Do yourself a favour, bet less but bigger !
 

Bank: $10
Goal: To increase bank to $30

Odds needed to win $20: 2/1 against.

Scenario A.
Possible results of a single $10 bet event at 2/1 against: 2
1. win: collect $20 win + $10 bet amount [bank $30]
2. Lose: No return [bank $0]
[1 in 2 chances of creating a $30 bank].

Scenario B.
Possible results of 2 x $5 single bet events: 3.
Odds of winning $10 in first bet: 2/1 against.
Odds of wining $10 in the 2nd. bet 2/1 against.

Odds of winning $20 with the above 2 x $5 bets: 3/1 against, or 1 chance in 3 !

Why ?

Because by having 2 x $5 bets we have now created 3 possible outcomes.
1. a win with both the 1st and 2nd. $5 bets [bank $30]
2. a loss with both bets [bank $0]
3. a win with one bet and a loss with one bet [bank $15]

1 chance in 3 [3/1] of creating a $30 bank instead of 1 chance in 2 [2/1].

What's that you say?
"If we get No. 3 result above we still have 3 x $5 bets left"!
OK, each bet has 3 possible outcomes, which means a 12/1 chance [9+3 poss. outcomes] from scenario B. of getting our $30 bank.
Or, you could have a 1 x $15 bet, which means 5 possible outcomes [2+3 possible outcomes] or a 5/1 chance from scenario B with this bet of scoring our $30 bank. The other alternative is a $10 x $5 bet [2+3+3 possible outcomes] at 8/1 odds of getting out $30 bank from scenario B.


The more we divide the original $10 bank into smaller and smaller bets, the more tragic and distant the original goal becomes of creating the $30 bank.

Naturally, as punters we are not going to whack a years betting money all on one bet or our hobby and interest is dead until next year [if we lose], but I'm just pointing out it's [hopefully] sound maths. to bet bigger, within a comfort zone of course and less.

Due to probability, when we consider long term goals for the year [profit hopefully], each bet we have when the odds are against us, are not all separate entities unaffected by our other bets as is commonly believed.

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Crash we might have 1/3 chances of winning with 2 X $5 bets. But we also have 1/3 chances of losing our total bank.

Compare that to a single $10 bet.

1/2 chance of winning
1/2 chance of losing

We have a greater chance in losing on our single bet, but we also have a better chance of winning.

But for me to end out with the same profit from both scenarios, one would only need to win once, against one needed to win in both selections.

I am looking through these scenarios from a point of view which involves "which is the safest option". The outcome of losing is greater in the $10 single bet than the 2 X $5 bets.

But don't worry crash, I can see the theory and the practical side behind the $10 bet instead of 2 X $5 bets. It does make sense. But it is probably something I won't do for quite some time.

I'm not sure how to make sense of all that.

If there is:
1/2 chance of winning
1/2 chance of loosing

How can losing have: ' a greater chance in losing' [?]
and winning have: 'a better chance of winning' [?]

Both winning and losing chances in the above single $10 bet is [surely] identical isn't it?

What I was trying to say, is compare the chance of winning and losing to the situation where you have the 2 X $5 dollar bets.

You said there were 3 outcomes possible.

"""Because by having 2 x $5 bets we have now created 3 possible outcomes.
1. a win with both the 1st and 2nd. $5 bets [bank $30]
2. a loss with both bets [bank $0]
3. a win with one bet and a loss with one bet [bank $15]"""""

Effectively that is 1/3 chances of losing.
Compare that to the single bet of 1/2 chances of losing.

There is also 1/3 chances of winning.
Compare that to the single bet of 1/2 chances of winning.

So 33% V 50% in both departments, showing that the single bet method, has a higher potential in BOTH win and loss strike rates.

I see. You were referring to the 2x$5 bets, not the single $10 bet.

OK, by following through on your logic, [2x$5 are better than 1 $10 bet], 20 x 50c bets would naturally be even better still.

I think we are losing sight of the original purpose of the $10 in this exercise: to try and turn it into $30 with the best mathematical chance, not protect it from risk [no bet at all will do that nicely].

If you are considering my original $10 bank in any other light except a mathematical scenario [like it was all the money we had to bet with per. year], by having 2x$5 we are risking 50% of our entire bank with each bet !

I'd be having 2c bets with it if I could. Not because it's a great way maths. wise to turn it into a $30 bank, but just because I enjoy punting and if I lose the 2x$5 bets, I'm up the creek as far as betting interest goes for the rest of the year! :-)

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Crash, your original premise is incorrect. If you are backing a (true) 2/1 shot then you have a 33% chance of winning and a 66% chance of going bust, not 50/50.

I get your point but no cigar.
We are talking bet outcomes, not the odds of a horse winning or losing.
There is no such thing as 'true odds' anyway in horse racing except after the event.

There are only 2 possible 'outcomes' of the 2/1 bet I used as an example. We win or we lose. As stated, I was referring to 'possible outcomes' [only].

Accepting a wager at 100/1 for a $10 bet there is also only 2 outcomes: [1]Win $1000 or [2] lose $10.

Crash, only two possible outcomes (barring dead heats) but the chances of each outcome occuring are not equal.

Wunfluova

p.s. I don't smoke :)

You either win or you lose. So your chances of winning are 50% and losing is 50%. But the probability of you winning in a field of 10, is 1/10.


In the end as crash said, there are only two possible outcomes. Win or lose.

But we are not having just one bet, we know we are having 100 bets plus, and we know that if we back 100 x 2-1 shots about 30% will win and 70 % will lose on average (that's if they are being picked by mere mortals) so in fact we are 100% certain of losing (about 10% of total stakes)