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View Full Version : Change Odds to Percentage

09-25-2002, 09:35 PM
How does one change an odds statement, say 4 to 1, to the percentage that it represents, like 75%? I knew at one time but have forgotten.

lorinda
09-25-2002, 09:40 PM
Lets see if I can do a better job of this post /forums/images/icons/laugh.gif
You take the second number and add it to the first one, then you divide the second number by your new number.
examples:
4 to 1 becomes 1/5 = .20 = 20%
7 to 2 becomes 2/9 = .22 = 22%
1 to 1 becomes 1/2 = .50 = 50%
5 to 4 becomes 4/9 = .44 = 44%

BruceZ
09-25-2002, 11:36 PM
4-1 underdog means 4 losses to every 1 win, so there is 1 win out of 5 total tries or 1/5 = 20%. 4-1 favorite means 4 wins to 1 loss or 80%. If you win 75% of the time you are a 3-1 favorite. If you lose 75% you are a 3-1 dog. Note that being a 3-1 favorite is the same as being a 1-3 dog, and visa versa.

If the pot is 4 bets, and you have to put in 1 more bet to win it, then the pot is offering you 4-1 pot odds. You would put in this bet if you thought you were better than a 4-1 dog to win without investing any more money.

sucka
09-26-2002, 12:48 PM
This is a very good explanation from BruceZ.

I once thought that using %'s at the table was better than using the odds - but with a few 'on the fly' calculations its actually more effective and easier to use.

Let's say I know that there's 6 SB's in the pot, and it's my turn to act (call a single bet) so I know that the pot is laying 6-1 for me.

I look down at my hand and see that, on the flop, I have an open ended straight draw giving me 8 good outs.

I use this formula: 50-(# of outs) / (# of outs) = odds to make my hand by the river

50-8 = 42/8 = 5.25 to 1

Notice, that this is the odds required to make my hand by the river.

I've got 6 BB's in the pot and I'm a 5.25 to 1 dog so I'm getting correct pot odds to call in this situation.

Mike Haven
09-26-2002, 05:14 PM
"Notice, that this is the odds required to make my hand by the river."

No, you have calculated incorrectly. It's much better than you think!

You have seen 5 cards once you have seen the flop, so you only need (47-8)/8 = 39/8 = 4.875 to 1 pot odds to try to catch your straight on your next card, and, if you miss, you only need (46-8) = 38/8 = 4.75 to 1 pot odds to try to catch on the river.

If you are able to go all in on the flop and see both cards for just one bet you only need a bit less than half these pot odds to call, roughly 2.2 to 1.

09-26-2002, 10:33 PM

I'm a bit lost, though. Why does this calculation not work:

If I have 8 outs after the flop, then 47 remaining cards divided by my number of outs gives me 47/8=5.875. What am I missing? Why must the outs be subtracted first?

lorinda
09-26-2002, 10:51 PM
you have 39 bad cards vs 8 good cards (thats what we are doing when we are dividing, we are making a ratio)
The word "to" in the odds 39 to 8, really is the english word (39 bad to 8 good)
when we divide by 8, we are just reducing this to get a term for "to 1" because that is the way we think better, your odds are 39 to 8, but this is hard to grasp when you are putting one chip in a pot, so we simplify to 4.9 to 1
Take the extreme case of two cards unknown, and you have one out. you have a ratio of 1:1 (one good, one bad) which is 1 to 1 odds.
With your method,you would get 2/1 which would severly undermine your thinking.

09-26-2002, 11:22 PM
That last part of your post cleared it up for me, lorinda. Thanks for your help.