View Full Version : technical question: one holdem hand as an "x-to-1" favourite

09-19-2003, 12:57 AM
When calculating the odds by which one hand is favoured over another, are you supposed to compare wins to losses (ignoring ties), or do you just compare pot equity?

For example, consider 98s against 98o. They will tie 90.67% of the time, and 98s will win 7.16% of the time (i.e. 76.7% of the time they don't tie). 98s has pot equity of 0.525 (http://twodimes.net/h/?z=85200).

So do you say 98s is a 3.3-to-1 favourite (using the first method), or an 11-to-10 favourite (using the second method)?

Another example is Ah-Th vs. Td-7d on a board of Tc-9c-8c-7c (http://twodimes.net/h/?z=85214). You could say the T7 is either a 2.6-to-1 favourite or a 1.8-to-1 favourite.

09-19-2003, 01:28 PM
I don't know of any `convention' but surely it is most natural to consider 98s as an 11-to-10 favourite over 98 rather than 3.3-to-1.

If pot equity is 0.525 (hot and cold) then this should be the crucial probability that you convert to odds in the standard way, that is prob p is equivalent to odds of p/(p-1) to 1.

In other words use Prob(win) + (1/2)Prob(tie).

09-20-2003, 11:01 AM
Think about what the numbers mean and it should lead to your answer. In playing 98o do you expect to win or tie nearly half the time, or less than 1/4 of the time?

When ties have a significant probability they enter the EV calculation just like wins and losses. The all in equity is:

.9067*0 + .0716*+1 + .0217*-1=.049