ODDs math question

Here's the way the coin flip works.

If we flip a coin twice we can get 4 outcomes:
2 heads probability .5*.5 = .25
2 tails .5*.5 = .25
1 heads 1 tails = .5*.5 = .25
1 tails 1 heads = .5*.5 = .25

3 of these outcomes gives us our required 1 head result. So the odds of getting at least 1 head out of 2 coin tosses is:

.25+.25+.25 = .75 = 75%

In our pocket pair/trips example we have the following possible results for the 3 flop cards, excluding the possibilities with 2 3rd cards because they would give us 4 of a kind (3rd card = the card that completes the 3 of a kind):

1=other 2=other 3=other Prob=.96*.96*.96 = .885
1=3rd 2=other 3=other Prob= .04*.96*.96= .037
1=other 2=3rd 3=other Prob = .037
1=other 2=other 3=3rd Prob = .037

Total = 3.7%+3.7%+3.7% = 11.1%

This gets me closer to the 10.8% from the web site.

Thanks for the example of the coin toss to stimulate my thinking. Perhaps one of the experts here can fill in the last few blanks to correct my errors.

Thanks to all for your help.

Eric

The probabilty of 'other' occuring is not 0.96 every time. You are assuming the events are independent. But the probabilty of the events depend upon what happened before, so you need to account for this. In your coin flip example the events are independent, the coin has no 'memory' so you were correct in using 0.5 each time.

For exmaple if 1=3rd then P(2=other) = 48/49 = 0.98 (not 48/50)

The 3 answers I get are all 0.038367
So 3*0.038 = 11.5%

This is still wrong because the 'other' cards can still produce a full-house. So 11.5% - P(full-house) = 11.5 - 0.74 = 10.76% to hit only a set.
Please correct me if you think this method is wrong.
Thanks,
Mike

That method looks correct to me, Mike.

Now I want to understand how to get the full-house value [img]/images/graemlins/smile.gif[/img].

I'll work on that for now.

Thanks, Mike.

Eric