[aguahombre]

I have been trying to figure out how the percentages are derived on the distribution charts which have been widely accepted. I didn't realize how dumb I really was until:

The chart shows that with an 8-card fit, the outstanding cards will divide 3/2 20 times out of 32. The chart gives the probability as .68. I always thought 20/32 was .625.

the chart shows seven outstanding cards will divide 4/3 70 times out of 128 and states the probability to be .64. My calculator keeps giving me .546.

What am I doing wrong?

[hotShot]

I think you are missreading the table, there are 20 different 3-2 combinations (distributing the 5 cards on a 3 and 2 card stack), and I don't know where you get the 32 from.
There are 70 4/3 combinations.

[pooltuna]

aguahombre, on Jan 19 2010, 01:36 PM, said:

I have been trying to figure out how the percentages are derived on the distribution charts which have been widely accepted. I didn't realize how dumb I really was until:

The chart shows that with an 8-card fit, the outstanding cards will divide 3/2 20 times out of 32. The chart gives the probability as .68. I always thought 20/32 was .625.

the chart shows seven outstanding cards will divide 4/3 70 times out of 128 and states the probability to be .64. My calculator keeps giving me .546.

What am I doing wrong?

The probability for any given 32 split is greater than any given 4-1 split. Think of it in terms of number of open spaces which is somewhat similar.

[barmar]

hotShot, on Jan 19 2010, 01:45 PM, said:

I think you are missreading the table, there are 20 different 3-2 combinations (distributing the 5 cards on a 3 and 2 card stack), and I don't know where you get the 32 from.

There are 32 different ways that 5 cards can be distributed: 2 5-0 breaks (either opponent can hold the void), 10 4-1 breaks (there are 5 possible singletons, and it can be in either opponent's hand), and 20 3-2 breaks.

[hotShot]

You can draw the same 3 cards in 3 different orders, so this triple has a population of 3. The question is, is 20 the number of different triples (thats what I think) or does it include the population of each triple.

[bb79]

68% probability comes from the constraint that two hands have 13 cards each. Once you give 3 cards to one side you can give 10 of the remaining 21 cards to that side.

[bb79]

The mathematical calculation, for those interested,
8 card fit , 3-2 distribution of a suit


C(5,3)*C(21,10)/C(26,13) + C(5,2)*C(21,11)/C(26,13)=0.678

C(n,k) is combination operator, choosing k cards from n cards, (n-k)!k!/n!

[Siegmund]

As bb79 said - the fact the opps are constrained to 13 cards each means that the 32 five-card distributions and the 128 seven-card distributions are not equally likely. Working out the 5-card case all the way:

Opps have (say) 5 spades and 21 non-spades.

There are two 5-0 breaks, for each of which the non-spades can be dealt 21!/(8!13!) 203490 ways.
There are ten 4-1 breaks, for which the non-spades can be dealt 21!/(9!12!) = 293930 ways.
There are twenty 3-2 breaks, for which the non-spades can be dealt 21!/(10!11!) = 352716 ways.

That is, the 3-2 breaks are more likely than the 4-1 breaks by a factor of 12:10, and the 4-1s more likely than the 5-0 breaks by a factor of 13:9 (so a 3-2 is more likely than a 5-0 by a factor of 12X13 : 9x10.)

You can now calculate 20 / (20 + 10 / 1.2 + 2 / 1.733 ) ~ 20/29.487 ~ 67.8%.

Similarly, in the 7-card case, the seventy 4-3 breaks are more likely than the 5-2 breaks by a factor of 11:9, the 5-2s more likely than the 6-1s by a factor of 12:8, and the 6-1s more likely than 7-0s by a factor of 13:7, and you can now calculate

70 / (70 + 42 / 1.222 + 14 / 1.833 + 2 / 3.405) ~ 70/122.588 ~ 62.2%.

[dcohio]

I still say the probability of a 3-2 split is 50/50. It either is or it isn't...

[aguahombre]

thank you bb, and sieg for the answers. Knew I would get some silliness, too.

[dcohio]

After all that math I had to inject some levity lol