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Old 04-21-2009, 04:46 PM   #1
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How to add odds together?

I tried searching both the forum and google, but couldn't find a simple answer.

If I have 2 sets of odds on two events, how do I add them together to get the total odds of either event happening?

Example: Event 1 is 30:1 odds, Event 2 is 19:1 odds - The combined odds of either event happening is X:X?
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Old 04-21-2009, 05:21 PM   #2
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Re: How to add odds together?

If the events are independent, then the probability that either event occurs is

30/31 + 19/20 - 30/31*19/20 = 619/620

The odds in favor of either event occuring is 619 : 1

This uses the well known formula

P(A or B) = P(A) + P(B) - P(A and B)
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Old 04-21-2009, 06:55 PM   #3
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Re: How to add odds together?

Maybe I asked the question wrong, but I don't understand. If one event happens 1 in 30 times, and another happens 1 in 19 times, how can the chance of either occuring be over 1 in 600?

Shouldn't it be at least less than 1 in 19?
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Old 04-21-2009, 07:03 PM   #4
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Re: How to add odds together?

Quote:
Originally Posted by Fuzzball View Post
Maybe I asked the question wrong, but I don't understand. If one event happens 1 in 30 times, and another happens 1 in 19 times, how can the chance of either occuring be over 1 in 600?

Shouldn't it be at least less than 1 in 19?
You weren't clear on the question.

When you say 30:1 in favor of event 1, then the probability that event 1 occurs is 30/31.

When you say 30:1 against event 1 occurring, then the probability that event 1 occurs is 1/31.

1/31 + 1/20 - 1/31*1/20 = 5/62

which is equivalent to odds "against" of 57 : 5
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Old 04-21-2009, 08:40 PM   #5
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Re: How to add odds together?

Ah ok, that makes sense now. Thanks so much for the answer.
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Old 04-22-2009, 05:15 AM   #6
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Re: How to add odds together?

jay_SHARK, u r a math stud...might need some help later on few probablity questions...sleeping now..thanks in advance
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Old 04-22-2009, 05:19 AM   #7
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Re: How to add odds together?

jshark, math stud, I like
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Old 04-23-2009, 06:08 AM   #8
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Re: How to add odds together?

Quote:
Originally Posted by Fuzzball View Post
I tried searching both the forum and google, but couldn't find a simple answer.

If I have 2 sets of odds on two events, how do I add them together to get the total odds of either event happening?

Example: Event 1 is 30:1 odds, Event 2 is 19:1 odds - The combined odds of either event happening is X:X?
You have been given the answer for independent events (meaning that the occurrence of one event does not affect the probability of the other event). Here is the general answer with special cases.


Always:

P(A or B) = P(A) + P(B) - P(A and B)

P(A and B) = P(A)*P(B|A)
P(A or B) = P(A) + P(B) - P(A)*P(B|A)

where P(B|A) = the probability that B occurs given that A occurs


A and B Independent:

P(A and B) = P(A)*P(B)
P(A or B) = P(A) + P(B) - P(A)*P(B)


A and B Mutually Exclusive:

P(A and B) = 0
P(A or B) = P(A) + P(B)


Converting between Odds and Probability

Odds of X:Y against = Probability of Y/(X+Y)

e.g. 30:1 against = 1/31

Probability of X/Y = odds of (Y-X):X or (Y/X - 1):1 against

e.g. 1/21 = odds of 20:1 against
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Old 04-23-2009, 01:41 PM   #9
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Spade Re: How to add odds together?

Quote:
Originally Posted by jay_shark View Post
If the events are independent, then the probability that either event occurs is

30/31 + 19/20 - 30/31*19/20 = 619/620

The odds in favor of either event occuring is 619 : 1

This uses the well known formula

P(A or B) = P(A) + P(B) - P(A and B)
I tested this formula but came up with a very different result.
30/31 + 19/20 - 30/31*19/20 = 619/620
30/31 +19/20=for a total of 49/51 for the first part of the equation
or another result 30/31=.968 and 19/20=.95 (.968+.95)=1.918 not sure which approach is correct.

the second part is 30/31 * 19/20 which I multiplied to get a result of 30*19=570 and 31*20=620 making the 2nd fraction 570/620

the way the problem appears to be set up is (a+b) - (a * b) but how does one come up with the top numerator of 619? 30*19=570, only when one adds the 30 and 19 to the 570 does one come up with a numerator of 619, however going by the formula shouldn't I be subtracting 570/620 from 49/51?
to make the equation look like this 49/51-570/620 ??

30/31 + 19/20 - 30/31*19/20 = 619/620 <-- how does one solve this math problem??? Or how does one set it up to come up with that result?

Appreciate any help
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Old 04-24-2009, 04:13 PM   #10
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Re: How to add odds together?

It's important that you understand how to add fractions with different denominators.

For instance, a/b + c/d = a*d/(b*d) + c*b/(b*d) = (ad + cb)/(bd)

We know that the fraction a/b is the same as a*d/(b*d) since multiplying the numerator and the denominator by a non-zero constant, in this case d, preserves the same ratio. If we multiply a/b by d/d and c/d by b/b, then we obtain an equivalent fraction with denominator (b*d) and numerator a*d + c*b.

i.e., 1/2 = 1*5/(2*5)

If the denominators of the fractions are the same, then we can simply add the numerators, without adding the denominator.

When we're multiplying fractions, we can multiply the numerator and denominator of the fractions directly.

i.e., 1/2*3/4 = 1*3/(2*4)

so we have,

30/31 + 19/20 - 30/31*19/20

= 30*20/(31*20) + 19*31/(31*20) - 30*31/(31*20)

= 619/620
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Old 04-24-2009, 09:53 PM   #11
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Spade Re: How to add odds together?

Wow thank you very much for taking the time to answer my post so thoroughly. I'm going to ask you a follow up question hopefully you will see this soon.
I played at a with 4 players. NLH a Friend had K9 suited clubs I had A5 suited clubs, we both made a flush and she asked me what are the odds that I would have an ace high flush since there were only 4 players.
From Harrington on Holdem I already knew a King high Flush is usually good and the odds it will be beat by an Ace high flush are 15.67-1. So I tried to figured based on a 15-1 longshot hand what were the odds that of the 4 players at the table I would be dealt that hand. I figure I had a 1 in 4 chance since there were 4 players, but how does one come up with the odds? 1 in 4 times 15-1? I'll try to send you a private message with the same question. Thanks again sincerely.
ransom
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Old 04-25-2009, 02:00 AM   #12
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Spade Re: How to add odds together?

so we have,

30/31 + 19/20 - 30/31*19/20

= 30*20/(31*20) + 19*31/(31*20) - 30*31/(31*20)

= 619/620
in the above problem it appears that there was a typo in the 2nd part. which should read - 30*19/(31*20) not - 30*31/(31*20) only then was I able to come up with the correct anser of 619/620 can that then be expressed as 619-1?
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Old 04-25-2009, 07:29 AM   #13
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Re: How to add odds together?

Quote:
Originally Posted by txsransom04 View Post
so we have,

30/31 + 19/20 - 30/31*19/20

= 30*20/(31*20) + 19*31/(31*20) - 30*31/(31*20)

= 619/620
in the above problem it appears that there was a typo in the 2nd part. which should read - 30*19/(31*20) not - 30*31/(31*20) only then was I able to come up with the correct anser of 619/620 can that then be expressed as 619-1?
Yup, you have the right idea.
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Old 04-26-2009, 02:32 PM   #14
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Re: How to add odds together?

Can someone please help me with this simple question.

If there are 2 cards you want in a deck of 50, what's the probability that 1 of those cards will appear when you pick 3 cards out of the 50?

What's the probability that 2 will appear when you pick 3 cards out of the 50?
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Old 04-26-2009, 03:00 PM   #15
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Re: How to add odds together?

I have one elaboration to add to Jay "Math Stud" Shark and BruceZ. Suppose one event has probability p. the other event has probability q and the probability of both occurring is s. Then the correlation of the two events is defined as:

r = (s - p*q) / [p*(1-p)*q*(1-q)]^0.5

We can manipulate that formula into:

s = p*q + r*[p*(1-p)*q*(1-q)]^0.5

r can range from -1 to 1. If it's zero, we get the result s = p*q, which is correct for independent events. If r > 0 then the probability of both events in higher than p*q, if r < 0 then the probability of both events is less than p*q.

I don't quite understand txsransom04's flush question. If you know the odds are 15 to 1 that one of four players will have an Ace-high flush, and the players are equally likely to have it, then the odds that a specific one of the four has an Ace-high flush is 63 to 1. The reason is these are mutually exclusive events, so if the probability of anyone having the Ace-high flush is 1/16, then the probability of any one player having it is that divided by 4, or 1/64.

If there are three clubs on the board, then there are ten among the remaining 47 cards of the deck. If you deal two of them at random, the chance is (10/47)*(9/46) = 90 / 2,162 or about 1 / 24 of getting two clubs. 20% of these are A-high flushes (assuming the suited Ace is not on the board) so that's 1 / 120 chance. With four players it's about 1 / 30 that someone has this. So I don't know what the 15 to 1 is.

When you pick three cards out of 50, there are 50*49*48 / (3*2*1) = 19,600 ways to do it.

Only 48 of those ways include both your cards, because there are 48 cards to pick for the third card.

There are 2*48*47 / 2 = 2,256 ways to pick exactly one of your cards. That means about 1 chance in 408 to get both, 1 chance in 8.7 to get exactly 1 and 1 chance in 8.35 to get at least 1.
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