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nameremorse
nameremorse
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August 14th, 2019 at 10:26:05 AM permalink
I previously read the WoO post on 'Monkey Baccarat', and there is a new game that will have a 200-1 side bet for all 6 cards being Monkey (includes 10's, J, Q, K). This bet would have a max of $25 but still at 200-1 that's $5k a pop.

In the Monkey Baccarat post, it says the 10's are not included and the bet is a max and min of $1, which leads to the huge HE... If this bet were as described above, $25 max and includes 10's, but is only 200-1, would this be a counters dream? Or would it take too long to realize any profit due to the reduced payout? On the surface, I see this as a viable opportunity to count down the side bet, only playing towards the end of the shoe when there are a lot of Monkey cards remaining...

TIA!
RS
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August 14th, 2019 at 1:02:20 PM permalink
This.....doesn't sound right to me. Maybe I'm mixed up here, but you basically need 6 faces (T-K) to come out in a row, right? No other way to get 6 monkeys.

If that's the case, that's an absolutely massive house edge (like 80%+).


I just googled it to verify now, and WOO site shows it pays 5000:1, but 10's aren't permitted. In that case, you'd need 24.183517935% remaining cards to be monkeys (J, Q, K) for the bet to be break-even. Actually, it'd be slightly higher than that, due to effect of removal (I THINK). I'm not sure how much higher, but I suspect it depends on the total # of cards remaining, since if there are fewer cards remaining, one taken out has a larger effect than if there are many cards remaining. I suspect an opportunity would be present, but with a 5000-to-1 payout, it's likely it's a $1 min/max bet, in which case your EV is going to, practically, $0.
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Mission146
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nameremorse
August 14th, 2019 at 1:52:50 PM permalink
Quote: RS

This.....doesn't sound right to me. Maybe I'm mixed up here, but you basically need 6 faces (T-K) to come out in a row, right? No other way to get 6 monkeys.

If that's the case, that's an absolutely massive house edge (like 80%+).


I just googled it to verify now, and WOO site shows it pays 5000:1, but 10's aren't permitted. In that case, you'd need 24.183517935% remaining cards to be monkeys (J, Q, K) for the bet to be break-even. Actually, it'd be slightly higher than that, due to effect of removal (I THINK). I'm not sure how much higher, but I suspect it depends on the total # of cards remaining, since if there are fewer cards remaining, one taken out has a larger effect than if there are many cards remaining. I suspect an opportunity would be present, but with a 5000-to-1 payout, it's likely it's a $1 min/max bet, in which case your EV is going to, practically, $0.



I did some work to try to figure this one out and the long and short answer is that you want the percentage of ten-value cards to be a little north of 40%. The exact percentage can swing things because you can have the percentage of ten cards be lower the more total cards remain (effect-of-removal). The base house edge, with eight-decks, is over 80% as you guessed:

(128/416)*(127/415)*(126/414)*(125/413)*(124/412)*(123/411) = 0.00078125028

(200*0.00078125028) - (1-0.00078125028) = -0.84296869372

Anyway, the percentage of ten cards remaining needs to be a little to the north of 40%. I think with two-decks (104 cards remaining) you needed 45 to be at an advantage:

(45/104)*(44/103)*(43/102)*(42/101)*(41/100)*(40/99) = 0.00536783898

(0.00536783898*200) - (1-0.00536783898) = 0.07893563498

Forty-four to be 0 value cards would be a disadvantage of around 6%.

For the OP, as far as effect-of-removal, think about it this way: Five of nine cards (55%) could be ten-value cards, but it would be impossible to win this bet because there are not six ten-value cards. The north of 40% rule really starts to break down once you get past two decks remaining because the effect-of-removal becomes too great. Even 11/24 cards doesn't help you:

(11/24)*(10/23)*(9/22)*(8/21)*(7/20)*(6/19) = 0.00343249427

(0.00343249427*200) - (1-0.00343249427) = -0.31006865173

It would be very difficult to count because the percentage of cards that need to be 0's is a moving target, so you would need to be able to count off the 0's, total cards and then do the division in your head to figure the percentage while having general guidelines for what I would call, "Trigger points." Your trigger point at two decks, for example, would be 45 cards. One card either way makes a huge difference, so you'd really need to know what you are doing.

I don't know the probability of arriving at 45/104 zero value cards remaining with two decks, but I don't imagine it's very high. Much counting and not much in the way of actual betting. More than that, any advantage you do have is going to be subject to a ton of variance.

Eliot would have figured this out, created a program to count it perfectly and then come back and told you what your value per 100 hands would be. I can't do any of that, but I've read enough of his stuff to say that I don't think it would be really high. Most shoes you would never make a single bet.

ADDED STUFF:

1.) You'll really stand out only playing the side bet and only playing it towards the end of the shoe.

2.) The house could cut off a few decks if they deem it necessary.
Vultures can't be choosers.
nameremorse
nameremorse
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Mission146
August 15th, 2019 at 3:28:03 PM permalink
I miss Eliot and his posts. Its just not the same reading the old posts on that other site.

I figured it would be a tough bet, like you say, only playing the bet on a few hands towards the end of a shoe, and skipping most shoes... based on the files WoO has (1,000 shoes with cards listed) this event (6 card monkey) happened 75 times in 1000 shoes... I know that's not exact science i just did right there, but got my brain to thinking...

I am told this bet has a total house edge ~10%, and the specific 6 card monkey HE would be ~15%... Per a math-brain-guy report. I think your numbers might be so different because i left out that there are other payouts for total monkey cards in the end result... but if you don't think this adds up, please say so!

3 monkey is 2-1, 4 monkey is 5-1, 5 monkey is 20-1, 6 monkey 200-1. 2 or less monkey is a loss.
Mission146
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nameremorse
August 15th, 2019 at 3:46:54 PM permalink
Yes, the other payouts were extremely relevant! Iíll try to look at this again a bit later, it might already be done for those on WoO.
Vultures can't be choosers.
RS
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August 15th, 2019 at 6:21:44 PM permalink
Quote: Mission146

Yes, the other payouts were extremely relevant! Iíll try to look at this again a bit later, it might already be done for those on WoO.


Donít bother (IMO). Obviously the other payouts are important, heís just yanking our chains at this point.
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Wizard
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August 15th, 2019 at 9:12:02 PM permalink
My own page on Monkey Baccarat says the minimum and maximum bet is $1. That is why I never did a card counting vulnerability study.
It's not whether you win or lose; it's whether or not you had a good bet.
nameremorse
nameremorse
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August 16th, 2019 at 9:27:17 AM permalink
Quote: RS

Donít bother (IMO). Obviously the other payouts are important, heís just yanking our chains at this point.



I admittedly do not know enough about statistical math to do any kind of real analysis; I would do it myself if I could! I wouldn't have even thought about how the other payouts would change the math until the above post by Mission, where I can see when it is written out, that it does make a big difference.

The other payouts in Monkey Baccarat (the one already done on WoO) are based on a different method... The only one that is similar is the top payout (Grand Monkey) which is why I brought that particular bet up. The difference of allowing a range of bets instead of a flat $1 bet was what caught my eye.

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