Intuitive Probabilities – Blackjack and Loss Rebates

I was recently discussing an article with some friends where the author described a guy who was a big winner at blackjack in the Atlantic City Casinos.  The article mentioned that the casinos offered this player a 20% rebate on his losses.  In other words, If he won $1 mm, he got to keep the million, but if he lost $ 1mm, he only had to pay $ 800k.  Now, obviously, this rebate is on the entire session – the overall loss total – not on each individual losing hand.   One of my friends suggested that this was why the guy was a big winner, but I thought otherwise.

It seemed intuitive to me that most people would overestimate the value of this rebate for longer sessions of gambling play – ie, more trials.  I thought that the casino would be able to afford to offer the loss rebate and still have positive expected value (EV) as long as they required the gambler to play a certain number of hands. This isn’t any sort of “EUREKA!” moment – it should be pretty obvious that when you’re playing a game where each trial has negative expected value, if you play enough trials, you will still have negative expected value even if some % of your losses are forgiven – the question is, how many trials does it take?

Obviously, if a casino offers you a 20% loss rebate and allows you to play only one hand, that’s a great gig for you.  You should jump at the chance to do this every single day.  If we assume that you have a 49% chance of winning in a hand of blackjack (and a 51% chance of losing – and in reality, it’s probably more like .495 vs .505 if you play perfectly), then we can easily calculate the EV of a one hand session where we get back 20% of our losses:

1) We’ll win 1 unit 49.5% of the time:  EV: .495

2) We’ll lose 0.8 units (because we get a 20% rebate on our session, which is only 1 hand) 50.5% of the time:  EV: -.404

3) Total EV:  .091 units!

On the other hand, the more hands you play, the more likely the casino is to have positive EV – but how exactly do we quantify this?  Well, I did it by brute force, using an Excel spreadsheet and the included “BINOMDIST” function to calculate the full binomial distribution outcomes for a given number of trials (10, 20, 50, 100, 250, 500, 1000, 2000).  In other words, for the 10 hand trial, I calculated the probability of each of the 11 possible distribution outcomes (0 wins, 1 win, etc… up to 10 wins).  Then I took the weighted sum of the total winnings, and did it again assuming a 20% rebate on losing sessions.

Assuming a 49% chance of winning each hand, which probably corresponds to a decent basic strategy player who makes the occasional mistake, and assuming a flat bet of 1 unit per hand, we get:

 

P (Player wins each hand) 49%
Number of hands EV for Player (in units) EV for Player with 20% loss Rebate
10 -0.2 0.0666
20 -0.4 -0.0062
50 -1 -0.3330
100 -2 -0.9882
250 -5 -3.1772
500 -10 -7.0413
1000 -20 -14.9891
2000 -40 -31.0930

 

Interestingly, after only 20 hands, the gambler has lost the expected value edge that came from his 20% loss rebate.

If we bump up the player’s odds of winning each hand to 49.5%, the data looks like this:

 

P (Player wins each hand) 49.50%
Number of hands EV for Player (in units) EV for Player with 20% loss Rebate
10 -0.1 0.15622
20 -0.2 0.17275
50 -0.5 0.11278
100 -1 -0.10013
250 -2.5 -0.97397
500 -5 -2.67237
1000 -10 -6.35241
2000 -20 -14.08124

 

You can see that it now takes almost 100 hands for the Casino to overcome the 20% loss rebate and get back to being EV positive.

I would guess that many people would find it counter-intuitive that the casino starts to accrue an EV benefit in this situation after such a “small” number of trials.

-KD

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