Probability Of Winning Craps On First Roll
- Probability Of Winning Craps On First Roll Hall Of Fame
- Probability Of Winning Craps On First Roll Ups
- Probability Of Winning Craps On First Roll Dough
- Probability Of Winning Craps On First Roll Poker
For example, since a 7 or an 11 is a winner on the first roll and their probabilities are 6/36 and 2/36, the probability of winning on the first roll is 6/36+2/36=8/36. Probabilities of Winning, Losing, or Getting a Point on First Roll. For example, if you want to calculate the probability of rolling a total of 2 on 2 dice, you would multiply the probabilities of rolling a 1 on the first die by the probability of rolling a 1 on the second die. 1/6 X 1/6 = 1/36, which can be represented as odds of 35 to 1.
In the game of craps you roll two dice. If on the first rollthe sum is 7 or 11 you win. If it is 2, 3, or 12 you lose. With anything elsethat becomes your “point”. If you roll your point again before your roll a 7you win.
The first step in calculating the probability you will is tofind the probability that you will “make your point”, i.e., roll that numberbefore a 7.
Probability Of Winning Craps On First Roll Hall Of Fame
Suppose that your point is 5. There are 4 ways to roll a 5versus 6 to roll a 7, and 26 other outcomes which continue the game. I claimthat the probability you roll a 5 before a 7 is 4/(4+6). To see this thinkabout a bag with 36 marbles: four are marked with 5, six are marked with 7, andthe other 26 are blank. If we draw repeatedly from the bag with replacementuntil we get a non-blank marble then when we do the choice will be one of theten non-blank marbles with equal probability. The probability that marble has a5 is then 4/10.
Probability Of Winning Craps On First Roll Ups
In the same way we can compute the probabilities for theother points. Taking into account what happens on the first roll we have thefollowing table:
Roll2
Prob.1/36
P(win)0
Total probability of winning is
Probability Of Winning Craps On First Roll Dough
anemone
MHB POTW Director
- Feb 14, 2012
- 3,766
Probability Of Winning Craps On First Roll Poker
P( winning a game of craps by rolling a sum of 6)
=P(getting 6 the first time and 6 the second time) or
P(getting 6 the first time, get some number other than 6 and 7 the second time and get 6 again the third time) or
P((getting 6 the first time, get some number other than 6 and 7 the second and third time and get 6 again the fourth time)...
the sequence continues in such a manner infinitely (this is to say that we're having an infinite sum of a geometric sequence)
= $((frac{2!}{36}times2)+frac{1}{36})+ S_infty$
=$frac{5}{36}+ frac{(frac{5}{36})^2}{1-frac{25}{36}}$
=$frac{5}{36}+ frac{25}{396}$
=$frac{20}{99}$
But a quick goggle search shows that I don't have to consider the infinite sequence and the probability of winning a game of craps by rolling a sum of 6 for the third time and so on is simply
$frac{P(winning ;a ;game ;f ;craps ;by ;rolling ;a ;sum ;of ;6 }{P(winning ;or ;losing)}=frac{frac{5}{36}}{frac{5}{36}+frac{1}{36}}=frac{5}{11}$
Therefore,
P( winning a game of craps by rolling a sum of 6)
=P( winning a game of craps by rolling a sum of 6 the second time) and P(winning a game of craps by rolling a sum of 6 on the third time and so on)
=$frac{5}{36}timesfrac{5}{11}=frac{25}{369}$
Finally,
P( winning a game of craps by rolling a sum of 6)=$frac{5}{36}+ frac{25}{396}=frac{20}{99}$
I can't quite get my head around that fact. It has confused rather than enlighten me.
Could someone please explain to me why this works?
Thanks.