Straight Bet or Parlay (Math Problem)

**pokernut9999** · 04-10-08, 03:57 PM

I would think in this case if you had to bet this example a $500 parlay would come out ahead in the long run.

**Ganchrow** · 04-10-08, 04:05 PM

Originally posted by pokernut9999

I would think in this case if you had to bet this example a $500 parlay would come out ahead in the long run.

I think you may have misunderstood the question:

If you could only bet the two straight bets OR the corresponding parlay (but not both, in other words you may NOT bet the singles AND the parlay) how high would the edge on each bet need to be for the parlay to represent a superior choice for a full-Kelly bettor?

**pokernut9999** · 04-10-08, 04:17 PM

Originally posted by Ganchrow

I think you may have misunderstood the question:

If you could only bet the two straight bets OR the corresponding parlay (but not both, in other words you may NOT bet the singles AND the parlay) how high would the edge on each bet need to be for the parlay to represent a superior choice for a full-Kelly bettor?

I live in a trailer and never got past 5th grade , give me a break.

**donjuan** · 04-10-08, 04:53 PM

Anything over 9.91%.

**pokernut9999** · 04-10-08, 04:58 PM

Originally posted by donjuan

Anything over 9.91%.

I thought 5.5 % so I must be wrong.

**Ganchrow** · 04-10-08, 05:14 PM

Originally posted by donjuan

Anything over 9.91%.

While an edge of 9.91% would be sufficient for preferring the parlay to the straight bets, the indifference point is still a bit lower.

But close.

**HedgeHog** · 04-10-08, 05:14 PM

Are we to assume the straights are at -110 or is it -105? Also are the parlays at the standard 13/5?

**pokernut9999** · 04-10-08, 05:15 PM

Originally posted by Ganchrow

While an edge of 9.91% would be sufficient for preferring the parlay to the straight bets, the indifference point is still a bit lower.

So 5.5 % may be close ?

**durito** · 04-10-08, 05:18 PM

Originally posted by HedgeHog

Are we to assume the straights are at -110 or is it -105? Also are the parlays at the standard 13/5?

.

let's say you're considering two uncorrelated bets each at -500, each with the same edge.

**HedgeHog** · 04-10-08, 05:20 PM

Originally posted by durito

.

Oops my bad. Thanks, Durito.

**HedgeHog** · 04-10-08, 05:24 PM

Roughly 3.5% ???

**Ganchrow** · 04-10-08, 06:12 PM

As previously warned:

Originally posted by Ganchrow

Kind of a pedantic question question, but someone brought it up with me in an e-mail.

Given two uncorrelated binary bets at equivalent odds with equivalent win probabilities, the no-parlay Kelly stake on each bet is given algebraically by:

K_np = (1 - p^2 - 3*w + 2*p*w - 2*p^2*w + 2*p*w^2 - p^2*w^2 + sqrt(8*(-1 + w)*w*(-1 + p + p*w) + (-1 + 3*w - 2*p*w*(1 + w) + p^2*(1 + w)^2)^2))/(4*(-1 + w)*w)

where p is the win probability on each bet, and w the decimal payout odds minus 1 on each bet (so for odds of -500, w=0.2).

Similarly, the only-parlay Kelly stake is given algebraically by:

K_op = (-1 + p^2*(1 + w)^2)/(w*(2 + w))

The no-parlay Kelly utility is given by:

U_np = p^2*ln(1+2*w*K_np) + 2*p*(1-p)*ln(1+(w-1)*K_np) + (1-p)^2*ln(1-2K_np)

The only-parlay Kelly utility is given by:

U_op = p^2*ln(1+(w^2+2*w)*K_op) + (1-(1-p)^2)*ln(1-K_op)

To find the indifference point we set U_np = U_op and solve for p given w = 0.2. THis yields a value for p ≈ 90.90%, implying edge ≈ 9.08%.

Hey, I did warn it was pedantic ...

**TheLock** · 04-10-08, 07:19 PM

My brain just blew up trying to follow that.

LOL

Good stuff though Ganchrow.

**jtuck** · 04-10-08, 08:55 PM

I wish i was smart

**donjuan** · 04-11-08, 01:21 AM

I suck at math. Thanks, Ganchrow.

**HedgeHog** · 04-11-08, 07:46 AM

Originally posted by Ganchrow

As previously warned:

Given two uncorrelated binary bets at equivalent odds with equivalent win probabilities, the no-parlay Kelly stake on each bet is given algebraically by:

K_np = (1 - p^2 - 3*w + 2*p*w - 2*p^2*w + 2*p*w^2 - p^2*w^2 + sqrt(8*(-1 + w)*w*(-1 + p + p*w) + (-1 + 3*w - 2*p*w*(1 + w) + p^2*(1 + w)^2)^2))/(4*(-1 + w)*w)

where p is the win probability on each bet, and w the decimal payout odds minus 1 on each bet (so for odds of -500, w=0.2).
Similarly, the only-parlay Kelly stake is given algebraically by:

K_op = (-1 + p^2*(1 + w)^2)/(w*(2 + w))

The no-parlay Kelly utility is given by:

U_np = p^2*ln(1+2*w*K_np) + 2*p*(1-p)*ln(1+(w-1)*K_np) + (1-p)^2*ln(1-2K_np)

The only-parlay Kelly utility is given by:

U_op = p^2*ln(1+(w^2+2*w)*K_op) + (1-(1-p)^2)*ln(1-K_op)

To find the indifference point we set U_np = U_op and solve for p given w = 0.2. THis yields a value for p ≈ 90.90%, implying edge ≈ 9.08%.

Hey, I did warn it was pedantic ...

Two questions:

A) Could you explain w=.2 to me? Since the line is -500, the decimal payout % is .83333? Subtract that from 1.0 to get .16667 for w? Obviously I'm a little confused.

B) Say a +Ev better has the option of betting straight at -105 versus two team parlays at 13/5. Assuming all bets have the same edge, what's the indifference point?

**Ganchrow** · 04-11-08, 08:02 AM

Originally posted by HedgeHog

Two questions:

A) Could you explain w=.2 to me? Since the line is -500, the decimal payout % is .83333? Subtract that from 1.0 to get .16667 for w? Obviously I'm a little confused.

B) Say a +Ev better has the option of betting straight at -105 versus two team parlays at 13/5. Assuming all bets have the same edge, what's the indifference point?

1) converted to decimal odds would be . Subtract 1 and you get 0.2. 83.33% refers to the implied probability.

2) The indifference point would be edge of 60.06% on the straights, which at -105 corresponds to a win probability of 82.0%.

**HedgeHog** · 04-11-08, 08:11 AM

Got it. Thanks, Ganch.

PS I'll stick with the straight bets as I'm a little shy of that 82% average.

**MiamiBoy86** · 04-11-08, 10:03 AM

Ganchrow, where did u go to school for math??? and exactly how did u figure out the equations to begin with??? probably a question that needs it's own thread lol

**billmunny** · 04-11-08, 10:43 AM

Originally posted by jtuck

I wish i was smart

You mean you wish you were smart.