GWAE
Joined: Sep 20, 2013
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March 22nd, 2017 at 10:06:33 AM permalink
I think OP is confusing odds and EV. The odds of hitting the slot on a fair wheel is 1/38, however the EV is completely different since the multiplier is random, right? Not sure how the multiplier works, does it just pop up and that is random? If that is the case the key is 1/38 but the multiplier would be weighted. The 20x might hit 95% of the time and the 100x hit .10 % of the time. Just like most slots it would be near impossible to figure the EV since you don't know what the weights are. Game could still be +ev or it could be 75% return rate, not real sure how you would ever tell unless they tell you the odds of hitting each multiplier.
beachbumbabs
Administrator
Joined: May 21, 2013
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March 22nd, 2017 at 10:25:05 AM permalink
Quote: GWAE

I think OP is confusing odds and EV. The odds of hitting the slot on a fair wheel is 1/38, however the EV is completely different since the multiplier is random, right? Not sure how the multiplier works, does it just pop up and that is random? If that is the case the key is 1/38 but the multiplier would be weighted. The 20x might hit 95% of the time and the 100x hit .10 % of the time. Just like most slots it would be near impossible to figure the EV since you don't know what the weights are. Game could still be +ev or it could be 75% return rate, not real sure how you would ever tell unless they tell you the odds of hitting each multiplier.

Beat me to it. I think this is exactly what's happening, with maybe a pyramid of values that are then chosen randomly.

Let's say, out of 10 choices, there are seeded

4 @20 x
3 @35x
2@50x
1 @ 100x

The mean return on hitting this would be about 38.5x rtp, assuming equal weight on those 10 options.

Hitting the key slot would have the 1 in 38 chance, but the subroutine would have the additional variance.

No way of knowing if this is their framework, but it's logical to think so. They could do many more than 10 options in whatever proportion they wanted, to adjust the rtp. In itself, it could be +ev (the 40.xx rtp the op mentioned), but diluted by the 1/38 chance of hitting it, the game would remain -ev.
"If the house lost every hand, they wouldn't deal the game."
Romes
Joined: Jul 22, 2014
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March 22nd, 2017 at 10:58:44 AM permalink
Quote: Canyonero

...What kind of gambler's fallacy is that?...

Not a gamblers fallacy at all, in fact not even close to a similar idea? The topic in regard is the weighted slots on a roulette wheel where it's very common tongue to equate a "fair" wheel (in which each slot has the same likely hood of hitting) to a "random" wheel.
Playing it correctly means you've already won.
CrystalMath
Joined: May 10, 2011
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March 22nd, 2017 at 11:14:28 AM permalink
Do the rules state the RTP of the other bets?
I heart Crystal Math.
charliepatrick
Joined: Jun 17, 2011
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March 22nd, 2017 at 3:24:27 PM permalink
I am guessing the following
(i) The Key Bet is a straight up bet that if won will pay one of the eleven payouts listed. If it's like normal roulette you retain the original chip. So the total payout is 41.45... (not 40.45..). This makes the probability of winning 1/ (41.45.../.9534) = 1/43.4807 which by co-incidence is almost exactly 2.3%. (If you assume 2.3% and the average payout is 41.4545 you get 95.3454%.)
(ii) Other numbers have an equal chance of coming in. Thus their chance is (100-2.3%)/37 = 2.6405%.
(iii) Other numbers pay standard 35 to 1, so payback is 36*2.6405% = 95.06% (which is a tad better than double zero roulette).
Methinks this is consistent with a UK bookmaker machine (which is where I suspect it can be found - cf http://games.ladbrokes.com/en/games/table-games/KeyBetRoulette ).
If the key bet is FOR 1, then 2.357%, 2.639%, 95.00%.

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