Question for math people

[FreeRoulette]

I have 10 units. Which is better?

1. Bet 10 units on 10 numbers for one spin.
2. Bet 1unit on one number for 10 spins.

I was told this has the same odds, but I'm not convinced. I would rather have two guesses on one flip of a coin. So maybe there is a bell curve and it would be best to bet 3 numbers for 3 spins. Anyone have the math on this?

[TwoCatSam]

Can't give you the math, but think on this example...............

You have 37 chips.  If you place all 37 on the numbers, you are certain to have a chip under the puck.  If you place a single bet on any number for 37 spins, you are not certain to have the winner. 

Sam

[Turner]

Maths doest help much because of varience. I d think in terms of playing for a stop loss of 10....and having 10 goes and staying there 20 mins...not 2

[MrJ]

I'm the one who came up with this years ago.....I still the love question though. It was in my top 10 best threads ever done (imo), interesting answers.

Ken

[iggiv]

place 10 chips on 10 splits

[Azim]

place 10 chips on 10 splits

I would do it as 10 corners. If I was prolonging my torture.

[FreeRoulette]

I'm the one who came up with this years ago.....I still the love question though. It was in my top 10 best threads ever done (in my opinion), interesting answers.

Ken

Hey Ken,
Great minds think alike. Did  you come to any conclusions?

[Steve]

1. Bet 10 units on 10 numbers for one spin.
2. Bet 1unit on one number for 10 spins.

With #1 at least you have potential to win on every spin (but unlikely).

[ati]

What are the odds for three numbers coming up in order? For example 1-2-3. I'm unable to work it out.

edit: Now that I'm thinking, it doesn't really matter, because I wanna know after 1 and 2 shown, what is the chance that the next number will be 3, but that's 2.7% for any number I guess.

[vladir]

This again?

Ok, the odds are the same for both scenarios:

(play 10 numbers at once) 10/37

and

(play one number 10 times) 1/37+1/37+1/37+1/37+1/37+1/37+1/37+1/37+1/37+1/37 = 10/37

In case of a loss, it's also the same and we loose 10 units.

However, let's assume we have a win on both scenarios and see the difference.
If you have a win on the first scenario, you always end up with  36-10 = 26 units.

If you have a win in the second scenario, you end up with at least a profit of 26. But it can be more, depending on what play you had your win.

This is the reason why I think it's better to play less numbers. (Unless you have enprison rule in an european table).

[warrior]

I would bet 10 numbers using visual ballistic and try to predict where the ball will land.