What is wrong with this scenario?

[Survivor]

I am relatively new to sportsbetting and gambling in general so bare with me if this is a stupid idea.

What would be wrong with randomly picking a dog in mlb and for example risking $10 on that team. If that team looses pick the next available mlb dog and risk $20. If that looses, pick the next available dog and risk $40 and so on. And once you get a winner, you start over at $10 again.

With this pattern, you would be risking $1920 after your 8th loss in a row. This is the most I would want to risk. What are the odds that I would loose 9 plays in a row randomly picking dogs over the course of many days...

I am thinking about doing this, but if someone can talk me out of it please do!

[REALITY]

SURVIVOR,

PROFESSIONAL SPORTS TEAMS CAN LOSE 10 GAMES IN A ROW.

TRUST ME, YOU PICKING 9 LOSERS IN A ROW IS NOT THAT FAR FETCHED DESPITE THE 512 TO 1 ODDS.

IS A MAXIMUM RISK OF 1,92O TO WIN 10 WORTH IT?

I THINK NOT...

REALITY




[This message has been edited by REALITY (edited 07-09-2000).]

[AussieVamp2]

If picking random underdogs it is perhaps also better odds than 512 to 1 as well

[REALITY]

512 TO 1 ARE THE ODDS OF PICKING THE WINNING TEAM REGARDLESS OF THE PRICE IN THIS SCENARIO.

REALITY

[Survivor]

"IS A MAXIMUM RISK OF 1,92O TO WIN 10 WORTH IT?"

If the random mlb dog selection had moneline of +130 and and I was risking $1,920, that would be a payoff of 2,496, or a profit of $575. Not $10. Am I missing something?




[This message has been edited by Survivor (edited 07-09-2000).]

[bbuster]

what you are missing is the 2486 or so you would have lost on the 1st 8 losers, so by the time you finally win a game your profit would not be comparable to your risk if you never win a game before you go broke.

[This message has been edited by bbuster (edited 07-09-2000).]

[zippy]

Most beginning gamblers come upon this doubling strategy, along with many variants, sooner or later. Does it work? Not really. What such money management strategies do is increase your chances of winning while proportionately decreasing the value of the win. You’ll almost always win with the strategy you’ve outlined, but the wins will be tiny and the rare loss you suffer will be huge. How huge? Just huge enough to knock your long run expectation back to precisely where it would have been with any other betting strategy.

Interestingly enough, although I frequently hear this strategy talked about, the opposite type of strategy seems to be far more common, at least based on many years of observations at blackjack tables. Most people prefer to “double up” after a win, rather than after a loss. That way, almost every run results in a one unit loss, but once in awhile, you break through with a huge win. Again, in the long run, it evens out to whatever your expectation would have been anyway.

The latter strategy just seems to be more appealing to the typical player. It’s as if you were planning a Vegas trip, and I told you: “I have two strategies you can use. One virtually always results in your winning about $500 for your trip, but once in a blue moon you’ll drop $50,000. The other virtually guarantees you’ll lose about $500 for your trip, but once in a blue moon you’ll win $50,000.” Apparently most people would opt for the latter strategy. The average non-pro gambler pretty much expects to lose anyway, so a small loss is quite acceptable to him, especially if it comes with a tiny chance of getting rich. A virtually guaranteed win, but with a small chance for a catastrophic loss, is much more stressful.

Actually I had a friend who tried just the strategy you’re suggesting. He was playing blackjack, and after a few successful runs where he won the $10 each time, he hit a losing streak. One hand after another he lost, and one hand after another he pushed out a stack of chips twice as high as the last one. As the losses mounted, he was sweating profusely, and watching the cards more and more intently. He reached the point where he had nothing left in reserve. With his shaking hands, he had placed the remainder of his money on the table for one last hand. As he told me later, “As I watched the cards coming out for that last hand, it dawned on me that I had about a 50-50 chance where I could either have a monstrous loss that would force me to drop out of college and wreak untold havoc on my life, or I could win $10.”

He won the hand.

He never used that strategy again.

[wintermute]

Trust me. Don't do it. Betting dogs in baseball is a good idea because the favourite is almost always overbet. If you shop around for lines, bet only dogs and have even a smidgeon of handicapping skill you will almost surely win money over the long haul if you bet a fixed amount per game.

If you do make money you can gradually increase your bets over time.

Good luck.

'mute

[Survivor]

Thanks for the input guys. I looked over my figures again and noticed I made an error. With this strategy, a person would be risking $1,280 on their 8th bet. Last question. Reality(or anyone that knows the answer), what are the odds of loosing 8 bets in a row on randomly selected mlb dogs over the course of many days?

[zippy]

Survivor:

The answer to your question depends in part, of course, on what degree of underdogs they are.

Assume for the sake of argument that you are betting on underdogs that have a 40% chance of winning. Your chances are:

Lose first pick: 60%
Lose first two: 36%
Lose first three: 21.6%
Lose first four: 12.96%
Lose first five: 7.776%
Lose first six: 4.6656%
Lose first seven: 2.79936%
Lose first eight: 1.679616%

So approximately once in each sixty runs, you’ll lose eight in a row before you get that one win.

[Survivor]

I am going to assume mlb dogs of +130 win 40% of the time. If this is not correct please let me know(If anyone has the breakdown on the various moneylines and their winning %'s that would be great.)

risk 10 - chance of loosing 60%
risk 20 - chance of losing 36%
risk 40 - chance of losing 21.6%
risk 80 - chance of losing 12.96%
risk 160 - chance of losing 7.776%
risk 320 - chance of losing 4.6656%
risk 640 - chance of losing 2.79936
risk 1280 - chance of losing 1.679616
risk 2560 - chance of losing 1.10077696%

the average risk amount is $284 @ +130 = $369, or $85 profit.

If I do this 60 times I will:

lose $2560 once and
win $85(avg.) 59 times
59 * $85 = $5015

$5015
-$2560
-------
$2455 profit(avg.) every 60 times through?

At the point that I lose the $2560, I would just start over at $10. Is there a flaw in this system that I am not recognizing?

I would appreciate any feedback. Thanks!




[This message has been edited by Survivor (edited 07-10-2000).]

[zippy]

I went through the math very quickly, so I apologize in advance for any calculation errors I might have made, but here’s what I come up with for your scenario:

40% of the time, you’ll win the first bet, for a profit of $13.

24% of the time, you’ll lose the first bet and win the second, for a profit of $16. (-$10, +$26)

14.4% of the time, you’ll lose the first two bets and win the third, for a profit of $22. (-$10, -$20, +$52)

8.64% of the time, you’ll lose the first three bets and win the fourth, for a profit of $34. (-$10, -$20, -$40, +$104)

5.184% of the time, you’ll lose the first four bets and win the fifth, for a profit of $58. (-$10, -$20, -$40, -$80, +$208)

3.1104% of the time, you’ll lose the first five bets and win the sixth, for a profit of $106. (-$10, -$20, -$40, -$80, -$160, +416)

1.86624% of the time, you’ll lose the first six bets and win the seventh, for a profit of $202. (-$10, -$20, -$40, -$80, -$160, -$320, +832)

1.119744% of the time, you’ll lose the first seven bets and win the eighth, for a profit of $394. (-$10, -$20, -$40, -$80, -$160, -$320, -$640, +$1664)

1.679616% of the time, you’ll lose the first eight bets, for a loss of $2550. (-$10, -$20, -$40, -$80, -$160, -$320, -$640, -$1280)

40% of +$13 = +$5.20.
24% of +$16 = +$3.84
14.4% of +$22 = +$3.17
8.64% of +$34 = +$2.94
5.184% of +$58 = +$3.01
3.1104% of +$106 = +$3.30
1.86624% of +$202 = +$3.77
1.119744% of +$394 = +$4.38
1.679616% of -$2550 = -$42.83
Expected average result of each run is about -$13.22.

That rare $2550 loss more than wipes out all your small wins.

Of course, under the specified conditions, you’d lose betting a flat amount each time also. Assume you made 100 bets of $100 each. You’d win 40 of them, for a profit of $5200, and lose 60 of them, for a loss of $6000, putting you down $800 overall.

The moral of the story is, if you’re betting with the worst of it, you can’t change that by manipulating the amounts of your bets.

If you bet those +130 dogs when you have good reason to believe they have a better than 43.5% chance to win, and you don’t bet them otherwise, then that’s what really matters. Concentrate on that, rather than on these clever ruses to get around the odds. They never turn out as good as they look.

[SLAM DUNK]

Looks like Zippy beat me to it, but double confirmation that Zippy and I got the math correct.


Survior, the correct math is as follows assuming +130 on dogs that win at a 40% rate. Fairline +150. The take in Baseball is a lot less, so you would do better, but there is not a positive expectation as you thought.
I have listed the chance of losing game # x
the chance that you will win that game,
multiplied that by the outcome if you win that game(eg-$10 to win game 1 at +130 is $13). Zippy said it well. The bottom line is for each game you have a positive or negative expectation, depending on the strength of your handicapping. No matter how you bet, you cannot make a series of bad bets profitable in the long run, no matter what strategy you use. I do not think there is anyone that has gambled seriously for a long time that would advise progression. Like said above, once you get to the last game and sweat it out,(if you have the cool to objectively pick the game) I doubt you will ever do it again. Anyways here is the math, if you want to substitute +140 or +150 etc. just recalculate the profits.(eg game 1 at
+140 will be +14 and the math would be .4 *14
or 5.6...)

The chance of winning game

#1 .40 *13=5.2
#2 .24 *16=3.84
#3 .144 *22=3.17
#4 .0864 *34=2.938
#5 .05184*58=3.00
#6 .0311 *106=3.29
#7 .0186 *202=3.75
#8 .0112 *394=4.41

total expected profit is $29.60

there is a 1.68% chance you will lose all 8 and lose $2550.
thus .0168*-2550= -$42.84

Net expected LOSS $13.24 per 9 game trial.

My hope is that after reading this you will reconsider any thoughts you had of doing progression, because who is to say that your 1st 9 game trial will not be the losing one.



[This message has been edited by SLAM DUNK (edited 07-10-2000).]

[This message has been edited by SLAM DUNK (edited 07-10-2000).]

[pay2play]

I believe the win rate for +130 MLB bets is 43%, not 40%. If you shop for dog prices, the bookie's hold should basically be zero because you can almost always find 5 cents worth of price difference. And fave's eat most of the juice in this sport. But assuming a dime line and a 1% hold on the dogs, .43 is the expected win rate.

[pay2play]

Not that that really changes Zippy and Slamdunk's basic point. It just brings you back to what Wintermute said.