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Ask the Wizard #207

The Wynn invited me to a slot tournament with the following prize structure.

1st place: $1,000,000
2nd place: $150,000
3rd-6th place: $25,000
7th-8th place: $20,000
9th-50th place: $5,000

The cost is $25,000, and the tournament is limited to 50 players. It is easy to see the expected win is $30,000. However, it is a huge long-shot. What would be the required bankroll for entry to be a sound bet under the Kelly Criterion?

Anonymous

The Kelly Approximation is the advantage divided by the variance. The possible outcomes are a win of 39, 5, 0, -0.2, and -0.8 times the bet amount. The advantage is (1/50)×39 + (1/50)×5 + (4/50)×0 + (2/50)× -0.2 + (42/50)×-0.8 = 0.2.

The variance is Expected(win2) - (Expected(win))2 = (1/50)×392 + (1/50)×52 + (4/50)×02 + (2/50)× -0.22 + (42/50)×-0.82 − 0.22 = 31.4192

So, the approximate optimal Kelly bet is 0.2/31.492 = 0.0063655 times the bankroll. For a full entry of $25,000, the required bankroll would have to be 25,000/0.0063655 = $3,927,400.

However, for large bets like this, I think it is worth the time to find the exact optimal Kelly bet. Next, find the bet size b, which maximizes the expected log of the bankroll after the tournament, as follows.

Log of bankroll after tournament = (1/50)*log(1+39×b) + (1/50)*log(1+5×b) + (4/50)*log(1) + (2/50)*log(1-0.2×b) + (42/50)*log(1-0.8×b)

There is no easy way to solve for b. Personally, I recommend the "Goal Seek" feature in Excel. The answer will come out to 0.0083418. So, the exact Kelly bet should be 0.0083418 times your bankroll. To justify the $25,000 entry fee, your bankroll should be $25,000/0.0083418 = $2,996,937.

I absolutely love your site. I enjoy the strategies and probability discussions as much as, or more than, the actual gambling! I was playing six-deck Blackjack in a St. Louis casino recently. After playing a shoe, the cards were returned to the auto shuffler, which indicated a card was missing. The dealer proceeded to deal the next shoe while the floor person inspected the returned set of cards. Upon completion of this shoe, the missing card from the previous shoe (a king) was found in the un-dealt portion of the second shoe.

Assuming this King was the bottom card and was left in the shuffler, it would have been in play in this first shoe (the cut was in rear portion of the deck). How much of an additional advantage did the house gain on me with this mistake?

Justin from St. Louis, MO

Thank you for the kind words. I’m going to assume the dealer hits a soft 17, and double after a split is allowed. According to table D17 in Blackjack Attack by Don Schlesinger, removing one ten per deck increases the house edge by 0.5512%. Dividing that by six, for the six-deck game, the effect is an increase in house edge of 0.09%.

I'd like your advice on a blackjack coupon. As I understand the rules, the coupon doubles any win, up to $25, and can be presented any time. If I bet $16.50, and wait for a blackjack to use it, the coupon will double the blackjack win of $24.75. Or should I bet $25, and use it on the first win of any kind? What is the expected loss both ways? Please assume a house edge of 0.64%.

Jim from Dallas, Texas

First let's calculate the expected loss if you bet $16.50, and wait until a winning blackjack to use the coupon. The probability of a player blackjack is the number of aces × number of tens / combinations of ways to choose two cards out of the 312 in the shoe. That comes to 24×96/combin(312,2) = 0.0474895. If both of you have a blackjack, the coupon does you no good. Assuming the player has a blackjack, the probability of a dealer blackjack is 23 × 95 / combin(310,2) = 0.045621. So, the probability of the player having a winning blackjack is 0.0474895 * (1-0.045621) = 0.045323, or once in 22.06 hands. So, your way of playing 22.06 hands at $16.50 each would have an expected loss of 22.06 × $16.50 × .0064=$2.33.

Next, let’s calculate the expected loss if you bet $25, and wait until the first win to use the coupon. The probability of any win is 42.42%, as found in my blackjack appendix 4. This is not exactly the applicable statistic for this situation, due to complications in splitting, but close enough. So, the expected number of hands to play to have a winning hand is 1/0.4242 = 2.36. The expected loss of betting 2.36 hands of $25 each is 2.36 × $25 × .0064=$0.38, which has a cost 84% less than waiting for a blackjack.

On two recent visits to the baccarat tables the results were definitely player biased. Please tell me if these results would be considered within two standard deviations of the expected results for bank and player. I have eliminated tie hands.

Session I
Player wins: 282
Banker wins: 214

Session II
Player wins: 879
Banker wins: 831

Arthur from Wayne, New Jersey

From my baccarat page, we see the probabilities in the usual 8-deck game are:

Banker: 45.86%
Player: 44.62%
Tie: 9.52%

Skipping the ties, the probabilities for the banker and player are:

Banker: 45.68%/(45.68%+44.62%) = 50.68%.
Player: 44.62%/(45.68%+44.62%) = 49.32%.

The total number of hands in session I was 282+214 = 496. In session I the expected number of player wins is 49.32% × 496 = 244.62. The actual total of 282 exceeds expectations by 282-244.62 = 37.38.

The variance for a series of win/lose events is n × p × q, where n is the number is the sample size, p is the probability of winning, and q is the probability of losing. In this case, the variance is 496 × 0.5068 × 0.4932 = 123.98. The standard deviation is the square root of that, which is 11.13. So, the total player wins exceeded expectations by 37.38/11.13 = 3.36 standard deviations. The probability of results that skewed, or more, is 0.000393, or 1 in 2,544.

Using the math method for sample II, the probability is 0.042234. If you combine the two samples into one, the probability is 0.000932. About 0.1% is not enough to be "definitely player biased." If you still think the game isn’t fair, I would collect more data, for a larger sample size.

A fellow employee swears his mom is on a 25-year video poker winning streak. She makes four trips a year to Vegas and always wins at least a $1000 with a $400 buy-in. He says she usually wins $10,000. He is upset at my lack of faith in her luck. He wants to bet me his mother will be ahead after a four-hour session. Should I take this even-money bet?

Anonymous

As long as she is flat betting at a steady rate, yes, by all means take the bet. Either she is using some kind of worthless progression, or this is second-hand exaggeration. This got me to thinking, what would be the optimal number of hands for your friend’s side. Assuming 9/6 Jacks or Better, and optimal strategy, the probability of being ahead is maximized at 136 hands, with a probability of 39.2782%.