Casino gambling, the ultimate strategy
What is the best way to lose money at a casino? This paper will give the answer.
A very common gambling plan is to play a game with a fixed probability of winning by taking $ n to the casino, betting the same amount on each play of the game, and leaving when you have lost all your money or have attained your goal of $N. If you bet $1 per play, your chance of ruin is (see, for example [1, ch. XIV])
The larger the size of the bet, the smaller the probability of being ruined. Suppose that you have a stake of $9000 and wish to win $1000. This can be modeled with $1 bets by taking n = 9000 and N= 10000, with $10 bets by taking n = 900 and N= 1000, with $100 bets by taking n = 90 and N= 100, and so on. If you play craps, where p = .4929, for $1 bets, the probability of ruin is 1 (to 14 decimal places), .941 for $10 bets, and only .262 for $100 bets. We will reduce the probability of ruin further by using what might be called an ultimate strategy. The ultimate strategy is always to bet the maximum possible amount needed to achieve the goal of $N.
With n = $9000 and N= $10000, the ultimate strategy would require a first bet of $1000. If that bet lost, the second bet would be $2000. If that also lost, the third bet would be $4000 and if that were also lost, the next bet would be all that is left, $2000. If this bet is lost, the player is ruined; if it is won the next amount bet is $4000, and so on.
The possibilities are shown schematically in Table 1. From the table, when using the ultimate strategy the probability of being ruined is, where alpha =p^sup 2^q^sup 2^,
In Table 2, we show how a craps gambler could expect to fare using some constant bet sizes and the ultimate strategy. The game of craps has an average of 3.4 tosses of the dice per game. The time per game varies, depending on the croupiers, the number of players, and the variety of bets placed. For the table we assumed the average time per game was 1 minute.
An 89.47% chance of winning $1000 is the very best mathematics has to offer for this $9000 vacation. If we could get a 90% chance of winning, the game would be fair, and no casino offers fair games.
For initial stakes and goals other than $9000 and 51000, the ultimate strategy would lead to schemes different from that in Table 1, but the same methods would produce the minimum probability of being ruined.
The ultimate strategy should be used sparingly. If a player were to spend just 93 minutes in the casino, perhaps accumulated over a period of years on repeated holidays, the player’s expected loss would be $2437. If the player bet in fixed amounts of $1000, the expected loss would be only $1300. To lose $2437 making constant $10 bets would take, on the average, around 290 hours, the equivalent of more than seven weeks of full-time work, so the ultimate strategy has the advantage of minimizing boredom as well as the probability of ruin.
1. W. Feller, Introduction to Probability, vol. 1, Wiley, 1958.
Dennis Connolly (firstname.lastname@example.org), B.Sc. University of Sydney, Australia, came to Canada and signed on at the University of Lethbridge for two years only, planning to return to Australia in 1969. Two years led to four, to…thirty-two and counting at the U of L with a Ph.D. in measure theory from the University of York, England, along the way. To become a Canadian citizen, he learned to skate but was disappointed not to be drafted into the NHL. He loves the Rockies, skating on ponds, and all that empty Alberta space. But when digging out of snow drifts and scraping ice from the windshield he dreams of retiring to Coonabarabran or Cairns on the Great Barrier Reef where it’s always summer.
Copyright Mathematical Association Of America Sep 1999
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