Expected return of money.

I’m trying to figure out how to get an expected return of a wager on a point number. Ex. I wager $10 on pass line and a point of 8 hits. Should it be an expected return of .8333 (5/6 the odds of 8 hitting before a 7 ) or 45% (the percentage of the 8 winning). One should be losing avg $1.7 and the other $1 if I’m not mistaken. Or am I getting this stuff confused and I am completely wrong on how I should expect the return to be? Thanks if anyone could clear this up.

Quote: Farha123

I’m trying to figure out how to get an expected return of a wager on a point number. Ex. I wager $10 on pass line and a point of 8 hits. Should it be an expected return of .8333 (5/6 the odds of 8 hitting before a 7 ) or 45% (the percentage of the 8 winning). One should be losing avg $1.7 and the other $1 if I’m not mistaken. Or am I getting this stuff confused and I am completely wrong on how I should expect the return to be? Thanks if anyone could clear this up.

5 ways to win, 6 ways to lose, total of 11 outcomes. Your return is:

5*$20 / 11 = $100/11 = $9.0909


Other way to do it is 5/11 you win $10 and 6/11 you lose $10.

$50/11 - $60/11 = -$10/11 = -$0.90909


Losing $0.90909 is the same as getting $9.0909 back.

The house edge of a pass line bet is 1.41%.
Therefore, the expected return is .9859 per dollar bet.

You really need to ignore the part about it being an 8. After all, any other number could be rolled...

Thanks, Appreciate the response. The second way makes it a lot clearer to me on understanding the math. The $20 on the first problem is for $10 on the 7 and 8?

So on the 4 the math would be: 3*20/9= 6.67?

Thanks, Appreciate the response. The second way makes it a lot clearer to me on understanding the math. The $20 on the first problem is for $10 on the 7 and 8?

So on the 4 the math would be: 3*20/9= 6.67?

Point	Win	Lose	Flat Loss	Loss/Point	Odds	Total Odds	Total Line	Total Bet	Loss Percent
4, 10	3/9=$30	6/9 = $60	$30/9	$3.33	$25	$225	$90	$315	1.05714
5, 9	4/10 = $40	6/10 = $60	$20/10	$2.00	$10	$100	$100	$200	1.00000
6, 8	5/11 = $50	6/11 = $60	$10/11	$0.9091	$5	$55	$110	$165	0.55100

That last column might be wrong. I used Loss/Point divided by Total Bet instead of Flat Loss divided by Total Bet. I might have some time to re-edit later.

Quote: DJTeddyBear

The house edge of a pass line bet is 1.41%.
Therefore, the expected return is .9859 per dollar bet.

You really need to ignore the part about it being an 8. After all, any other number could be rolled...

He means the house edge after a point of 8 is established. In the case, the house edge is 9.09%.

Quote: BlackjackLover

He means the house edge after a point of 8 is established. In the case, the house edge is 9.09%.

I knew that. I was pointing out that it’s a very bad way to look at things since you can’t insure that the point will be 8, or even that a come out roll with produce a point at all!

Quote: DJTeddyBear

I knew that. I was pointing out that it’s a very bad way to look at things since you can’t insure that the point will be 8, or even that a come out roll with produce a point at all!

I don’t think so, as I can imagine at least one reason why this would be good information to have.

Quote: Farha123

Thanks, Appreciate the response. The second way makes it a lot clearer to me on understanding the math. The $20 on the first problem is for $10 on the 7 and 8?

$20 is what is returned to the player on a win for your $10 example.

Quote: Farha123

So on the 4 the math would be: 3*20/9= 6.67?

why not just use a $1 bet, that way you can just multiply out what total bet on the pass line, say $25 or $6 or $13
RTP
for a POINT
6&8 = 2*5/11 = $10/11 (round to $0.91 per $1 bet)
5&9 = 2*4/10 = $8/10 = $4/5 ($0.80 per $1 bet)
4&10 = 2*3/9 = $6/9 = $2/3 (round to $0.67 per $1 bet)

Come out roll
do not forget that the pass line can win on the come out roll. no point required
8 ways to win and only 4 ways to lose
2*8/12 = $16/12 = $1 1/3 (round to $1.33 per $1 bet)

why the question as some want to know