Any math people? Please let me know if this is correct.
I want to see the expected returns if I bet $8 per number on 12 numbers for 15 spins.
Expected return = (26/38) x $0 + (12/38) x $288 + (-12/38) x $8 x 15
(https://gcdnb.pbrd.co/images/dekwfG6xqmeE.jpg?o=1)
I was not able to edit, but on B, it is $8 per bet not $12.
Quote from: FreeRoulette on Apr 21, 2023, 09:39 PMI was not able to edit ...
Please feel free to make a reply with all the edits you need to make it a correct post and I'll copy/paste it as the first message.
(Then I'll delete these ones of course)
This is how I understand expected returns to work. Please correct me if I am wrong.
This is calculated on a 38 number roulette wheel.
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Expected Return
The expected return tells us the average amount of money a player can expect to win
or lose per bet over a long period of time.
Example:
Bet $8 on 12 straight numbers for 15 spins.
Expected return = (12/38) x $288 + (-26/38) x $96 x 15
Breaking the above formula down into sections
A: (12/38) x $288
B: (-12/38) x $96
C: 15
How to use the formula
A: The probability of winning
Probability: We bet on 12 numbers out of 38 numbers (12/38)
Amount on win: $8 bet x 35 payout = $280 + $8 (because we get the winning bet back) = $288
B: The probability of losing
Probability: We did not bet on 26 out of 38 numbers (-26/38)
Amount on loss: $8 per bet x 12 numbers that we bet on = $96
C: Spins
The calculation to this point is the average win/loss per spin.
So if we want know how much won/loss over 15 spins, we multiply by 15.
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Probability of winning over X amount of spins
Example:
What is the probability of winning if I bet 2 numbers for 4 spins?
1 - (1 - 2/38)^4 = 0.1884 = 18.84%
We bet 2 numbers so the probability to win this spin is 2/38.
Because the win plus loss always equals 1.
We take the win away from the 1 to get the loss.
The probability that we lose is (1-2/38).
We lose that for 4 spins, so (1-2/38)^4.
Now we take that loss away from 1 to get the win.
1 - (1 - 2/38)^4 = 0.1884
Then we multiply it by 100 to change it from decimal to percent.
0.1884 x 100 = 18.84% chance to hit over 4 spins.
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Probability of winning over X amount of spins continued
We bet for a total of 10 spins
In the above example, we bet 2 numbers for 4 spins
1 - (1 - 2/38)^4 = 0.1884
Now we want to bet 3 numbers for an additional 6 spins
1 - (1 - 3/38)^6 = 0.2924
We multiply the two probabilities together to get a 5.51% chance of hitting at least once in the entire 10 spins.
0.1884 x 0.2924 = 0.0551 or 5.51%
Quote from: FreeRoulette on Apr 22, 2023, 04:09 AMThis is how I understand expected returns to work. Please correct me if I am wrong.
Hi FreeRoulette,
I'm here to correct you. O:-)
QuoteThe expected return tells us the average amount of money a player can expect to win
or lose per bet over a long period of time.
That's right, so I was a little puzzled why in your first calculation you had multiplied by 15 (a number of spins). This doesn't really make sense because the expectation is an average over a (theoretically) infinite number of spins, not 15 spins. If you multiply the basic expectation formula by a number of spins or trials as you did, you will get the wrong answer. You just need to find probability of a win × profit and add this to the probability of a loss × loss. That gives you expectation, job done!
QuoteExample:
Bet $8 on 12 straight numbers for 15 spins.
Expected return = (12/38) x $288 + (-26/38) x $96 x 15
Your mistake here is to forget about the 11 losing numbers; only one number can win! The probabilities are ok.
When you win, your profit is $8 × 35 = $280 minus the $8 on each of the 11 losing numbers which amounts to $88. Therefore the profit on a win is $280 ─ $88 = $192. Your numbers for the expected loss are correct.
So the expectation is 12/38 × $192 + 26/38 × (─$96) = ─$96/19. If you divide by the total staked ($96) you get ─$1/19 or ─$0.0526, about 5 cents lost for every dollar bet. This is the standard result for the double zero wheel so it serves as a check on the calculation.
Glancing over your remaining calculations I see you've made further errors, but I don't have time right now to correct them. I'll be back later.
Sorry! not trying to make you look bad, but you did ask...
...continued
Your second calculation is ok, although I get a slightly higher value.
1 ─ (1 ─ 2/38)
4 = 0.1945
In the first part of your third calculation of 3 numbers for 6 spins I also get a higher value.
1 ─ (1 ─ 3/38)
6 = 0.3893
QuoteWe multiply the two probabilities together to get a 5.51% chance of hitting at least once in the entire 10 spins.
0.1884 x 0.2924 = 0.0551 or 5.51%
This can't be correct, because it's a smaller value than what you calculated for betting on just 2 numbers over 4 spins. If you're betting 3 numbers for an additional 6 spins the probability of at least one win over the 10 spins must be considerably higher than the probability of a win just betting the 2 numbers for 4 spins. What you have calculated here is the probability of getting at least one hit in the first 4 spins (2 numbers)
and at least one hit in the remaining 6 spins (3 numbers).
This is asking much more of the wheel that at least one hit over 10 spins, and hence the probability is smaller (0.1945 × 0.3893 = 0.0757 or 7.57%).
To get the probability of at least one win over the 10 spins you add the separate probabilities, using the rule that P(A or B) = P(A) + P(B) for mutually exclusive events. P(A or B) means at least one of the events A and B occurs.
0.1945 + 0.3893 = 0.5838, or about 58%.
Hey FreeRoulette,
Now it's my turn to make a mistake. :'(
I thought that the final probability of ~58% was a little too high, then I realized that the events are not mutually exclusive because you may get a win in both events.
In that case you have to use the formula P(A or B) = P(A) + P(B) ─ P(A)×P(B).
ie, P(a win in the 2 spins or a win in the 3 spin) = 0.1945 + 0.3893 ─ 0.0757 = 0.5081, or about 51%.
I confirmed this from a simulation. It's always best to double check a probability calculation by another means if possible. Probability can be tricky!
Hello Smith,
Thank you for looking over the calculations and for the corrections. I really needed to get those formulas correct because it is too easy, once the program is coded, to accept the answers as good.