Calculating Craps Cost Per Hour

Quote: unJon

Do you think you win 1/2 or 50% of the time after a point of 4 or 10 is established?

I know I win half the time. So CO, there are 8 ways to win and 4 ways to lose right? Accordingly 4+8=12 or 8/12 or (2/3) two-thirds. Do you believe you win the CO 67% of the time? BTW the payout is irrelevant to the percentage of wins vs losses.

tuttigym

Quote: tuttigym

Quote: unJon

Do you think you win 1/2 or 50% of the time after a point of 4 or 10 is established?

I know I win half the time. So CO, there are 8 ways to win and 4 ways to lose right? Accordingly 4+8=12 or 8/12 or (2/3) two-thirds. Do you believe you win the CO 67% of the time? BTW the payout is irrelevant to the percentage of wins vs losses.

tuttigym
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Wait. You really truly think you make the point 50% of the time when the point is 4/10?

Do you think you make the point 2/3 of the time when it’s a 5 or 9? 5/6 of the time when it’s 6 or 8??

Do you think you win coin flips 100% of the time?

This is fascinating.


And I’ll answer your question but only if you keep answering mine.

I already answered your question in my original post.

The CO has 36 outcomes. 8 winners, 4 losers, 24 “continue playing”.

Quote: unJon

Do you think you make the point 2/3 of the time when it’s a 5 or 9? 5/6 of the time when it’s 6 or 8??

Yes

Quote: unJon

Do you think you win coin flips 100% of the time?

No.

The number of possible rolls of a particular point does not relate to the win/loss percentiles. Actual outcomes of a point are based on the number of ways to win vs the number of ways to lose not the number of ways a number can be rolled. Why is that a foreign concept to anyone?

tuttigym

Quote: tuttigym

Quote: unJon

Do you think you make the point 2/3 of the time when it’s a 5 or 9? 5/6 of the time when it’s 6 or 8??

Yes

Quote: unJon

Do you think you win coin flips 100% of the time?

No.

The number of possible rolls of a particular point does not relate to the win/loss percentiles. Actual outcomes of a point are based on the number of ways to win vs the number of ways to lose not the number of ways a number can be rolled. Why is that a foreign concept to anyone?

tuttigym
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Because it’s wrong. And wrong math is foreign. Your chance of winning is the number of ways to win divided by (the number of ways to lose plus win).

So the point is 6. You claim you make the point 5/6 of the time? So you claim that you will roll a 6 before a 7 5/6 of the time. Well more than half. You think you win that often? 83% of the time you roll a 6 before a 7??

How often do you not make the point? 1/6 of the time? Or 7/6 of the time? Lol

Quote: unJon

So the point is 6. You claim you make the point 5/6 of the time? So you claim that you will roll a 6 before a 7 5/6 of the time. Well more than half. You think you win that often? 83% of the time you roll a 6 before a 7??

How often do you not make the point? 1/6 of the time? Or 7/6 of the time? Lol
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No. It is not how often one wins the point. It is the odds of winning the point. One does not win the 6 point 5 out of 6 times; one has an 83% chance of winning the 6 point. Further one has a 67% chance of winning the 5 or 9, and one has a 50% chance of winning the 4 or 10. Your illustration above misleads and misstates the outcomes. Are these values correct? If they are, then my calculations are correct.

tuttigym

Quote: tuttigym

Quote: unJon

So the point is 6. You claim you make the point 5/6 of the time? So you claim that you will roll a 6 before a 7 5/6 of the time. Well more than half. You think you win that often? 83% of the time you roll a 6 before a 7??

How often do you not make the point? 1/6 of the time? Or 7/6 of the time? Lol
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No. It is not how often one wins the point. It is the odds of winning the point. One does not win the 6 point 5 out of 6 times; one has an 83% chance of winning the 6 point. Further one has a 67% chance of winning the 5 or 9, and one has a 50% chance of winning the 4 or 10. Your illustration above misleads and misstates the outcomes. Are these values correct? If they are, then my calculations are correct.

tuttigym
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No they are not correct. One wins the 6 point 5 out of 11 times. One has a 45.45% chance of winning the 6 point. The correct odds to pay for such an event are 6:5.

Too many posts to read em all. This is an easy calculation. Every time you bet the DP line, you lose the house percentage times your bet. Multiply that figure by the number of times you bet the DP line per hour.

The rest is just variance.

Quote: tuttigym

No. It is not how often one wins the point. It is the odds of winning the point. One does not win the 6 point 5 out of 6 times; one has an 83% chance of winning the 6 point. Further one has a 67% chance of winning the 5 or 9, and one has a 50% chance of winning the 4 or 10. Your illustration above misleads and misstates the outcomes. Are these values correct? If they are, then my calculations are correct.

tuttigym

These values are completely incorrect. You are overlooking the fact that the bet on a point of 6 does not resolve about 69% (i.e. 25/36) of the times you try to resolve it. The odds of winning the 6 point are 45.454545% at the instant you throw the dice to try to resolve it.