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View Full Version : Once and for all probability question(s)

09-12-2002, 12:06 PM
There must be someone here who is knowledgable enough to explain to an idiot like me a SIMPLE way to calulate the probability of:

a). hitting one of your outs over the next two cards

b). hitting runner-runner (i.e. hitting one card AND another card) over the next two cards.

I know there are two ways to do this. One has something to do with figuring combinations. Another has to do with figuring out the probability of it NOT happening first. PLEASE.... Can someone help a mathematically challenged person here and just tell me how to do it for myself?

Thanks VERY MUCH...

BruceZ
09-12-2002, 09:50 PM
a) Let's say you are trying to complete a 4 flush. You have 9 outs. The easiest way to do this is to compute 1 minus the probability of not getting it on either card. The probability of not getting it on the turn is 38/47 since there are 38 cards that don't complete the flush out of 47 remaining. The probability of not getting it on the river after not getting it on the turn is 37/46 since now there are only 36 remaining cards out of 46 which will not give you the flush. So the probability of not getting it on either is (38/47)(37/46) = 65%. The probability of getting it is 1 minus this or 35%.

If you want to do this with combinations, you note that there are C(47,2) ways to get the turn and river cards, and the number that do not give you a flush are C(38,2). So 1 - C(38,2)/C(47,2) = 1 - (38*37/2)/(47*46/2) = 35%. Note that this is the same computation as before. Or we could have equivalently done [C(47,2)-C(38,2)]/C(47,2) = 35% where now the numerator is the number of ways to get the flush. This is just another way of writing 1 minus.

If you want to use combinations to consider the number of ways of making the flush directly instead of thinking about the number of ways to not make it, this is slightly more difficult because you have to separate the case of getting exactly 1 flush card from 2 flush cards. For this reason I don't recommend this method in this case, but since you asked here it is: The number of ways to get exactly 1 flush card is 9*38 since there are 9 flush cards that can be combined with 38 non-flush cards. The number of ways to get exactly 2 flush cards is C(9,2) = 9*8/2. So the total probability of making the flush is (9*38 + 9*8/2)/C(47,2) = 35%.

b) Probability of catching runner-runner is (9/47)(8/46) = 3.3% since there are 9 cards out of 47 to catch on the turn, and 8 out of 46 ways to catch on the river after you have caught on the turn. Note this is similar to the calculation of not making the 2 card flush above except there is no reason to do 1 minus here.

To do this with combinations, there are C(9,2) = 9*8/2 ways to get runner-runner out of C(47,2) total ways to get the turn and river, so the probability is C(9,2)/C(47,2)= 3.3%.

We get the same answers to these problems in different ways, so we can be sure they are all correct.

BruceZ
09-12-2002, 10:49 PM
For the runner-runner case, there are actally 10 flush cards, so here is a rewrite of that case. Doing it two different ways didn't guarantee correct results if we get this part wrong /forums/images/icons/blush.gif

b) Probability of catching runner-runner is (10/47)(9/46) = 4.2% since there are 10 cards out of 47 to catch on the turn, and 9 out of 46 ways to catch on the river after you have caught on the turn. Note this is similar to the calculation of not making the 2 card flush above except there is no reason to do 1 minus here.

To do this with combinations, there are C(10,2) = 10*9/2 ways to get runner-runner out of C(47,2) total ways to get the turn and river, so the probability is C(10,2)/C(47,2)= 4.2%.

09-13-2002, 11:20 AM