| Probability trees
Maybe this approach will help you.
Assume we've seen 7 cards (KQ, JJ, flop XXX). So there are 45 unseen cards.
If any K or Q will help you, you have 6 outs. You have 39 nonouts.
There are four possible outcomes to think about when wondering whether you will catch one of your outs on the turn or on the river or both. The four outcomes are very much like the coin flip outcomes of HH, HT, TH, and TT. But instead of success being heads, think of a success as hitting one of your outs.
Case 1: P(out, out)
Case 2: P(out, nonout)
Case 3: P(nonout, out)
Case 4: P(nonout, nonout)
For each case you have to do a multiplication problem.
Case 1: P(out, out) = (6/45) x (5/44) =0.015 = 1.5%
This is the case where you are lucky enough to get outs on both the turn and the river. Chance of first card being an out is 6/45. Then if that happened, chance of second card also being an out is 5/44.
Case 2: P(out, nonout) = (6/45) x (39/44) = 0.118 = 11.8%
This is the case where you hit on the turn but not on the river.
Case 3: P(nonout, out) = (39/45) x (6/44) = 0.118 = 11.8%
This is the case where you do not hit on the turn but hit on the river.
Notice it is the same probability as Case 2.
Case 4: P(nonout, nonout) = (39/45) x (38/44) = 0.748 = 74.8%
This is the case where you don't hit at all.
To find the chance of hitting on the turn or river or both, just add up Case 1 + Case 2 + Case 3. You get 1.5% + 11.8% + 11.8% = 25.2%
The faster way is to just take (1 - Case 4). You get 1 - 74.8% = 25.2%.
In my earlier post, I combined Case 1 + Case 2 as the first fraction, and added Case 3. You can see that Case 1 + Case 2 simply add up to 6/45.
Case 1 + Case 2
= (6/45) x (5/44) + (6/45) x (39/44)
= (6/45) x (5/44 + 39/44) by the distributive property
= (6/45) x 1
= 6/45
Then I added Case 3 = (39/45) x (6/44)
That is how I got (6/45) + (39/45)(6/44) as my expression in the earlier post. It's just a compact way of expressing Case 1 + Case 2 + Case 3.
If you understand tree diagrams, you can do it that way too. Draw two branches. One will have probability (6/45). The other has probability (39/45). Then at the bottom of each branch, draw two more branches. Label them with the probabilities (5/44), (39/44) and (6/44), (38/44), respectively. The four cases represent the four possible pathways down the tree. Each probability is simply a multiplication of the probabilities you encounter as you travel down the tree. That will generate the four probabilities for Cases 1, 2, 3, and 4. Then you look at those probabilities and see that what you want to happen is Case 1, 2, or 3. So you add those probabilities. Or you instead take (1 - Case 4).
-Will
P.S. All of the above presumes you understand that to find the probability of two or more events, you MULTIPLY their respective probabilities.
For example, the probability that if you toss three coins, you will get HHH is:
P(HHH) = (1/2) x (1/2) x (1/2) = 1/8 = 0.125 = 12.5%
In that case, the fractions are all the same because the events are independent (they don't influence each other). In the above calculations, the fractions vary because the events are dependent (the numbers of good cards and available cards change as you proceed), but you still multiply them to get the probability of the "compound" event.
Last edited by wijwij : 03-31-2006 at 04:31 PM.
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