Probability of 1 corner bet hitting over 3 spins

Great site and seems the perfect place to try and get an answer to this question.

Been having a debate with some collegues about the probability of a corner bet hitting on one of 3 spins. Eurpoean roulette so single zero wheels. Assume that you place a bet and you have to let it ride on all three spins. If thats the case I work it out as around 1 in 11.6 chance of hitting the corner bet once and then missing on the second and third spin. Are my calculations correct?

First spin hits 4/37 chance Second spin misses 33/37 and third spin misses 33/37

so 1 in 9.25 first spin you hit the corner, 1 in 1.121212 chance second spin misses, and 1 in 1.121212 chance the third spin misses.

Multiple that up 9.25*1.121212*1.121212 = 11.6283 or 1 in 11.62 chance of hitting 1 corner bet over 3 spins.

So if a casino offered this bet they could perceivable offer odds of 11 to 1 giving a return to player % of 94.60 and have a house edge of 5.40%

Aim I right to think this?

Your math is wrong.

You are calculating that it hits on the next spin AND MISSES ON THE FOLLOWING TWO SPINS.

You have to add the possibility that it misses then hits then misses, or misses twice then hits, as well as hits more than once.

The easy way to calculate that is to calculate the odds of missing all three times (33/37)^3=0.709. Then subtract from 1 making the chance that it hits at least once 1-0.709=0.291

29.1%

So lets say we select the corner bet 1,2,4,5. what are the odds of hitting this bet once in any 3 spins?

Then the same bet whats the chance of this repeating on any two of three spins?

Finally same bet, whats the chance of hitting this bet repeating over 3 consecutive spins?

Im eager to undestand how you calulate this over the 3 spins and like to know the steps involved.

Thanks

Note that the probability of winning on one spin (assuming a double-zero wheel) is 4/38, or 2/19, and the probability of losing is 17/19.

The probability of it hitting at least once in three spins is 1 minus the probability that it does not hit at all in three spins, which is 1 - (17/19)³ = 1946/6859, or about 28.37%.

The probability of it hitting exactly once in three spins is the sum of the probabilities of (winning just on the first spin) + (winning just on the second spin) + (winning just on the third spin). Each of these has probability 2/19 x 17/19 x 17/19, so the total probability = 3 x 2/19 x 17/19 x 17/19 = 1734 / 6859, or about 25.28%.

There are three ways to win on exactly two spins (first and second; first and third; second and third); each has probability 2/19 x 2/19 x 17/19, so the total probability is 3 x 2/19 x 2/19 x 17/19 = 204 / 6859, or about 1 in 33.62.

The probability of hitting all three times is 2/19 x 2/19 x 2/19 = 8 / 6859, or about 1 in 857.

Note that the probability of hitting at least twice = the probability of hitting exactly twice + the probability of hitting exactly three times = 204 / 6859 + 8 / 6859 = 212 / 6859, or about 1 in 32.35.

im struggling to understand how there is a difference between hitting it at least once and hitting it exactly once. Surley they would be the same?

what i really want to know is the full calulations broken down so i can understand each part of the equation. preferably in excel format so i can play around with it!!

Maybe best for me to re phrase the original question.

single zero wheel. 3 spins. the player selects a single corner bet. lets say 1245.

The player places the same bet on all 3 spins.

What are the odds that the player hits this bet on any 1 of the 3 spins? either spin 1 or 2 or 3
What are the odds that the player hits this bet on 2 of the 3 spins? either spin 1+2 or 1+3 or 2+3
What are the odds that the payer hits this bet on 3 of 3 spins?

Quote: thejarv

im struggling to understand how there is a difference between hitting it at least once and hitting it exactly once. Surley they would be the same?

It might be easier if you're struggling with the maths to look at the same question based on three rolls of the dice and how many sixes there are. Then you can also confirm the numbers in a spreadsheet but having all the 6x6x6=216 combinations of three rolls.

The difference between exactly one roll and at least can be seen by considering rolls such as 655 665 and 666, the former is in both while the last two are also in at least. There are 91 combinations with at least one six (216-125) while only 75 of these have exactly one six.

0 rolls are sixes	125	No rolls are a six consists of the first roll being 1 to 5, second 1 to 5, third 1 to 5 - so 125.	e.g. 111 112 113 114 115...555
1 roll is a six	75	(i) 6XX First roll is a six then , second 1 to 5, third 1 to 5 - so 25; (ii) X6X (iii) XX6.	e.g. 611 612 613 ... 161 162 ,,, 116 126 ... 556
2 rolls are a six	15	(i) 66X First two rolls are a six then, third 1 to 5 - so 5; (ii) 6X6 (iii) X66.	e.g. 661 662 663 664 665 616 ... 656 166 ... 566
3 rolls are a six	1	Only all three are six works	666

The same logic is use for your roulette question.

Quote: thejarv

im struggling to understand how there is a difference between hitting it at least once and hitting it exactly once. Surley they would be the same?

what i really want to know is the full calulations broken down so i can understand each part of the equation. preferably in excel format so i can play around with it!!

Maybe best for me to re phrase the original question.

single zero wheel. 3 spins. the player selects a single corner bet. lets say 1245.

The player places the same bet on all 3 spins.

What are the odds that the player hits this bet on any 1 of the 3 spins? either spin 1 or 2 or 3
What are the odds that the player hits this bet on 2 of the 3 spins? either spin 1+2 or 1+3 or 2+3
What are the odds that the payer hits this bet on 3 of 3 spins?

"At least once" is "exactly once" plus "exactly twice" plus "exactly three times" - that's the difference.

Look at it this way - instead of roulette, toss a coin three times.
There are eight possible results: HHH, HHT, HTH, HTT, THH, THT, TTH, and TTT (where H is heads and T is tails). Each one is equally as likely as the other seven.
Only three of them - HTT, THT, and TTH - have heads coming up exactly once.
On the other hand, seven - all but TTT - have heads coming up at least once.

Back to your roulette problem...
First, note that the probability of a particular spin winning is 4/37, and the probability of that spin losing is 33/37.
The probability of winning on all three spins is 4/37 x 4/37 x 4/37 = 64 / 50653 = about 1 / 791.
The probability of winning on exactly two spins is 3 x 4/37 x 4/37 x 33/37 (that is, the probability of each of the two winning spins winning is 4/37, the probability of the one losing spin losing is 33/37, and there are 3 ways for this to happen - 1 & 2 win, 1 & 3 win, or 2 & 3 win) = 1584 / 50563 = about 1 / 32.
The probability of winning on exactly one spin is 3 x 4/37 x 33/37 x 33/37 = 4356 / 50563 = about 1 / 11.6.

On the other hand, the probability of winning on at least one spin = (exactly one spin) + (exactly two spins) + (exactly three spins) = 4356/50563 + 1584/50563 + 64/50563 = 6004 / 50563 = about 1 / 8.4.
The probability of winning on at least two spins = (exactly two spins) + (exactly three spins) = 1584/50563 + 64/50563 = 1648 / 50563 = about 1 / 31.