Roulette Ideas

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

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.

__________________________________________________________________________________
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.

_____________________________________________________________________________________
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. 

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.