statistical question

bogwalrus

I am helping a friend with a business venture.

He is creating a pub game based on wheel roulette but I am not certain he has thought out the statistical odds of a person winning etc.


Basically each player pays €10 to play.

The player makes three guesses on a card:

(the gamer rolls the roulette wheel)

1st guess is whether it is red/black - odd/even

If the player guess right that it is black and odd game continues.

(gamer rolls the wheel again)

2nd guess is whether it is 1st,2nd or 3rd tier, top middle or lower

If the player guess its the 3rd middle the game continues for them.

(gamer rolls a final time)

3rd guess is for three numbers

If the player gets one of the numbers right he/she wins a weeks holiday for two to a sunny destination incl flights and accommodation.



How exactly can you calculate the odds of winning?

I think the game is not attractive enough so they should be able to guess five numbers at the end as I think the odds of winning are still around 1/1000 at least


Sure any help with the odd's calculating would be great.:D

locum-motion

Is the game based on a roulette wheel or on a pack of cards? Coz as far as I can see, you mention both.

bogwalrus

Its based on a roulette wheel with the table.

locum-motion

bogwalrus wrote: »

...The player makes three guesses on a card:
...

So where does the card come in to it?

Yakuza

It just means the player makes 3 guesses and records them (in this case on a card) so there can be no squabbling over what they picked and what they say they picked .

There are 3 independent events that take place, the probabilty of 3 successes is the product of all three individual events.

Roughly speaking, (ignoring what happens when the ball lands in 0) I think it should be 1/4 * 12/36 * 3/36 or 36/(4*36*36), or 1/144. A tenner stake with those odds to win a prize of, what, €500 value...I don't know if I'd take that bet...

I'm a bit stuck for time right now, will take a proper crack at this later. Also, I take no responsibility for any figures above!

BeanbagBallbag

Assuming there is no green 0 in your game for simplicity, there are 4 possible outcomes for the first guess, so that's 1/4.

There are 9 possible outcomes for the second guess, so 1/9.

And for the third guess, you have a 3/36 chance, which is 1/12.

For all those things to happen you multiply each probability which is 1/432.

Yakuza

^^^

I read the middle game as guessing where the ball falls in the ranges 1-12, 13-24 or 25-36 (essentially a 1 in 3 probability)

BeanbagBallbag

^^^^
I thought by tiers he meant the "columns" from the picture, 1 4 7 10..34 etc, if not then just divide my answer by 3 OP!

locum-motion

Does each player have one guess/bet on each of three separate spins of the wheel?

Standard European roulette wheel, and standard roulette layout?

Guess 1/1st spin:
Red v Black OR Odd v Even.
Player's chance of success is 18/37.
(Another possible thing to bet on here is 1-8 v 19-36; same odds)

Guess 2/2nd spin:
I presume here you're referring to the 'Dozen' bets. These are:
1st Dozen v 2nd Dozen v 3rd Dozen OR 1st Column v 2nd Column v 3rd Column
Player's chance of success is 12/37.

Guess 3/3rd spin:
What you call 'three numbers'. Correctly called a Street. These are 1 2 3 v 4 5 6 v 7 8 9 etc down to 34 35 36.
Player's chance of success is 3/37.

18/37 = 0.486 approx.
12/37 = 0.324 approx.
3/37 = 0.081 approx.

Overall player's chance of success is 18/37 * 12/37 * 3/37
which is 0.0127929 (1 in 78 or 77 to 1)

So, for a €10 stake, if the prize is worth more than €780, it is an attractive bet for the punter, and a loser for the guy running the game. If the prize is worth less than €780, it's a winner for the guy, and a less attractive bet for the punter. The bigger the difference between the price of the prize and €780, then the bigger the difference between the attractiveness of the bet and the profitability of the game. That is: Tiny prize = massive profit for the guy but difficult to persuade punters to play, while massive prize = easy to get punters, but potentially huge losses for your mate.

The only way to balance these two opposites is if the 'perceived value' of the prize to the punter and the 'actual cost' to your friend of buying the prize are very different. Essentially, your mate has to buy the prizes cheap (at a 'wholesale' price, shall we say) and they must appear to be expensive to the punter. So if your mate can buy a prize for €680, but it would cost the punter €880 if he were to book it himself, then there is both profit for your mate and attractiveness to the punter.

In that situation, your mate makes €100 profit for every 78 people he can persuade to play the game; he buys a thing for €680, and 'sells' it for 78 x €10. On the other side of the table, you have 78 punters who have put in a tenner, and they know that for their €780 they have 'bought' something that would cost them €880, so in theory they are happy - they've got out more than they put in. In actuality, though, you've got 77 guys that are mumbling and grumbling about being ripped off, and one guy that's absolutely over the moon. (Their 'happiness' is not evenly distributed!)

So, that's the answer to your question.


Now, as far as I'm concerned, there's a much bigger question: Why all this messing around with a card and a marker?

I mean, you've got a table in front of you with the roulette layout on it, so why not just give the guy three little tokens like for example three Monopoly dogs, and just let him indicate his three choices by placing his three Monopoly dogs on the relevant part of the layout. He can record his choices that way. Then, after the spin, put some kind of marker on the number that came up, so that everyone around the table can immediately see if none, one, two or three of his guesses were right. That makes it much quicker for you to 'reset' for the next punter and the next spin too, so you can play the game more times per hour.

In fact, if you had 3x Monopoly cars and 3x Monopoly hats and 3x Monopoly irons too, you could allow four punters to take part in the game for each spin of the wheel, and get through more bets per night that way.

Wow, your friend is really getting through his 78 punters really quickly now!

But, hang on a second! Why go to the bother of buying loads of Monopoly sets to get your hands on the dogs and the cars and the hats and the irons? Why not let the first guy get three red plastic discs for his tenner, and the second guy three blue discs and so on. That's much easier, isn't it?

Then, wait a minute! One of the punters has an idea: "Here, mate!" he says, "why is it that I have to get all three parts of my guess right to get one big prize? Why can't you just give me a smaller prize if I get one bit of it right?" Then all the rest of them join in, all wanting to be able to bet for smaller prizes.

The point I'm making here (in a roundabout and probably somewhat facetious way), is that it would be a whole lot easier just to play roulette in the first place! I mean, he's gone to the trouble of buying himself a roulette wheel and layout, and I'm sure they're not cheap. Of course, he's also gone to the trouble of getting whatever license he needs to operate a gambling establishment too, and they're definitely not gonna be cheap (what with the whole 'must-be-a-private-members'-club thing, and whatnot). So, if he's going to go to all that expense and hassle, he might as well just play the game properly, and not go messing about trying to get a good deal on holidays from his local travel agent (or maybe he'd act as a travel agent himself, in which case he needs to be bonded, a member of IATA etc etc etc)

I think that what your friend has done here with his 'pub game' is the closest thing I have ever heard to reinventing the wheel (excuse the pun!).

TL;DR: Tell your mate to forget it!

bogwalrus wrote: »

I am helping a friend with a business venture.

He is creating a pub game based on wheel roulette but I am not certain he has thought out the statistical odds of a person winning etc.


Basically each player pays €10 to play.

The player makes three guesses on a card:

(the gamer rolls the roulette wheel)

1st guess is whether it is red/black - odd/even

If the player guess right that it is black and odd game continues.

(gamer rolls the wheel again)

2nd guess is whether it is 1st,2nd or 3rd tier, top middle or lower

If the player guess its the 3rd middle the game continues for them.

(gamer rolls a final time)

3rd guess is for three numbers

If the player gets one of the numbers right he/she wins a weeks holiday for two to a sunny destination incl flights and accommodation.



How exactly can you calculate the odds of winning?

I think the game is not attractive enough so they should be able to guess five numbers at the end as I think the odds of winning are still around 1/1000 at least


Sure any help with the odd's calculating would be great.:D

kingsenny

I could see locum-motion's point coming a mile away... still made me laugh! I was thinking the exact same thing tbh...

locum-motion

BeanbagBallbag wrote: »

...there are 4 possible outcomes for the first guess, so that's 1/4...

Nope, not four outcomes. On my understanding of what the OP has said, the player is going to guess either on whether red or black will come up or on whether even or odd will come up. As I said, there is also a third possible pair of 50:50 options to guess on, which is 1st 18 vs 2nd 18. So if he chooses to bet on red v black, there are two possible outcomes. If he chooses to bet on even v odd, there are two possible outcomes. If he chooses to bet on 1st 18 v 2nd 18, there are two possible outcomes.


Of course, maybe my interpretation is wrong and yours is right, in which case what he's betting on is one of the following four options: RedEven v RedOdd v BlackEven v BlackOdd. However, all 4 options are not equally likely in this case: there are 10 RedOdd and 10 BlackEven possibilities, and only 8 RedEven and 8 BlackOdd possibilities on the layout the OP has provided above.

kingsenny

My interpretation would've been:

RedOdd vs RedEven vs BlackOdd vs BlackEven (being in reference to the left most columns and NOT the actual numbers)

thus still a 1 in 4 chance

locum-motion

kingsenny wrote: »

My interpretation would've been:

RedOdd vs RedEven vs BlackOdd vs BlackEven (being in reference to the left most columns and the actual numbers)

thus still a 1 in 4 chance

So you were interpreting it in the same way as BeanBallBag (or whatever his name is), not in the same way I was. But: read my 'However...' in the second paragraph of my reply immediately above yours. If your (and BBB's) interpretation was what the OP did in fact mean, the 'However...' makes a nonsense of it.

bogwalrus

BeanbagBallbag wrote: »

^^^^
I thought by tiers he meant the "columns" from the picture, 1 4 7 10..34 etc, if not then just divide my answer by 3 OP!

Sorry my explaining was not that great.

I do mean the three columns that are verticle. then there are three tiers top middle and bottom. Usually in European Roulette you make bets on different outcomes.

Yakuza

With that said, bbbb's 1/432 probability is more correct than mine (ignoring the 0), makes the game even less worthwhile. For those odds and that stake, I'd want to be winning a 10-day cruise to bother trying.

bogwalrus

Just to clear a few things up.

The betting is not odd/even or black/odd it is both.

So player could bet black & odd for first guess or red & even.

second guess Could be first column & middle or second column & top

If the player has got the first two right then now he/she has their numbers to possibly win.


I really appreciate the work. The statistics part is quite confusing but I follow what you are all saying.

Thanks for the help.

BeanbagBallbag

Just to amend my ESTIMATION, as I am not a degenerate and I did not study the picture closely at first, if you chose even&black or odd&red, then the final odds would be 1/388.8, and if you chose odd&black or even&red they would be 1/486.

bogwalrus

And how about the odds in the players second guess now that it is:

1st column & 2nd Tier or 2nd Column & 1st Tier?

Thanks again

bogwalrus

Were just looking into the attractiveness now.

I think loco motion is right about his post. Thanks for that:)

locum-motion

bogwalrus wrote: »

...
The betting is not odd/even or black/odd it is both...

OK, so my interpretation was wrong and BBBB/Kinsenny's interpretations were right.

Therefore, on the first guess/spin, there are four possibilities:

RedEven: chance of success = 8/37
BlackEven: chance of success = 10/37
RedOdd: chance of success = 10/37
BlackOdd: chance of success = 8/37

The fact that the four options do not have the exact same odds makes a mockery of the game, tbh. No punter in their right mind would ever choose RedEven or BlackOdd.

locum-motion

bogwalrus wrote: »

...second guess Could be first column & middle or second column & top...

Similarly, I thought you were talking about Columns OR Dozens, not Columns AND Dozens.

Therefore, there are 9 possibilities:

1stDozen/Left Column
1stDozen/Middle Column
1stDozen/Right Column
2ndDozen/Left Column
2ndDozen/Middle Column
2ndDozen/Right Column
3rdDozen/Left Column
3rdDozen/Middle Column
3rdDozen/Right Column

Each one of these has the same probability: 4/37

locum-motion

bogwalrus wrote: »

...If the player has got the first two right then now he/she has their numbers to possibly win...

So having got through the above two obstacles, the player then chooses a 'Street' of three numbers for their last spin, is that right?

There are 12 possibilities:

1,2,3
4,5,6
7,8,9
10,11,12
13,14,15
16,17,18
19,20,21
22,23,24
25,26,27
28,28,30
31,32,33
34,35,36

The probabilities here are all identical: 3/37

locum-motion

Then take all three sets of probabilities I outlined in my last three posts to get the total probability:

EITHER: 8/37 * 4/37 * 3/37 = 0.002369 = 1/422

OR: 10/37 * 4/37 * 3/37 = 0.001895 = 1/528

So, you can take everything I said last night about 78 people and holidays worth €680-880 and change it!
You're now going to have about 500 unhappy people, not 77. And your prize has to be worth at least €5,000 or nobody will ever have a go.
And then of course the whole process is made a mockery of by the simple fact that not all choices have an equal probability of coming up, and the player can improve their chance by 25% simply by virtue of choosing BlackEven over BlackOdd at the first step.

locum-motion

And my "Why not just play Roulette?" point still applies.

Yakuza

I would imagine such a venture would require a licence too.

bogwalrus

locum-motion wrote: »

So having got through the above two obstacles, the player then chooses a 'Street' of three numbers for their last spin, is that right?

There are 12 possibilities:

1,2,3
4,5,6
7,8,9
10,11,12
13,14,15
16,17,18
19,20,21
22,23,24
25,26,27
28,28,30
31,32,33
34,35,36

The probabilities here are all identical: 3/37

No this bit is any three random numbers like a lotto. That was the idea anyway.

bogwalrus

locum-motion wrote: »

And my "Why not just play Roulette?" point still applies.

Because of the gambling laws. The way we wanted this game so it would be a lottory of sorts.

bogwalrus

locum-motion wrote: »

Then take all three sets of probabilities I outlined in my last three posts to get the total probability:

EITHER: 8/37 * 4/37 * 3/37 = 0.002369 = 1/422

OR: 10/37 * 4/37 * 3/37 = 0.001895 = 1/528

So, you can take everything I said last night about 78 people and holidays worth €680-880 and change it!
You're now going to have about 500 unhappy people, not 77. And your prize has to be worth at least €5,000 or nobody will ever have a go.
And then of course the whole process is made a mockery of by the simple fact that not all choices have an equal probability of coming up, and the player can improve their chance by 25% simply by virtue of choosing BlackEven over BlackOdd at the first step.

The idea was to make the first two choices seem easy enough to guess right. So lets say the player gets to the final spin then that is the hard part.


The Prize would reflect how hard the game is to win. So maybe yes the prize should be 5k but really there should be a nice balance to make it attractive and worth a gamble.

Thanks for all your contributions.

I think the game is flawed but early stages I suppose.

locum-motion

bogwalrus wrote: »

No this bit is any three random numbers like a lotto. That was the idea anyway.

OK, so it doesn't have to be one of the 'Streets'. Even so, the odds are the same. The player's still looking for one of three out of 37.

locum-motion

bogwalrus wrote: »

Because of the gambling laws. The way we wanted this game so it would be a lottory of sorts.

It's gambling either way.

bogwalrus

locum-motion wrote: »

RedEven: chance of success = 8/37
BlackEven: chance of success = 10/37
RedOdd: chance of success = 10/37
BlackOdd: chance of success = 8/37

The fact that the four options do not have the exact same odds makes a mockery of the game, tbh. No punter in their right mind would ever choose RedEven or BlackOdd.

This is something that I missed when looking at it. I will say it to my buddy because he thinks the odds are the same.

This is a big flaw.

kingsenny

bogwalrus wrote: »

This is something that I missed when looking at it. I will say it to my buddy because he thinks the odds are the same.

This is a big flaw.

The Red/Black Odd/Even refers to the left most column, not the actual numbers

locum-motion

kingsenny wrote: »

The Red/Black Odd/Even refers to the left most column, not the actual numbers

When you're playing Roulette, the left-most column is where you place your chips to indicate that you're betting Odd or Even, this is true. But what you are betting on is whether the actual number will be an even one or an odd one. So I'm not sure I understand you.

kingsenny

Sorry I'm not sure if I'm explaining it correctly.

In normal roulette, it's possible to simply place a chip on the Odd/Even section without choosing a specific number. It's (obviously) a 1 in 2 chance and i think it pays out 1/3

locum-motion

kingsenny wrote: »

Sorry I'm not sure if I'm explaining it correctly.

In normal roulette, it's possible to simply place a chip on the Odd/Even section without choosing a specific number...

Yes, absolutely correct.

kingsenny wrote: »

It's (obviously) a 1 in 2 chance and i think it pays out 1/3

It's not a 1 in 2 chance, it's 18 in 37 (because there's a 0 on the table*)
And it doesn't pay 1/3. It pays evens. Put a euro on, and you win a euro if it comes up (you also get your own euro back, of course.)

* American tablets have a 0 and a 00, so they have 38 slots on the wheel, not 37. Therefore the odds become 18 in 38, not 18 in 37. Payout remains the same.

Check here
http://en.wikipedia.org/wiki/Roulette#Bet_odds_table for the actual odds of all possible outcomes for both European and American tables, and what the payouts are.

kingsenny

Ha... sh*t. I just realized where I was going wrong. My bad! Yeah, it seems you're right

CJC86

locum-motion wrote: »

And my "Why not just play Roulette?" point still applies.

You have to pay out all winning bets on the spot. Casinos win in the long run because they have loads of capital.

locum-motion

CJC86 wrote: »

You have to pay out all winning bets on the spot. Casinos win in the long run because they have loads of capital.

And?
You do realise that the OP's mate is going to have to have enough capital put aside to pay for at least two prizes before he starts the game; even though the odds are 500 to one, there's no reason why two of the first 10 players couldn't win.

Also, casinos win because of the existence of the 0 on the board. That's the 'house margin'. The actual odds on a single number bet are 36/1, and the odds they pay are 35/1. The actual odds on a red/black bet are 19/18, but they pay evens (1/1). Of course, having loads of capital helps, because even if they lose to you at the beginning, they have enough capital to let you keep playing until you eventually lose, but the point is that you will eventually lose, and the reason that you will eventually lose is that the basic mathematics of the game are in their favour, not yours.

CJC86

locum-motion wrote: »

And?
You do realise that the OP's mate is going to have to have enough capital put aside to pay for at least two prizes before he starts the game; even though the odds are 500 to one, there's no reason why two of the first 10 players couldn't win.

Also, casinos win because of the existence of the 0 on the board. That's the 'house margin'. The actual odds on a single number bet are 36/1, and the odds they pay are 35/1. The actual odds on a red/black bet are 19/18, but they pay evens (1/1). Of course, having loads of capital helps, because even if they lose to you at the beginning, they have enough capital to let you keep playing until you eventually lose, but the point is that you will eventually lose, and the reason that you will eventually lose is that the basic mathematics of the game are in their favour, not yours.

Of course I know how casinos actually win. Now that you've pointed out how much he'd have to have for the current game he's offering (2 trips, maybe 3 if he's really unlucky), please offer me a quick back of the envelope calculation for how much he'd need to cover an actual game of roulette?

locum-motion

CJC86 wrote: »

Of course I know how casinos actually win. Now that you've pointed out how much he'd have to have for the current game he's offering (2 trips, maybe 3 if he's really unlucky), please offer me a quick back of the envelope calculation for how much he'd need to cover an actual game of roulette?

Performing such a calculation for Roulette is complicated by the fact that there are different types of bet available and that punters may choose to place bets of different sizes, but let's just perform the calc for the 'single number' bet (as that the one with the highest payout, so a bad run early on could be disastrous), and assume that the maximum stake has been set at €1.

The odds on a single number bet are 36/1. For every 37 bets that are placed, 1 of them is a winner and 36 of them are losers. The payout from the winning bet is €35 and the income from the losing bets is €36.

If 1 bet wins out of every 37 placed, then 10 win out of every 370, 100 out of every 3,700 etc. What we are trying to calculate is the size of the cushion of capital required to cover payouts if there's a really bad run in the early stages of the game. What constitutes a really bad run? Let's say a really bad run is that, out of the first 3,700 bets, the expected 100 winners all happen in the first 100 bets (and then, of course, we could expect 3,500 losing bets in a row after that!) The capital cushion needed to cover that eventuality is 100x €35 or €3,500. Now I'm afraid I'd need to bow to someone like Yakuza to calculate the odds of that actually happening, but I'm sure that the chances are much slimmer than the odds of the OP's mate getting 2-3 winners in his first 10 players.

But, regardless of the mathematical chance of 100 bets in a row being winners (or losers if you're looking from the casino's POV), it would never actually occur, because of the way Roulette is actually played. It's not just one player/one bet/one spin of the wheel. For each spin, there could be up to 6 or 8 people playing, and each of those can place as many or as few chips on the table as they wish. Therefore, for each spin on a busy roulette table, virtually all of the numbers will have at least one chip on them. Even if we take the case that the table's not so busy and only half the numbers have bets placed on them, that still means that there'll be about 17 losing bets for each winning one. The only way to have 100 bets in a row win would be if all of those 6-8 people decided to place only one bet per spin, and for them all to choose the same number, and for that number to come up, and for that to happen multiple times in a row. (If that happens, then it's time to book your table for breakfast at Milliway's.)

kingsenny

1/(37^100) I think. Ridiculously large at any rate