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1. Which of these bets do you like?

Let's say that there are 100 different bags of marbles and each bag has exactly 100 marbles in it. Some of the marbles are blue and some are red, there are no other colors. Each bag has exactly 100 marbles in it. I'll tell you exactly how many blue marbles are in each bag and then I will give a price on whether or not you can blindly pull out a blue marble. E.g. If the first bag had 50 blue marbles in it(and 50 red) and the price was -110 for you to blindly pull out a blue marble... would you take that bet?

Tell me which bets below you would take. The first number is the number of blue marbles in the bag(each bag has 100 marbles in it) the 2nd number is the price you are being offered to blindly pull out a blue marble. Which play(s) below would you take?

A. 51 -105
B. 30 +225
C. 87 -750
D. 23 +360
E. 80 -450
F. 12 +675
G. 73 -290
H. 16 +590
I. 63 -175
J. 95 -2000
K. 3 +3800
L. None of the above

2. A and I for me.

3. and G also

4. K..

5. Basic method to eval +EV.

A
B
C
E
F
G
I
J

Many of these have barely a positive return.
The method for eval is:

If a + line use: 100/(+line+100) = Win %
If a - line use: - line/(-line -100) = Win %

Compare win % to odds of blue marble in bag. If win % > odds --> take the bet.

I've had a few so my math could be off.

6. D, H and K
Points Awarded:
 JohnGalt2341 gave Optional 1 Betpoint(s) for this post.

7. S#!T... didn't think it through.
Win % < odds for blue is a good bet.

8. republican - “ L since I don’t trust communists”

9. Nice going Opti!

On \$100 units I have

D = + 1080
H = + 1040
K = + 1700

Now, I know the math is with you on these dogs but will they bark?
You lose if you only get 2 on K
win peanuts if D only gets 22
lose if H only gets 14

My question is would you bet these even with the math behind you as it is very possible to lose real fast with only a little change in the expectancy. I see it as risking too much to win too little.

10. Originally Posted by BuckyOne
Nice going Opti!

On \$100 units I have

D = + 1080
H = + 1040
K = + 1700

Now, I know the math is with you on these dogs but will they bark?
You lose if you only get 2 on K
win peanuts if D only gets 22
lose if H only gets 14

My question is would you bet these even with the math behind you as it is very possible to lose real fast with only a little change in the expectancy. I see it as risking too much to win too little.
If someone offered you those sort of odds on an ongoing basis you should bet them hard every single time.

As if you could do this forever you have ZERO chance of losing.

But in a one off game just because you are getting paid 10% higher odds than you should be it does not mean you are any more likely to win or lose. So need toa ssess the risk against your bankroll still in real world.

It's a good thread because in reality you should be thinking about every single bet you make this way. The stumbling block comes in when it is not as clear cut as a bag of colored balls. We can only guess an expected probability in sports. But the people you see around here being called sharps who really are sharp are mostly basing their capping on this starting point.

BTW, the easiest way to work this out is to use the SBR Odds Converter tool

Just put the % of balls in the "Implied Probability" box and it will tell you the correct odds for that outcome.
Nomination(s):

11. Honestly, I’d probably bet all of them. Except the one that is seemingly one of the smarter ones in K. I would hammer J until the wheels fell off. Square as can be over here. As a wise man once said, “there ain’t no juice when you win!”

12. A at -105 is fine for me. I would take that with 51% chance of winning.

13. Actually, I would pound C instead of J.

14. i'd max bet them all

I have a lifetime experience pulling out blue balls

15. E for me

Let's say you could have 1000 turns at picking out a marble and were allowed to risk \$1 each time as your bet.

You would want to pick the option that gives the most profit in that circumstance?

Not the one that "feels" less risky, like Option A appears to for some people?

Just to compare the two extremes, A and K;

A. 51 -105

51 blue balls out of 100 in bag = 51% chance of selecting a winner.

51% probability converts to odds of -104

He is offering -105 which is 99.12% of reality.

So after 1000 bets you should be sitting at an average result of -\$8.76 loss

K. 3 +3800

3 balls = 3%

3% converts to +3233

Offer is +3800 which is 117.53% of reality

So after 1000 bets your average position would be +\$175.38 profit

There will be more variance in the result of option K, but the more bets the less that becomes. So if you are betting long term, 1000s of bets per year, would you really feel better (or safer) betting option A type bets as the norm?

17. Originally Posted by RudyRuetigger
or you can you just do it on a great feature provided by sbr

18. Originally Posted by Otters27
E for me
E. 80 -450

80% probability = -400

-450 offer is 87.5% of reality.

You would be down an average of -\$125 after 1000 x \$1 bets.

19. Originally Posted by RudyRuetigger
or you can you just do it on a great feature provided by sbr

Originally Posted by Optional
BTW, the easiest way to work this out is to use the SBR Odds Converter tool

Just put the % of balls in the "Implied Probability" box and it will tell you the correct odds for that outcome.

20. why are people still guessing answers?

when i am drunk i will bet any and all

21. D seems to be a good percentage odds bet.