Simple Probability Question

[closer2313]

Hi,
I'm trying to teach myself probability, and I'm not sure what im doing wrong here.

If two events are mutually exclusive, then the probability that one of the events will occur is

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

I was trying to use this formula to figure out the probability of making a hand by the river.

For example
2 outs = (2/47) + (2/46)= 8.6%

Here is my problem
21 outs = (21/47) + (21/46) = 90%

But I know that isn't the right answer. Could someone tell me what im doing wrong?

Thank You!

[RocketManJames]

The events are not mutually exclusive.

You have 21 outs. You can compute this two ways.

1) 1 - P(missing on both streets) = 1 - (26/47 * 25/46) = 69.935%.

2) Hitting on the turn and not caring about river + missing on turn and hitting on river. = 21/47 + (26/47 * 21/46) = 69.935%.

In Edit: Your calculation for 2 outs would be flawed in a similar fashion.

Also, mutually exclusive simply means that both things can't happen together. So, say you have a population of apes. 5% of them have no arms and 20% of them are holding a banana in their hands. The probability that you randomly pick an ape from this population that is armless OR is holding a banana is P(armless) + P(holding banana) = 0.25. We can add these because they are mutually exclusive. If you're armless you simply can't be holding a banana in your hand, and if you're holding a banana in your hand, you're not armless.

If we had a population of apes where 5% were blue and 20% were holding a banana. You can no longer add the probabilities. This is because it is possible that some blue apes are holding bananas.

Hope this clarifies some stuff for you.

-RMJ

[closer2313]

Yes it does. Thanks alot.

[gamble4pro]

You have to be more specific. Where your 21 outs come from? Maybe you are double-counting some situations (i.e your events to measure are not mutually exclusive). Or you are not correctly apply the formula. For example: if your expect J, Q or K (for turn and river), the "outs" are not cards anymore, but 2-card combinations, so the denominator should be C(47,2)=1081 and not 47.

[Duke]

[ QUOTE ]
You have to be more specific. Where your 21 outs come from? Maybe you are double-counting some situations (i.e your events to measure are not mutually exclusive). Or you are not correctly apply the formula. For example: if your expect J, Q or K (for turn and river), the "outs" are not cards anymore, but 2-card combinations, so the denominator should be C(47,2)=1081 and not 47.

[/ QUOTE ]

You better tell him to change his numerator too, if you want to do that.

~D