Help! EV for DP Craps Strategy with Hedge

Can someone help me calculate the EV for the following craps strategy.
1 unit on don't pass and 2 units hedge lay bet on 10.
Remove 10 lay bet once point is established.
Replenish DP bet if knocked off by 11
So, the 10 lay bet is a temporary hedge bet against the 7 during come-out roll.
I hope I explained it clearly.

I understand it.
If the comeout is:
...a 2 or 3 (probability 3/36), you win 1 on the DP; total EV = 3/36
...a 4 (probability 3/36), there is a 2/3 chance of rolling a 7 before a 4 (win 1 on the DP) and 1/3 of rolling a 4 first (lose 1), so that's 3/36 x (2/3 - 1/3) = 1/36
...a 5 (probability 4/36), there is a 3/5 chance of rolling a 7 before a 5 and 2/3 of rolling a 5 first, so that's 4/36 x (3/5 - 2/5) = (4/5) / 36
...a 6 (probability 5/36), there is a 6/11 chance of rolling a 7 before a 6 and 5/11 of rolling a 6 first, so that's 5/36 x (6/11 - 5/11) = (5/11) / 36
...a 7 (probability 6/36), you lose 1 on the DP but gain 19/20 on the lay bet; EV = -1/6 x -1/20 = -(3/10) / 36
...an 8 is the same as a 6; that's another (5/11) / 36
...a 9 is the same as a 5; that's another (4/5) / 36
...a 10, you lose the 2 you laid on the 10, and then it is the same as a 4, so that's -2 x 3/36 + 1/36 = -5/36
...an 11 (probability 2/36), you lose 1 on the DP; total EV = -2/36
...a 12 (probability 1/36), that is a push
The sum of the EVs is -(87/110) / 36, or about -0.02197

...1/3 of rolling a 4 first (lose 1)
...a 10, you lose the 2 you laid on the 10, and then it is the same as a 4, so that's -2 x 3/36 + 1/36 = -5/36

There is no lay 4 bet, so why lose with a 4?

Saw a bubble craps strategy yesterday where the player put $1 lay bets on the 4 & 10 on the come-out (it would pay $0.47 each on a come-out 7 winner or $0.94 total) and he would put $2 on the DP (which would lose on the come-out 7-11's). Then once the point was established he would move the lay bets to buy bets on the 4 & 10 and they would pay $1.95 each on a win. If the point was 4 or 10, he had the option to move the buy bet over to the sister number but didn't. He only put $20 into the machine and wasn't budging more than $5 from his buy-in over half an hour, so it just seemed like a waste of time.
He probably should have done $2 lay & buy bets and $4 on the DP so the half cent breakage on the lay bet win wasn't affecting him.

If I wanted to try it out on a real table, I'd need $41 lay bets (vig upfront) on the 4 & 10 and $80 on the DP, and $42 buy bets (vig upfront) on the 4 & 10; and I'd be pressing the 4 & 10 on a win a bit and setting the dice for that. See if a $1,000 buy-in lasts me long for that. With the vig upfront, the HA gets rather high, so it's probably better to do it on the bubble craps machine, but I'd rather set the dice, so maybe I shouldn't bother with this at all unless I'm gonna win big at this early.

Quote: roumin

...1/3 of rolling a 4 first (lose 1)
...a 10, you lose the 2 you laid on the 10, and then it is the same as a 4, so that's -2 x 3/36 + 1/36 = -5/36

There is no lay 4 bet, so why lose with a 4?
link to original post

The point is 4, so you lose the 1 you bet on DP

Thanks - That makes sense. so, the cost of the hedge is higher than the benefit, since EV for straight DP is -1.4%

Quote: roumin

Thanks - That makes sense. so, the cost of the hedge is higher than the benefit, since EV for straight DP is -1.4%
link to original post

Yes and you can know that without having to calculate the EV simply by knowing that the hedge bet itself is a bet with a house edge. So it by definition has negative EV.

Quote: ThatDonGuy

I understand it.
If the comeout is:
...a 2 or 3 (probability 3/36), you win 1 on the DP; total EV = 3/36
...a 4 (probability 3/36), there is a 2/3 chance of rolling a 7 before a 4 (win 1 on the DP) and 1/3 of rolling a 4 first (lose 1), so that's 3/36 x (2/3 - 1/3) = 1/36
...a 5 (probability 4/36), there is a 3/5 chance of rolling a 7 before a 5 and 2/3 of rolling a 5 first, so that's 4/36 x (3/5 - 2/5) = (4/5) / 36
...a 6 (probability 5/36), there is a 6/11 chance of rolling a 7 before a 6 and 5/11 of rolling a 6 first, so that's 5/36 x (6/11 - 5/11) = (5/11) / 36
...a 7 (probability 6/36), you lose 1 on the DP but gain 19/20 on the lay bet; EV = -1/6 x -1/20 = -(3/10) / 36
...an 8 is the same as a 6; that's another (5/11) / 36
...a 9 is the same as a 5; that's another (4/5) / 36
...a 10, you lose the 2 you laid on the 10, and then it is the same as a 4, so that's -2 x 3/36 + 1/36 = -5/36
...an 11 (probability 2/36), you lose 1 on the DP; total EV = -2/36
...a 12 (probability 1/36), that is a push
The sum of the EVs is -(87/110) / 36, or about -0.02197
link to original post

how would the EV change if I hedge with 1 unit on lay 5 or lay 6 instead of 2 units on 10?

From another website:
LAY BETS HOUSE EDGE
The best deal on lay bets is on 4 or 10, where the commission represents the smallest proportion of your bet.

Let’s start with casinos where you must pay the commission on all lay bets, win or lose.

If you lay 4 or 10, the house edge is 2.44 percent. In the long run, per $100 wagered, you would average $2.44 in losses
If you lay 5 or 9, the house edge is 3.23 percent.
If you lay 6 or 8, the house edge is 4 percent.

What if the house charges the commission only on winning bets? Then the house edge is reduced on all numbers.

On 4 or 10, commission on winners only, the house edge is 1.67 percent.
On 5 or 9, commission on winners only, the house edge is 2 percent.
On 6 or 8, commission on winners only, the house edge is 2.27 percent.
***********************************************************************************
Where I'm from, all the tables are in the first part, and all the bubble craps are in the second part. I like laying the 5 or 9 with a 2% HA using a progression, but I do tend to get knocked off extraordinarily and early.
A better way to play is $10 DP or DC with $30 odds on any number and that only has 1.4% HA on the DP or DC bet. But for a come-out roll plus one I'd go with the Lay 5 or 9 at 2%. At a table, my minimum Lay 5 or 9's would probably start at $75 and go up by $15 increments.

since you asked another question without working it out for yourself, you may be missing the essential fact that when you make a multitude of bets, the EV of each bet is added together to get the total EV. When they are negative and added together, the amount of negative increases. Minus 2 + minus 2 is minus 4

Bets that hedge do not cancel each other in terms of EV