4.68
Actual busts = 3

Hands busted by dealer with (8)
Times dealt = 15
Expected busts = 15*0.2386 = 3.58
Actual busts = 3

Hands busted by dealer with (10)
Times dealt 77-24 (not played because of player bust)=53
Expected busts = 53*0.2143= 11.35
Actual busts = 9

Except for the (5), which is within 12%, the dealer qualifies too much on all other hands.

Especially the the (6) and (7) is more than 50% too high.

All the numbers are in favour of the dealer here. Lets look at the players hands:

Player: Expected

A: 19 18
T: 86 72
9: 14 18
8: 16 18
7: 16 18
6: 17 18
5: 16 18
4: 24 18
3: 15 18
2: 13 18

Total 236

The numbers are fairly okay. The player received the TENS about 20% too much, and the (4) 35% too much.

With the above average number of TENS received the following numbers seems VERY strange:

Two-card count frequencies

Expected 236*0.048 = 11.32
Received 11 times

Hard standing (17-20)
Expected 236*0.3 = 70.8
Received 55 times

Even though the player received too many TENS, he should have had hard standing (17-20 on the first two cards) 22% more often than he did.

The player was expected to bust 32 times during the session, but actually busted 52 times.

This number is 62.5% too high.

I would be glad if someone could give me some input about this.
Was i unlucky, or deosn't this consist with a fair game.

Best regards


Dormant account
Mar 30, 2007
The bar
I understand your frustration, I lost once 50 bets in 250 flatbets of 1, but variance is part of the game. ~250 hands are just not enough hands to make a conclusion out of it.


Meister Member
Mar 9, 2006
Once I lost 33 units within 100 hands, you lost 30 units within 236 hands, so I can feel your pain. But yours is not an extremely big loss in terms of units.

The first cards show nothing special, the corresponding probabilities for aces and sixes are ~1/39 and ~1/13 respectively.

The dealer busts rates are ok considering the sample size.

Same applies for the hard standing hands.

It's not so trivial to analyze your bust frequency, because I don't know the hand composition and the number of hands for this statistic, because for instance drawing two 2s to 12 counts twice, and the number of trials is very important. For instance 52 busts from 68 trials is unlucky, 52 busts from 64 trials is extremely unlucky, while 52 busts from 58 trials is probably an unfair game.


Dormant account
Jun 11, 2006
Planet Earth
The player was expected to bust 32 times during the session, but actually busted 52 times.
This sounds odd. I don't know player bust stats off hand . Assuming the expected is indeed 32/236, then 52/236 is very unlikely to occur. The other stats seem reasonable individually, but combined they may be unlikely.

Two ways to evaluate such issues are using the return & variance calc on my site (will have return & variance simulator as well soon), and treating results as binary variables with SD = SQRT(n*p*(1-p)). If you are familiar with stats or Excel functions, there are some other useful options as well.

I am curious about with which software it occurred.

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