Flawed Lottery Systems by Steve Player

Posted by BigJer on July 20, 2000.

In Reply to: Martingale Pick-3 Lottery, Anyone? posted by Ion Saliu on July 19, 2000.

Thanks Ion,

I saw that in the banner ad the was on the USA Today lottery results page and also that

there was just a minor note about having to pick ONE digit for his system to work but

as any one who plays pick three knows the one digit can be very hard to do if your

whole system depends on it , that is why I like to play boxed back ups on any thing that I play straight.

It looks to me like road to riun to use a system that costs that much to play

BigJer

: • BigJer posted a while ago asking if I had heard of “Steve Player”. Later, I saw in rec.gambling.lottery a thread on “Steve Player’s pick-3 system”. I think I got it right. The foundation of such a lotto system is choosing one digit in the pick-3 game. Something like a favorite digit. I put the “system” to a test using LotWon lottery software.

I ran POWER-3.EXE without any filters. I pressed N to all the 7 prompt screens. Then I selected the option “1 favorite digit”. The program always generates 271 combinations with one favorite lottery digit. Now, you are supposed to play all those 271 combinations. Let’s say we play them for $1 each. The cost; $271. If we win the first time, we make a profit of $500 - $271 = $229. But what guarantees you get the “fixed” digit right from the first try? Let’s use here the fundamental formula of gambling (FFG). The probability p = 271/1000 = 0.271. In order to have a 50% degree of certainty (DC) that we pick the right digit we need 2.2 tries (drawings).

That is, there is a 50% chance we will lose more than 2 drawings in a row. Since we lost our first drawing, only a Martingale will recuperate the loss AND offer a profit. Now, the 2nd drawing, we double the amount: $271 x 2 = $542. We already lost $271, so the total cost is now $813. We played $2 a ticket, so the win will be $1,000. Profit: $187 over a period of 2 days; profit per day: $93.5. What happens if we lose 3 days in a row? We must play now $4 per ticket: $271 x 4 = $1,084. We add to this amount what we lost the previous 2 days. Total cost: $1,897. The win will be $2,000, the profit $103 or $34.34 daily profit. It is very plausible to lose one more time. The Martingale will be 8 x $271 = $2,168. Total cost for 4 drawings: $4,065. The win will be 8 x $500 = $4,000. From this point on, we always lose, even if we hit the winning combination! So, a power-2 Martingale is not sufficient to apply this “lottery system”. We probably need to apply a power-3 Martingale: 1-3-9-2.

Ion Saliu

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