playing 1 or 37 numbers

Hi, I have always had a questions, a silly probably one.

In roulette there is only one number that can hit

You play a single number at an european wheel , your odds are 1/37

As there is only one possible result, it seems a waste of money to play more than 1 number because you know for sure you will miss at least one of them

Suppose you decide to play 10 spins, you invest a tottal of 10 units in case you decide to play a single number. You have got an unlikely probability to hit each of the 10 spins.

When you play 2 numbers you know for certain that you miss at least 1 number you play

I know(guess), it might be a fallacy.

I would like you to clarify this item

kind regards
ybot

Well, yeah, I guess you're right that you'd be certain to lose all but 1 (assuming the straight up 35:1 bet). It doesn't change the house edge or nothing like that, though.

I play strait bet, you play 2 numbers. My probability is 1/37, yours is 2/37

House edge does not change 2,7%

There must be a math explanation to this fallacy

play all 37 numbers

you hand over 37 units and the dealer hands you back 35 + 1 [original wager on number that won]. I think I am doing this right.

edge of 1/37=2.7%

edge does not change, it's the variance that changes; as you bet more numbers, you are more likely to have a winner on occasion but are less likely to have a winning session. Playing 37 numbers guarantees a losing session.

So, playing 37 numbers guarantees -1 unit each spin. Variance=0

Playing a single number=Variance=maximum, same house edge

Quote: ybot

So, playing 37 numbers guarantees -1 unit each spin. Variance=0

Playing a single number=Variance=maximum, same house edge

correct

final factor, expected value as you bet more

the more numbers you cover, the more you are betting overall [most likely]. Each bet carries the edge.

So there are two things undesirable in my book with betting more numbers. You lower the variance, your only hope in negative expectation.* And you increase the amount you should expect to lose, assuming you are betting more.

*some players want less variance and are willing to lose more often to get it; a comfort level thing. That can be a legitimate thing if the player knows what they are really doing. As you note, you never want zero variance or close to it, though

By playing more numbers, you are more likely to win, but you will profit less when you hit. Expected value is the same.

Two example where you risk $5 each time.

Playing ONE NUMBER for FIVE DOLLARS:
You win 1/37 of the time, and when you do, you profit 175 dollars. (35*5)
You lose 36/37 of the time, and when you do, you lose 5 dollars.
EV is (1/37)*(175)+(36/37)*(-5) = -5/37 dollars

Playing FIVE NUMBERS for ONE DOLLAR EACH:
You win 5/37 of the time, and when you do, you profit 31 dollars. (the winner makes you 35, but you lose the other 4, so 35-4=31)
You lose 32/37 of the time, and when you do, you lose 5 dollars.
EV is (5/37)*(35)+(32/37)*(-5) = -5/37 dollars

As you can see, for the same amount of money bet, your EV is the same. With spreading your money around, you will have smaller wins but they will happen more often.

If you were deciding between $5 on a single number or $5 EACH on a lot of numbers, the single number would have less expected loss.

Quote: ybot

Hi, I have always had a questions, a silly probably one.

In roulette there is only one number that can hit

You play a single number at an european wheel , your odds are 1/37

As there is only one possible result, it seems a waste of money to play more than 1 number because you know for sure you will miss at least one of them

Think of it this way:

Bet 1 on each of the 18 red numbers.

Now, instead of that, bet 18 on Red.

You will notice that, in the first case, if the winning number is red, you will gain 35 for the winning number, and lose 17 for the 17 losing numbers, so you end up gaining 18, and if the winning number is not red, you lose the 18.
In the second case, if the winning number is red, you gain 18, and if it is not red, you lose the 18.
How about that - the two bets have the same result.

Treat betting more than one number at a time as a single bet with reduced odds. If you bet two adjacent numbers, it pays 17-1; if you make separate bets of half of that size on each of the two numbers, in effect it also pays 17-1 - so why does it matter if the two bets are on adjacent numbers? You can bet on 1 and 36, and you get 17-1 on your total amount bet if either number comes up.

On a single-zero wheel, any bet that wins if any of N numbers wins is the same as if you bet 1/N on each of those numbers separately. (On a double-zero wheel, there is one exception to this; the bet that combines 0, 00, 1, 2, and 3 should pay 6.2-1, but pays only 6-1, so you are better off playing the five numbers separately.)

Quote: MrGoldenSun

By playing more numbers, you are more likely to win, but you will profit less when you hit. Expected value is the same.

correct, if you spread the same amount of action around as in your example. My point was that many players would also increase the total amount bet. Perhaps that's not usual in roulette.

Quote: odiousgambit

correct, if you spread the same amount of action around as in your example. My point was that many players would also increase the total amount bet. Perhaps that's not usual in roulette.

I think you're correct that most players would be betting more in total. I tried to make that point about "total action" in the rest of my post--I agree with you it's important to understand.

Some time ago, I read John Haigh's book "taking chances"
He made some examples of players playing 1 to 18 numbers

He said a player had 53% chance to be at least +1 unit if he played Strait bet for 179 spins

Poisson fórmula confirms it

The same fórmula for 2 or more numbers lower the chance to be up after 179 spins

Is there Someone kind enough to explain it to me?

The quick version: 53.2685% of the time, in 179 spins, you will win at least five times. If you win five times, you gain 35 x 5 = 175 from the five wins and lose 174 x 1 = 174 from the 174 losses, so you will be +1. Each win past five puts you another +36 ahead.

However, the amount you lose with 4 or less is larger, to the point where, overall, you average result will be a loss.
18.25% of the time, you will win four times; that's 35 x 4 = 140 for the wins and 175 x 1 = 175 for the losses, for a total of -35.
14.94% of the time, you will win three times; that's 35 x 3 = 105 for the wins and 176 for the losses, for a total of -71.
9.11% of the time, you will win twice; that's 35 x 2 = 70 for the wins and 177 for the losses, for a total of -107.
3.69% of the time, you will win once, that's 35 x 1 = 35 for the win and 178 for the losses, for a total of -143.
1/135 of the time, all 179 spins will lose; that's -178.
Haigh could have just as easily said that the player had a 47% chance of being -35 units or worse in 179 spins.

It's sort of like saying, "I'll roll a die; if it is 1-5, you win 1, but if it is 6, you lose 10"; 5/6 of the time, you win, but overall, your average bet is a loss.

There's only a 50.331% chance of being ahead after 179 spins if you bet 2 numbers at once; this is because you now need 10 or more wins at a 2/37 chance (assuming a single-zero wheel) rather than 5 or more wins at a 1/37 chance.

Poisson says we could hit 5 or more times within 179 trials playing 1 number.
We can hit 5 times in a row at the beginning or anytime.

The 53% probability is to be up

We know house edge cannot be avoided. We lose the same % playing 1, 2 or 37 numbers

This topic is about different probabilities to win or lose in this short run when you play 1 or more numbers.

Suppose a 53% chance to succeed, we might think of a progression every 179 trials or more, just thinking outloud.

You must pick the number, you have got -2,7%, suppose a pro VB player picking one number belonged to a half best 18 number side

As the rule for unfauvorable games is bold play to be up for some time before long run.

Is timid play a rule when your game is favourable?

Quote: ybot

As the rule for unfauvorable games is bold play to be up for some time before long run.

Is timid play a rule when your game is favourable?

Sort of. The Kelly Criterion sets an ideal bet size, and it usually winds up being many small bets as a proportion of bankroll.

Quote: odiousgambit

Sort of. The Kelly Criterion sets an ideal bet size, and it usually winds up being many small bets as a proportion of bankroll.

When I said " bold play" I mean a bet with high variability as straitgh up as opposite to play 24 numbers(2 dozens).

When you have got negative expectation, you face the truth faster the most numbers you play. Just imagine playing 37 numbers, you see in each trail that you lose 1 unit(-2,7%) This is the least variability.

We know that no matter what we do, in the end , with negative expectation , we cannot escape from ruin.

But, with a positive one, we could play 1, 6 o 24 numbers.

Kelly criterion prevents a player`s bankroll with a positive expectation from bankrupcy

Supposse you have got 1 number with 10% edge, 10 units , or, 10 numbers yielding 1% each, 1 unit each. Which play have got less variabily and risk?

Both will end up successfully, both play 10 units, both win 10 units every 100 spins played, on average

kind regards

Hi thatdonguy
as regards as john haigh calculations we do know that in a 1/37 roulette any bet we play we are -2.7%and the bank is +2 7%

There are other calculations J haigh wrote

62% chance to be up in a Strait up within 35 spins
this % drops to 27% at spin 37 because we need 2 hitaor more

48.6% chance to double your Bank rol playing Red as a very nice option in a random wheel

bold play with high variability (strait bet) to avoid long term facts

For the bank, which has got the advantage prefers players who place bets in many places in the layout

bold play adds risk of a Lucky player winning too much although in the end they have got their 2 7% no matter what

what if you have got some Kind of physical advantage due to bias o visual balistics?

You might have from 1 to 10 % edge playing from 1 to 24 numbers.

As bank prefers lower variability having an edge in a rrandom we should play with the less variance, don 't we?

Suppose a 8% edge playing 3 numbers or 1% edge playing 24 numbers. Both yield the same units in the same time. Which is the best short term strategy?

How could we take john haigh writtings to win easier in a physics advantage scenario?

Kind regards
ybot