Pass line odds - Did Wizard of Odds make a Mistake?

My question is about the odds of winning on the pass line and subsequently winning on a point established before hitting seven.
Wizard of Odds shows the calculation for this on his website, but I believe he might have made a mistake. My question is am I wrong?

I was seeking the probability of winning before hitting a seven.

I made the following calculation: Wizard of Odds calculates the P(winning on pass line) and adds this to P(winning on a point established). All fine and good. The devil is in the details for me.

The answer I believe he derives is 49.29%, which is equal to the P(winning on pass line) = 22.22% + P(winning on point established) = 27.07%. Again all fine and good.

Except for this. I believe Wizard of Odds made a mistake in the first portion of this calculation. Here is why.

I recalculated this myself. Here is what I did. The probability of winning on on the Pass line is P(Hitting 7) + P(Hitting 11). This is equal to 6/36 +2/36 = 8/36. This is the same as 0.2222, or 22.22%. That is what Wizard of Odds says.

But you could also hit a 2,3 or 12 and then not win on the Pass Line. So you would have to subtract these probabilities in as well.

So the, P(hitting 2,3 or 12) = 1/36 + 2/36 +1/36. Thus there is a -4/36 probability which must be added to the .22222 calculation. This works out to 0.2222 - 0.11111. Thus I calculate the odds of winning on the Pass Line as 0.1111

Then, adding 0.1111 to the P(winning on the pass line): = 0.1111 + 0.2707 = 0.3818.

So I figure there is only a 38.18% chance of winning on the Pass Line or subsequently winning on a Point (all before a 7 hits in the latter case). This is not = 49.29% as Wizard of Odds says (and many others do, as well).

Needless to say, this is a significantly lower result. You should expect to lose much more often on roll where either a Pass win or Point established win occurs than what Wizrd of Odds says in his website.

Please - anyone - let me know what is wrong with my thinking, if at all.

Quote: mrh

My question is am I wrong?
Please - anyone - let me know what is wrong with my thinking, if at all.

Yes, you are wrong.

come out roll

2 1/36 win 0
3 2/36 win 0
4 3/36 win 3/9 of the time = .027778
10 same as 4 = .027778
5 4/36 win 4/10 of the time = .044444
9 same as 5 = .044444
6 5/36 win 5/11 of the time = .063131
8 same as 6 = .063131
7 6/36 win = .166667
11 2/36 win = .055556
12 1/36 win 0

Add the win totals= .492929

Quote: mrh

My question is about the odds of winning on the pass line and subsequently winning on a point established before hitting seven.

ok

your 1st post
and a
long
time
member!

what are those odds!?

Quote: mrh

Wizard of Odds shows the calculation for this on his website, but I believe he might have made a mistake. My question is am I wrong?

we shall see
as you seem to have a bunch of different questions here
or
am I wrong?

Quote: mrh

I was seeking the probability of winning before hitting a seven.

this sounds way different from the probability for ANY pass line win
as there are 8 paths that can take. The math is easy for that.
2 paths are during the come out roll round.

so you want the probability of winning
GIVEN
that a point is already made B4 a 7 rolls?

that is a different answer than the 244/495 value

that value, calculated, should be 201/495 (67/165)

or is your actual question still different?
Sally

Let's say I have a coin, it lands on heads I win and tails I lose. What's the probability that I win? It's 50%. Your logic would dictate that I have a 0% chance to win, because 50% of the time I lose, therefore 50% - 50% = 0%. The advantage in this case is 0%, but the % to win is 50%.

What you're calculating is the RETURN (or advantage), not the % TO WIN.

Quote: RS

Let's say I have a coin, it lands on heads I win and tails I lose.

a horrible example i due say
as the game (your coin flip game)

only has one round of play,

not like the pass line that can win during the come out round
or
the point round. (a second chance to win)

The OP seems to be asking about winning in the point round, to me, on the over
time may tell
and you may have scared him/her away

thank you for sharing

Sally

Quote: mustangsally

a horrible example i due say
as the game (your coin flip game)

only has one round of play,

not like the pass line that can win during the come out round
or
the point round. (a second chance to win)

The OP seems to be asking about winning in the point round, to me, on the over
time may tell
and you may have scared him/her away

thank you for sharing

Sally

Yew should try reading the OP next thyme, or at least a majority of the post. Just my too sense.

Soopoo

I responded at length, but somehow my response got not posted. I am not sure why.

I suppose I could try again but first I want to see if this gets posted.

MRH

To everyone who wrote me back, thanks for your responses.

This question has been bothering me a lot.

Here is how the question came up.

I was reading Mike Shackleford's page "How the House Edge for Each Bet is Derived" in the Wizard of Odds Appendix/1/

For the Pass/Come derivation, he derives the probablity of winning on a come out roll and then derives "The probability of establishing a point and then winning is ......"
and I get that formula. Then he says "The overall probability of winning is" ..... 244/495, (or 49.29%).

So in that case he has not deducted the event of losing on a 2,3 or 12 in come out roll.

Yes, he takes care of this in the next line by pointing out that the probability of losing is 1-244/495, or 50.7%. But that is just a math function. I believe the probability is much less, as stated above so the probability of losing is much higher.

It still does not deal with my fundamental question. The real probability of winning is not just the Gross Probability of winning. There is no accounting for hitting the 2,3 or 12 and losing on the come out on a NET basis in the formula for the probability of winning. I think he has made a mistake.

For example, in the case where there is a point established, hitting a 2,3 or 12 is not relevant since it results in a push (if you only have money on the pass line) - and so does not factor into the formula for that portion. But in the come out roll there has to be a negative probability function for 2,3, or 12 and there is not one.

I wish Mike Shackleford would respond to this, since this is a very practical question about winning either on the pass or point before a seven rolls.

Mark

The NET is -1.41% aka HE (House Edge).

I don't think you understand what you're asking/saying, or else you'd see the problem.

Quote: mrh

I wish Mike Shackleford would respond to this, since this is a very practical question about winning either on the pass or point before a seven rolls.

Mark

you are still fighting the obvious

you say this
"either on the pass or point before a seven rolls."

that is your error
the pass wins either on the come out roll or during the point round
it DOES win when a 7 rolls

why would it lose?
<<< >>>
here is an Excel in Google showing all the math
https://goo.gl/UYCR4G


what do you disagree with?

the
8 ways to win...
7 ways to lose...
counting the point round and the come out roll

The point is Four
Mark the Four

Quote: mrh

I was reading Mike Shackleford's page "How the House Edge for Each Bet is Derived" in the Wizard of Odds Appendix/1/

For the Pass/Come derivation, he derives the probablity of winning on a come out roll and then derives "The probability of establishing a point and then winning is ......"
and I get that formula. Then he says "The overall probability of winning is" ..... 244/495, (or 49.29%).

So in that case he has not deducted the event of losing on a 2,3 or 12 in come out roll.

Yes, he takes care of this in the next line by pointing out that the probability of losing is 1-244/495, or 50.7%. But that is just a math function. I believe the probability is much less, as stated above so the probability of losing is much higher.

It still does not deal with my fundamental question. The real probability of winning is not just the Gross Probability of winning. There is no accounting for hitting the 2,3 or 12 and losing on the come out on a NET basis in the formula for the probability of winning. I think he has made a mistake.

Mark, you're correct that come-out losses must be taken into account when figuring the edge, but so too must point resolution losses and you're not realizing that Michael has correctly accounted for both whereas you have not.

Some of your confusion likely stems from being sloppy with your terms. The expected value of a wager is the theoretical average outcome. It's calculated by summing all possible winning and losing outcomes. As you know, the Passline wager has two phases: a come-out phase and a point resolution phase. It can win or lose in either phase. Write down all the possibilities and add them up to discover the average outcome. In craps, it's common to use 1980 trials when doing this. For example:

#	# wagers	win	lose	total
2	55		55	-55
3	110		110	-110
4	165	55	110	-55
5	220	88	132	-44
6	275	125	150	-25
7	330	330		330
8	275	125	150	-25
9	220	88	132	-44
10	165	55	110	-55
11	110	110		110
12	55		55	-55
Sum	1980	976	1004	-28

You can see that out of 1980 outcomes, there are 976 wins and 1004 losses. So, the wins account for 976/1980 = 49.29% and the losses account for 1004/1980 = 50.71%. The expected value is -28/1980 = -1.41%.

You should be able to see how it all comes together in this table and you can also see that the numbers are borne out by running a simulation where you'll see that the actual outcomes will approach the theoretical outcomes as the number of rolls increase.

Still not convinced? Well then, if the facts don't fit your theory, change the facts! (Albert Einstein)

Steen

Quote: mrh

My question is about the odds of winning on the pass line and subsequently winning on a point established before hitting seven.
Wizard of Odds shows the calculation for this on his website, but I believe he might have made a mistake. My question is am I wrong?

I was seeking the probability of winning before hitting a seven.
...
But you could also hit a 2,3 or 12 and then not win on the Pass Line. So you would have to subtract these probabilities in as well.

Others have said this as well, but you're mixing apples and oranges.

If you care about the probability of winning, you simply add up the probabilities for all winning events. You don't subtract anything else because either the event is a winning one in which case you add the probability or it is not, in which case you ignore it because it's not a winning one. If I were to ask you "what's the probability of rolling a number less than 3 with a single die" your answer would be 2/6 because 1 and 2 are successes and 3,4,5,6 are failures. You wouldn't take 2/6 and then subtract the probability of failure (4/6) and end up with -2/6. There's no such thing as a negative probability.

In the case of the passline, the probability of winning is .4929292... The probability of losing is .5070707....
When you calculate the expectation or EV, which is a different concept than the probability of winning, you end up with:
0.4929292*1 + 0.5070707*-1 = -0.0141414.