The payout for this optional bet is 9:1
The House advantage is 3.24% on this side bet
WizardofOdds site also lists Probability and return numbers
How can I determine the odds of winning this bet (as in 1:?)
Are there any conversion fomulas from advantage % to odds?
Thank you,
Byron,
Since on this sidebet, all the winning hands pay the same amount, 9:1, we can write the following formula:
(9+1)*W -1 = EV
Since the WoO provided the EV as -0.0324, we can rewrite the above equation to solve for the unknown W, the winning fraction:
W = (1 + EV)/(9 + 1) = (1 - 0.0324)/10 = 0.09676.
This means that we will win, on average, once per 1/0.09676 = 10.33..., or once per 10.33 rounds.
Notice that if we were to win once every 10 tries, the EV would be exactly zero. To have a positive EV, we'd have to win the bet more often than once every 10 tries.
Hope this helps!
Dog Hand
As a rule you're not supposed to play side bets since they have almost always negative ev.
But even if an AP ignores the side bet, he cannot avoid its negative effect caused by other bettors: The slowing down of the game!
Francis Salmon
Thank you for that formula. You have helped my education immensely.
I understand that the odds of winning that side bet is 1:10.33
Can you now help me with the formula for figuring out the standard deviation.
For example: What percentage of the time might you go for 20,30, or 45 rounds without winning that bet?
I fully understand that this is not a bet to make, I would just like to know the math so I can apply it elsewhere when needed.
Thanks again,
They help pay the mortgage on the billion dollars casino properties.
Cobbson
When I first started counting in casinos, with all the distractions, I preferred tables with sidebets and sidebet junlies at them. I didn't play the side bets, but the extra time dealing with the side bets slowed the game enough to make the counting and thinking about the index tables easier. I highly recommend these tables to new counters.
Devising a betting strategy for this but it will not work in the long run. Tempting but a clear path to ruin.
Byron,
You asked what is the standard deviation (SD) of this sidebet. That's easy to calculate. As I mentioned in my last post, the winning probability is W = 0.09676, and a win pays 9:1. Since the SB has no pushes, that means the losing probability is L = 1 - W = 0.90324, and each loss "pays" -1:1. Finally, we saw that the overall EV of this SB is -3.24%, so that means the average (actually, the "mean", represented by the Greek letter "mu") outcome is -0.0324.
As any elementary statistics textbook (or, fo that matter, Wikipedia's "Standard Deviation" page: see link below) will tell you, the SD is the square root of the sum of the winning variance times its probability and the losing variance times its probability. For our case, we get:
SD = SQRT(W*(9-mu)^2+L*(-1-mu)^2)
SD = SQRT(0.09676*(9-(-0.0324))^2 + 0.90324*(-1-(-0.0324))^2)
SD = 2.956...
Thus, the SB has an EV of -0.0324 with a SD of 2.956...
Now, for your second question, you wanted to know the percentage of time you'd go 20, 30, or 45 rounds without winning the SB.
I assume here, you're counting ONLY those rounds on which you PLAY the SB, since for me, the probability is exactly 1 because I never play this SB ;-)
At any rate, the probability that you will lose the SB the next n times in a row is simply L^n, so for 20, 30, and 45 we get probabilities of 0.13064, 0.04722, and 0.01026, respectively.
Hope this helps!
Dog Hand
Sidebet is profitable at TC +7 in the long run
