Any Mathematicians out there?

[kill_em_all_tigers]

Does anyone know if there is a formula for the following problem?
And if so, what may it be?

If you have two different options for a bet and both are paying over $2 then it is possible to place a certain amount of money on both bets that will not only pay you for your desired return but also pays for the money invested on the losing bet.

E.g

Desired Return = $100
Bet 1 Odds = $5.50
Bet 2 odds = $2.75

If i put on $40 @ $5.50 it’ll return $220,
And if i put $80 @ $2.75 it’ll return $220.

$220 – $120 (both bet amounts) = $100 my desired return. Just as long as one of these two bets wins.

So what formula would be used to find out the amount to put on the two bets when only the odds and desired return are known?

 

[Tiger_Watto]

your taking the fun out of placing a wager… put your money in a high earning interest account!

 

[alexaki]

ehhhh 1+1=11

 

[kill_em_all_tigers]

haha

Half the fun (and pain) is all in the amount you put on.

The other half of the fun is devising plans to give you an advantage.
The other half of the pain is realising they don't work until after you put your money on. :slight_smile:

 

[Justo]

7
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[happy_tiger]

Probably depends whether John Elias or Ryan Tandy are involved

Mathematical X factor there

 

[Sabre]

You wouldn't find a bet where if there was only 2 options those would be the odds. The people who come up with these odds aren't dumb.
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[Saturnalia]

Does anyone know if there is a formula for the following problem?
And if so, what may it be?

If you have two different options for a bet and both are paying over $2 then it is possible to place a certain amount of money on both bets that will not only pay you for your desired return but also pays for the money invested on the losing bet.

E.g

Desired Return = $100
Bet 1 Odds = $5.50
Bet 2 odds = $2.75

If i put on $40 @ $5.50 it’ll return $220,
And if i put $80 @ $2.75 it’ll return $220.

$220 – $120 (both bet amounts) = $100 my desired return. Just as long as one of these two bets wins.

So what formula would be used to find out the amount to put on the two bets when only the odds and desired return are known?

Using the odds you used first you'd need to find the minimum amount you would need for each individual odd to return the amount you desire

in this case it would be:

$100 / $5.50 = $18 to win $100
and
$100 / $2.75 = $36 to win $100

This will return you $200 if they both win, but you're trying to offset the cost if you lose one.

So now you have to find the amount you'd need to bet on bet 1 if bet 2 loses.

$136 / $5.50 = $24 to win $100 and cover the loss from bet 2

And you have to find the amount you'd need to bet on bet 2 if bet 1 loses.

$118 / $2.75 = $42 to win $100 and cover the loss from bet 1

So it would be

Bet 1: $24 @ $5.50
Bet 2: $42 @ $2.75

If both wins you should return $242, if only one wins it should return $100(ish)

I may have my numbers wrong but it's late and I'm tired.

 

[underdog]

"The house always wins"

thats why when teams are dead heated, the are both at $1.90 - the 10 cents is the margin the sports tab make on even bets, and if you chuck a 20 on BOTH teams, you will lose.

The best options only come up when something affects the result before the game- something like this.

Tigers vs Newcastle - kick off sunday

* Tuesday - teams released - No injuries, cept for Gidley. - Tigers paying $1.55 - Newcastle - $2.70 - Put 25 Bucks on Newcastle @2.70 (bear with me here)

* Wednesday - Knights sign Willie Mason - fighting fit, rearing to go. Gidley is cleared to play on the weekend - average punters think Newcaslte might win, and back them into $2.10 - Tigers are paying $1.75

* Thursday - Benji Marshall rolls ankle at training and knocks Farah in the nose- both effectively ruled out by Dr Quah. Newcastle get money dumped on and firm into $1.45 favourites, Tigers out to $2.85

*Saturday - News on the forums that Marshall and Farah are cleared, nothing official yet - JUMP ON the tigers at 2.85 - 30 bucks.

* Sunday 1hr before kickoff - marshall names in jumper 19, farah in jumper 20\. tigers plunge back to 1.90.

now, as you can see you have put yourself in a win/win situation, however it takes a series of unfortunate events (or prior knowledge of an event) to really take full advantage. The old cliche of "the house always wins" applies here.

One betting tip i can give you revolves around the Tennis - in particular 3 set matches - anytime someone is unbackable (e.g - $1.15 or less) see what the odds are of a 2-0 straight set win - (generally out at around 1.40 or so) since most floggings in tennis are in straight sets, its most of the time a fairly easy win.

 

[monkey]

Like last week tigers paying $4.50 to win 19+ waited til Friday when the news of all the panthers injuries had to dump $50 on it
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[Juro]

Okay, time for some algebra.

Let x = the first bet $ amount
Let y = the second bet $ amount
Let r[x] = the return on the first bet ($5.50 in your example)
Let r[y] = the return on the second bet ($2.75 in your example)
Let P = the payout ($100 in your example)

Now, if x wins and y loses, you win x.r[x] and pay x+y.
And, if y wins and x loses, you win y.r[y] and pay x+y.

You want both of these equations to equal P.

x.r[x]-x-y=P (1)
y.r[y]-x-y=P (2)

Rearranging equation (1), x=(P+y)/(r[x]-1)
Putting this into equation (2), y(r[y]-1)-(P+y)/(r[x]-1)=P
Multiplying both sides by (r[x]-1), y(r[x]-1)(r[y]-1)-y=P(r[x]-1)
or, y=Pr[x]/((r[x]-1)(r[y]-1)-1)

Solving with the stated assumptions:
y=100*5.50/((5.50-1)(2.75-1)-1)
y=100*5.50/6.875
y=$80

Flipping the equation, x=Pr[y]/((r[x]-1)(r[y]-1)-1)
x=100*2.75/((5.50-1)(2.75-1)-1)
x=100*2.75/6.875
x=$40

You bet $40+$80 = $120
If only x wins, you win 40*5.50 = $220 and are up $100
If only y wins, you win 80*2.75 = $220 and are up $100
If both win, you win $440 and are up $320
If both lose, you win $0 and are down $120

 

[ryanda01]

Okay, time for some algebra.

Let x = the first bet $ amount
Let y = the second bet $ amount
Let r[x] = the return on the first bet ($5.50 in your example)
Let r[y] = the return on the second bet ($2.75 in your example)
Let P = the payout ($100 in your example)

Now, if x wins and y loses, you win x.r[x] and pay x+y.
And, if y wins and x loses, you win y.r[y] and pay x+y.

You want both of these equations to equal P.

x.r[x]-x-y=P (1)
y.r[y]-x-y=P (2)

Rearranging equation (1), x=(P+y)/(r[x]-1)
Putting this into equation (2), y(r[y]-1)-(P+y)/(r[x]-1)=P
Multiplying both sides by (r[x]-1), y(r[x]-1)(r[y]-1)-y=P(r[x]-1)
or, y=Pr[x]/((r[x]-1)(r[y]-1)-1)

Solving with the stated assumptions:
y=100*5.50/((5.50-1)(2.75-1)-1)
y=100*5.50/6.875
y=$80

Flipping the equation, x=Pr[y]/((r[x]-1)(r[y]-1)-1)
x=100*2.75/((5.50-1)(2.75-1)-1)
x=100*2.75/6.875
x=$40

You bet $40+$80 = $120
If only x wins, you win 40*5.50 = $220 and are up $100
If only y wins, you win 80*2.75 = $220 and are up $100
If both win, you win $440 and are up $320
If both lose, you win $0 and are down $120

Wow.
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