This week I tried out a little strategy, a small Martingale.
I programmed my computer to play a $1 5-spot, with a house percentage of 28 and payouts of $1, $9, and $811 for catches of three, four and five of five, respectively. With an initial bankroll of $100, each player did a $1 5-spot each game with a provision to stop if hitting a solid five, or going bust.
In addition, the player would Martingale a 2-spot played for a multiple of $1 that would regain the initial $100 bankroll if hit. If the player was down $25 with a remaining bankroll of $75, he would play his $1 5-spot, plus a $3 deuce giving him $75-$4+$36 or $107 if he hit the deuce.
I set the program for 100,000 players with the initial play $10,000,000 of keno. As we might suspect, the great majority went broke, but a higher proportion (1 in 23) hit a solid five.
Now, 1 in 23 doesn’t sound too bad, but a couple of things are wrong with this picture. I set up this demonstration so the player either goes broke or wins $811. With 1 chance in 23 of doing this, his mathematical expectation is $811/23, or $35.26!
Remember, the house percentage is about 28 percent, so our strategy has unfortunately doubled the house keep! If we ask the 100,000 players to play a straight 5-spot for 100 games, we’d expect to see some 6,447 solid fives on $10 million action, instead of only 4,314! And this is ignoring the extra money we won on pushes and four out of fives.
Why? Well, our average wager is $2.88 per game for 6,688,818 games and we spend 65 percent merely playing deuces. If we stick to straight 5-spots, all our wagers go towards hitting a solid five. The Martingale, which is supposed to protect our bankroll and lengthen staying power, has reduced the number of games by 33 percent. Thus, less chance of hitting.
The basic strategy of playing to win or bust might be faulty. We’ll look into that next week.
If you have a keno question please write to me c/o GamingToday or contact me by email at [email protected].