SteinMeister
SteinMeister
Joined: Apr 23, 2019
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February 15th, 2021 at 8:13:22 AM permalink
The equation for probability of p happening before q is: p / (p+q).

Letís start with analyzing the probability of a single Place 6 bet before a 7-out: p = 5/36, q = 6/36; so p/(p+q) = (5/36) / ((5/36) + (6/36)) = 0.4545. So the probability of a place 6 happening before a 7-out is 45.45% of the time; thereby making the probability of a 7-out happening before a Place 6 occurs is 1 Ė 45.45% = 54.55%

And the EV of a Place 6 before a 7-out is: [(0.4545 * 7 ) - ((1-0.4545) * 6)] / 6 = -1.515% of each dollar wagered.

Confirming this w/ WinCraps (1M rolls) results in -1.50% of each dollar wagered. Close enough, so the gambling statistics are copacetic.

Now to analyze betting on both Place 6 and Place 8: Probability of EITHER a Place 6 or Place 8 hitting before a 7-out: p = (5/36) + (5/36) = 10/36: q = 6/36; thus p / (p+q) = (10/36) / [ (10/36) + (6/36) ] = 0.625. So the probability of EITHER a Place 6 or a Place 8 hitting before a 7-out is 62.5%; making the probability of a 7-out before either place bet hits is 1 Ė 62.5%, or 37.5%.

Now, here comes my befuddlement: The EV of either a Place 6 or 8 before a 7-out is: [ (0.625 * 7) - (0.375 * 12) ] / 12 = -1.042% of each dollar wagered. Which, on the surface, appears that combining two -1.515% EV bets results in a lower EV of -1.042%. But this cannot be! This would violate the edicts of gambling statistics, and could add fuel to the gamblers fallacy misconception.

Furthermore, doing a sanity check w/ WinCraps, the -1.042% is nary to be found. Instead, placing both 6 and 8 bets w/ the same 1M dice roll of the previous Place 6-only analysis, nets a -1.49% (close enough to call it the expected -1.515%.)

SoÖ. What am I missing in the Place 6 and 8 EV analysis?
Mission146
Mission146
Joined: May 15, 2012
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February 15th, 2021 at 8:27:00 AM permalink
Quote: SteinMeister



Now, here comes my befuddlement: The EV of either a Place 6 or 8 before a 7-out is: [ (0.625 * 7) - (0.375 * 12) ] / 12 = -1.042% of each dollar wagered. Which, on the surface, appears that combining two -1.515% EV bets results in a lower EV of -1.042%. But this cannot be! This would violate the edicts of gambling statistics, and could add fuel to the gamblers fallacy misconception.

Furthermore, doing a sanity check w/ WinCraps, the -1.042% is nary to be found. Instead, placing both 6 and 8 bets w/ the same 1M dice roll of the previous Place 6-only analysis, nets a -1.49% (close enough to call it the expected -1.515%.)

SoÖ. What am I missing in the Place 6 and 8 EV analysis?



(Quote clipped, relevance)

Okay, so the probability of winning either/or v. losing both:

(10/16) = .625

(6/16) .375

This is precisely what you said and is correct.

However, the difference that you're getting in your numbers stems from the fact that only one bet or the other resolves on a win. In other words, if the six wins, then the bet on the eight is unresolved and could still lose. If the eight wins, same thing with the six.

Your formula above is treating it as if only one bet is being made, but two bets are being made.

If there was a separate SIX OR EIGHT bet that existed for which you could bet $12 to profit $7, then you would have:

(10/16 * 7) - (6/16 * 12) = -0.125

Which reflects an expected loss of 12.5 cents on a $12 bet, thus:

.125/12 = -0.01041666666

Which is what you are getting. That's because that is the formula you are doing. The reason it comes out this way when you treat it as one bet is because the other bet is not getting resolved on a win; you can either pull the bet at that point or it can be resolved later...but if the six hits, then the eight is unresolved. The House Edge on the Six and Eight refers to bets that are resolved.

CORRECT HOUSE EDGE

We must look at every sequence of events, individually, to arrive at the correct House Edge for this proposition. In order to do that, we are going to have both bets stay up until they are resolved. When you do this, you will get new probabilities:

BOTH LOSE: (6/16) = .375

ONE WINS, THE OTHER LOSES: (5/16) * (6/11) * 2 = 0.3409090909 (Notice the Six and eight are treated as separate events; this is crucial)

BOTH WIN: (10/16) * (5/11) = 0.28409090909

0.28409090909+0.3409090909+.375 = 0.99999999999

Okay, so now we come up with the expected value of each possible series of outcomes:

.375 * (-12) = -4.5

0.3409090909 * 1 = 0.3409090909

0.28409090909 * 14 = 3.97727272726

For our expected total result:

(3.97727272726+0.3409090909) -4.5 = -0.18181818184----An expected loss of 18.181818184 cents, thereby resulting in House Edge:

.18181818184/12 = 0.01515151515

What it should be!
Last edited by: Mission146 on Feb 15, 2021
Vultures can't be choosers.
SteinMeister
SteinMeister
Joined: Apr 23, 2019
  • Threads: 1
  • Posts: 3
Thanks for this post from:
Mission146
February 15th, 2021 at 8:55:40 AM permalink
Quote: Mission146

Quote: SteinMeister



Now, here comes my befuddlement: The EV of either a Place 6 or 8 before a 7-out is: [ (0.625 * 7) - (0.375 * 12) ] / 12 = -1.042% of each dollar wagered. Which, on the surface, appears that combining two -1.515% EV bets results in a lower EV of -1.042%. But this cannot be! This would violate the edicts of gambling statistics, and could add fuel to the gamblers fallacy misconception.

Furthermore, doing a sanity check w/ WinCraps, the -1.042% is nary to be found. Instead, placing both 6 and 8 bets w/ the same 1M dice roll of the previous Place 6-only analysis, nets a -1.49% (close enough to call it the expected -1.515%.)

SoÖ. What am I missing in the Place 6 and 8 EV analysis?



(Quote clipped, relevance)

Okay, so the probability of winning either/or v. losing both:

(10/16) = .625

(6/16) .375

This is precisely what you said and is correct.

However, the difference that you're getting in your numbers stems from the fact that only one bet or the other resolves on a win. In other words, if the six wins, then the bet on the eight is unresolved and could still lose. If the eight wins, same thing with the six.

Your formula above is treating it as if only one bet is being made, but two bets are being made.

If there was a separate SIX OR EIGHT bet that existed for which you could bet $12 to profit $7, then you would have:

(10/16 * 7) - (6/16 * 12) = -0.125

Which reflects an expected loss of 12.5 cents on a $12 bet, thus:

.125/12 = -0.01041666666

Which is what you are getting. That's because that is the formula you are doing. The reason it comes out this way when you treat it as one bet is because the other bet is not getting resolved on a win; you can either pull the bet at that point or it can be resolved later...but if the six hits, then the eight is unresolved. The House Edge on the Six and Eight refers to bets that are resolved.



Thank you... you've made it perfectly clear to me now. I get it.

Yet another case where simulation (so long as it's scripted correctly) prevails over ill-conceived mathematics.
Mission146
Mission146
Joined: May 15, 2012
  • Threads: 125
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February 15th, 2021 at 8:58:02 AM permalink
Quote: SteinMeister



Thank you... you've made it perfectly clear to me now. I get it.

Yet another case where simulation (so long as it's scripted correctly) prevails over ill-conceived mathematics.



I also added to my above post with well-conceived mathematics; no simulation necessary.
Vultures can't be choosers.
SteinMeister
SteinMeister
Joined: Apr 23, 2019
  • Threads: 1
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Thanks for this post from:
Mission146
February 15th, 2021 at 5:22:08 PM permalink
That is true. However, (as in my case) "ill-conceived mathematics" may initially be misconstrued as "well-conceived" until a sanity check is performed with simulation.
Last edited by: SteinMeister on Feb 15, 2021

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