ksdjdj
ksdjdj
Joined: Oct 20, 2013
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September 4th, 2019 at 7:27:50 PM permalink
There is a Multi bet promotion where you get $200 back (in bonus*** bets) if you have a "5-leg multi and miss by one leg, minimum price per leg $1.30, and you can take advantage of this offer as many times as you like (as long as you have at least one selection different on the next multi-bet)"

***: When using the bonus bet, if the bonus bet wins you would only keep the winnings (not the stake), so the $200 is probably worth a maximum of ~$170 in real value

Firstly, through trial and error,I have found out that for "miss by one leg" bonuses the best chance to take each leg are (see formula below):

(L-1)/L where L is the number of legs (see example)

So, if you had a 5 Leg multi, you would look for all the games to be about 80% chance per leg ((5-1 legs)/5 legs), in other words you would pick something that was about $1.1875, if the juice was ~5% with that book.

Note: If this bonus had no other rules, then 80% chance per leg would be the ideal amount to maximize this bonus, but since you have to take minimum odds of $1.30 per leg, the true max value of the bonus is ~11% (see below for working out)

Odds received for a 5 leg multi @ $1.3 per leg = $3.71293.
implied chance of winning one leg @ odds of $1.30 and juice of ~5% = 73.26%
chance of winning 5 legs, if all legs are $1.30 = 0.7326^5 =~0.211 or 21.1%
chance of "missing by one leg = (0.7326^4 x (1-0.7326)) x 5 = ~0.385 or 38.5%
chance of losing = 1- ("win all 5 legs" + "miss by one legs") = 1 - (0.211 + 0.385) = ~0.404 or 40.4%

Therefore, the "gross value of the promotion" is (see below):

(chance of winning 5 legs x (odds - 1) ) - (chance of losing) = 57.249...% - 40.385...% = ~ +16.86%.

Then to work out the max "net value of the promotion, you do (see below);

"Gross value" - ((200-170^^^)/200 x chance of missing by one leg) = ~ 16.86 % - 5.77% = ~11.08...%

Note: for the above formula, 200 is the gross value of the bonus and 170 is the max/net value of the bonus.

Therefore the maximum EV of this promotion is about +11%.

-------------------------
Question

Hope this was helpful , but does anyone know why "the chance per leg" should be (L-1)/L for any multi bet promotion that misses by one leg, to maximize value?

Note: i worked this out using pascal's triangle, and through trial and repeat (trial and error) I "know" this general formula^*^ is true from 2 to 6 legs (but i haven't tested it any further than this).

^*^: (L-1)/L (chance per leg) to maximize the chance of "missing by one leg"


Thanks for your time
Last edited by: ksdjdj on Sep 5, 2019
DogHand
DogHand
Joined: Sep 24, 2011
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Thanks for this post from:
ksdjdj
September 5th, 2019 at 5:02:27 AM permalink
Quote: ksdjdj

<snip>Firstly, through trial and error,I have found out that for "miss by one leg" bonuses the best chance to take each leg are (see formula below):

(L-1)/L where L is the number of legs (see example)

<snip>

Question

Hope this was helpful , but does anyone know why "the chance per leg" should be (L-1)/L for any multi bet promotion that misses by one leg, to maximize value?

Note: i worked this out using pascal's triangle, and through trial and repeat (trial and error) I "know" this general formula^*^ is true from 2 to 6 legs (but i haven't tested it any further than this).

^*^: (L-1)/L (chance per leg) to maximize the chance of "missing by one leg"


Thanks for your time



ksdjdj,

This question can be answered using calculus. If we let w = probability of winning each leg (so the probability of losing each leg is 1-w), and L = number of legs, then the probability of S = winning (L-1) legs is given by this:

S = L*[w^(L-1)*(1-w)] = L*[w^(L-1) - w^(L-1)*w] = L*[w^(L-1) - w^L]

Now we want to find the value of a that maximizes S, so we take the derivative of S with respect to w, set the result equal to zero, and solve for w:

dS/dw = L*[(L-1)*w^(L-2) - L*w^(L-1)] = 0

Divide by L*w^(L-2):

L - 1 - L*w = 0, so

w = (L-1)/L

as you found by trial and error.

If we substitute this into the S equation, we find the probability of success is then:

S = L*[w^(L-1)*(1-w)] = L*[((L-1)/L)^(L-1)]*[1-(L-1)/L] = L*[((L-1)/L)^(L-1)]*[1/L] = ((L-1)/L)^(L-1)

so for the 5-leg example, S = (4/5)^4 = 0.4096

Hope this helps!

Dog Hand
Last edited by: DogHand on Sep 5, 2019

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