Consecutive Winning Points in Craps

At the Mohegan Sun they have the Fire Bet and I have read many posts on this and basically understand the math there but was curious if there was any way of calculating the odds of a player marking and winning on the same point multiple times.....for instance, a player marks and wins on a point of 6 and on the next come out roll marks the same point of 6? Do they have the same chance of winning? Does it completly boil down to how many times they have thrown the dice. I would love to see some numbers on the 3rd, 4th, 5th time won on and so on. THANK YOU

There are 24 rolls that result in establishing a point: 4 & 10 will happen 3 times each, 5 & 9 will happen 4 times each, and 6 & 8 will happen 5 times each. The odds of establishing a specific point number are the number of times it occurs out of 24. Then the odds of making the point are based on hitting the point before the seven, so you only have to consider the rolls where the point hits or the seven hits: 3 out of 9 for the 4 & 10, 4 out 10 for the 5 & 9, and 5 out of 11 for the 6 & 8. For the odds of establishing and making a specific point you multiply. For your example of the 6, this is 5/24 * 5/11 = .095 (9.5%). To find the odds of n occurances, multiply this n times, or (5/24 * 5/11)^n. To establish and make the 6 twice this is .009 (< 1%); 5 times is 7.6e-6.

I have to add that usually one is not concerned about what point is established, unless you are calculating things like the fire bet. These results are a lot smaller than, say, the odds of hitting 5 points in a row, which is approximately 1%.

So I may completly screw this up but are you saying that establishing and winning on a 6 are basically 100 to 1 and only get worse on any other number. To do it 5 times would be at least 1000000 to 1?

The specific combination of establishing and winning the 6 just one time is about 10 to 1. Doing it twice is about 100 to 1.

But these numbers really aren't related to a real craps game because we shouldn't care which point is established. Every come out will eventually establish a point. Over time, you will win approximately 40% of the points. Hitting two in a row - about 16%.

The odds of establishing and hitting a 6 would be useful information if you were to bet a friend that the next point will be a 6 and the shooter will make it OR if you are being selective about which bets you place odds behind which is not recommended.

Im not worried about placing odds or anything I am actually betting a friend but how can numbers hit twice in a row 16% of the time when on a 6 or 8 its 9.5% chance of happening or 10 to 1. If the other points only have a smaller percent then how did you get 16% combined for any number to hit cons. times

You are mixing up two different types of events -- or I'm not explaining it very well.

The 9.5% for the six is based on the following type of bet - the next point will be a six and the point will be made. The point will be a 6 for 5 out of 24 rolls, and when the point is 6 it will be made 5 out of 11 times.

The 40% average for making a point is based on the chance of making whatever point happens to be thrown - we are not stipulating the first event as we did in the previous example. 10 out of 24 point rolls will be 6 or 8 and will be made 5 out of 11 times; 8 out of 24 point rolls will be 5 or 9 and will be made 4 out of 10 times; 6 out of 24 point rolls will be 5 or 10 and will be made 3 out 9 times. Sum the products for each number: 10/24 * 5/11 + 8/24 * 4/10 + 6/24 * 3/9 = .4061 which is 40.61%.

The 16% for making two points in a row is 40% * 40%. (I consider these numbers close enough for discussion and illustration.)

Have you checked out The Wizards info on craps? If my answers are not helping you, there's a lot of information there that might answer your questions.

This is not precisely your question, but here is a table of the percentages that the player throwing the dice will get to a certain number of rolls. You have a 100% chance of throwing twice. The average number of rolls to make a point or crap out is 3.7 (so the person who said that the odds of hitting 5 points is only 1% was incorrect).
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After the percentages went below 1%, it is easier to look at the inverse (i.e. 1 out of 109 times). The odds are 1 in a thousand that you will reach 49 rolls. The final number is the longest streak ever recorded (154 throws) at Borgata casino in Atlantic City in the last two years. The Time magazine article incorrectly listed the odds as 1 in 1.56 trillion. The writer did not know how to play craps and calculated the odds of never throwing a 7, even though it is permitted on a coming out roll. The actual odds are 1 in 5.59 billion (still formidable).
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Incidentally there is no algebraic formula to do this calculation or I would give it to you. The calculation must be done by Markov Transition Matrices or the equivalent Recursive formula.
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Roll Probability
1 100.00%
2 100.00%
3 88.89%
4 77.21%
5 66.74%
6 57.61%
7 49.72%
8 42.90%
9 37.02%
10 31.94%
11 27.55%
12 23.77%
13 20.51%
14 17.69%
15 15.26%
16 13.16%
17 11.35%
18 9.79%
19 8.45%
20 7.29%
21 6.28%
22 5.42%
23 4.67%
24 4.03%
25 3.48%
26 3.00%
27 2.59%
28 2.23%
29 1.92%
30 1.66%
31 1.43%
32 1.23%
33 1.06%

34 109
35 126
36 146
37 170
38 197
39 228
40 265
41 307
42 356
43 412
44 478
45 554
46 643
47 745
48 864
49 1,002

...

154 5,590,264,072

To PacoMartin: This is a great help but as I understand this table applys to making any point and Im more interrested in singular points repeated....so how does that change your numbers?

TO SeattleDice: You are helping so much and I am very appreciative. You completly cleared that up for me. I will admit that I love math but some of this is outside my usual level. If at all possible could you produce a table of odds that goes up to each number being repeated 6 times. I understand the math and you have explained it so well but me doing it would be like driving a Mack truck through an obstical course. Thank you so much

Quote: checker7799

To PacoMartin: Im more interrested in singular points repeated....so how does that change your numbers?

Do you determine what number you want before the first point is made? Or does the first point the player makes determine which one you want repeated.

Case I: You pick a 6 or either a (6 or 8) before the game begins.
Case II: The player runs a streak of 7, 11's and craps and then rolls a 6 . The #6 is determined.
Case III: The player must make his point before the number is determined.