Not blackjack, but a question about calculating EV.

Lonesome Gambler

Well-Known Member
#1
A person has a chance to spin a wheel that has the following values attached to it: $2, $3, $5, $10, and a 2X multiplier. There are 12 total results: (4)$2, (4)$3, (2)$5, (1)$10, and (1)2X multiplier. The person receives cash in the amount of the value they spin (anywhere from $2-20). How do you go about calculating the EV of each spin? The main thing throwing me off is the 2X multiplier. That, and the fact that I'm clueless about math!

Anyway, this is just an experiment that had me thinking. My hypothesis is that the EV is slightly above $3 per spin. How close am I?
 

Pro21

Well-Known Member
#2
What happens if you land on 2x? do you spin again and win double what you land on? And if you hit 2x again do you spin again and get 4x?
 

Sonny

Well-Known Member
#6
Lonesome Gambler said:
Great! Would you mind sharing the math on that? Thanks!
Sure!

So we have a wheel with 12 spaces. Four of them pay $2, another four pay $3, two of them pay $5 and one pays $10. So far we have a pretty straightforward EV calculation.

EV Excluding the 2x Multiplier:
(4/12)$2 + (4/12)$3 + (2/12)$5 + (1/12)$10 = $3.33

The only part we are missing is the 2x multiplier space (hereafter “Multi”). We know that it has a 1/12 probability so we’re almost done. All we need to know is the average payoff (EV) of the Multi space.

After we hit the Multi space we are only concerned with the spaces that give a payout (spaces $2, $3, $5 and $10). If we hit the Multi space again nothing happens so we can ignore that outcome. That leaves 11 spaces on the wheel. Of those 11 spaces, four of them pay $4, another four pay $6…you get the picture. The EV for the Multi space is:

EV of Multi Space:
(4/11)$4 + (4/11)$6 + (2/11)$10 + (1/11)$20 = $7.27

Including that in our original calculation gives us a total EV of $3.94 per attempt. I use the word ‘attempt’ instead of ‘spin’ because it may take several spins before the bet is resolved. In fact, the average length of a turn is 1.09 spins. If we divide $3.94 by 1.09 then we get an EV of $3.61 per spin, although that number is kinda pointless. I can’t imagine a scenario where you would quit playing, willingly or otherwise, after only a single spin, especially after your EV has more than doubled. Still, in the interest of pointless calculations, I included it in my earlier post. If anyone wants to see some more pointless math I’ll be happy to show how I calculated the average number of spins.

-Sonny-
 

Lonesome Gambler

Well-Known Member
#7
Excellent, thanks Sonny! That's pretty close to the way I was going about it, but I was having a hard time figuring the numbers after taking the 2X into effect. Anyway, much appreciated.
 

Canceler

Well-Known Member
#8
Sonny said:
If anyone wants to see some more pointless math I’ll be happy to show how I calculated the average number of spins.
Sure, go ahead! Is it some nasty calculation that takes into account the possibility of hitting 2X 100 times in a row?

Also, no one has mentioned how much it costs to play this game, which seems crucial to me.
 

Sonny

Well-Known Member
#9
Canceler said:
Sure, go ahead! Is it some nasty calculation that takes into account the possibility of hitting 2X 100 times in a row?
We don’t need to go that far. The probabilities are pretty skewed so they will converge quickly. In this case doing it “the hard way” is pretty easy. The probability of having the turn last 1 spin is 11/12 = 91.67%. The probability of the turn lasting 2 spins is (1/12)*(11/12) = 7.64%. For 3 spins we get (1/12)*(1/12)*(11/12) = 0.64% and for 4 spins we get (11/12)*(1/12)^3 = 0.00. Basically we’re done at that point since everything becomes insignificant. Using those frequencies gives us an average turn length of 1.0885 spins. We could easily write out a nice little algorithm for those calculations, but since it only requires 3-4 iterations it's actually faster to do it the long way.

-Sonny-
 
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