probability Q from someone who knows zilch about probability

[wufwugy]

is the probability of winning two 4 to 1 dog hands the same as winning eight 4 to 1 fav hands?

it makes sense to me that it is, yet i know it's highly likely that im completely off my rocker with my logic here since i have no clue how to figure probability.

just kinda wondering if i can expect to win sixteen 4 to 1 favs in a row as often as i lose four 4 to 1 favs in a row.

[swiggidy]

4:1 is odds, so it's win 4 - loose 1, or 80% - 20%

So chance of loosing 2 in a row is 0.2 * 0.2 = 0.04

96% : 4%
divide both sides by 4

24:1

Winning 8 4:1s is 0.80^8 = 16%, or 1:5.25

If you want me to explain more I will, I don't want to aimlessly ramble.

Cliff notes: no there isn't an easy conversion for odds, you have to use percents.

[wufwugy]

explain as much as you like.

im not that smrt.

[swiggidy]

fractions make more sense than percents (IMO)

winning as 4:1 dog:
1:4 dog is same as 1 out of five (as stated above) or 1/5

since they are independent events we can multiply for total probablility

2 times
1/5 * 1/5 = 1/25 (which becomes 1:24 odds)

3 times
1/5 * 1/5 * 1/5 = 1/125

n times
(1/5)^n

winning as 4:1 fav:
winning a 4:1 is the same as winning 4 out of 5 times, or 4/5

so winning 2 times
4/5 * 4/5 = 16/25

winning n times
(4/5)^n

There isn't a direct correlation between the two because winning them all and loosing them all aren't the only two options (e.g. in 2 games played you can also win one, and loose one)

Your initial question:
... expressed mathematically

(1/5) ^ 2 ?= (4/5)^8

so no that's not true

[Xianti]

roffle...

dunno about wufwugy, but I'm more confused now than I was when the question was asked.

[swiggidy]

explaining maths on the internets is

[Xianti]

No. It's not that. I just suck at reading abstract math expressions.

[GatorJH]

OK, so I may be asking a REALLY stupid question, but what the $%^&* does this question have to do with poker? When would I need to calculate all of that when playing a hand?

[wufwugy]

my math expertise ends at simple multiplication.

so the amount of times you win sixteen 4 to 1 favs is 256/625?

[wufwugy]

OK, so I may be asking a REALLY stupid question, but what the $%^&* does this question have to do with poker? When would I need to calculate all of that when playing a hand?

nothing, never.

im just curious for general perception purposes. as of now, many of my hands are all in when im a 4 to 1 fav. i've had a couple streaks where i lose four of those in a row, which is rather a lot, yet i haven't noticed any streaks where i win a shtiload of those in a row. im simply just wanting to know the probable equivalent of amount of losing favs to the amount of losing those same favs.

[flyingPenguin]

No. 4 to 1 dog means you have a 1/5 chance. 1/5, or 0.2, is then your probability of winning once.

For sixteen times you have to do:
0.2 * 0.2 * 0.2 * 0.2 * 0.2 * 0.2 * 0.2 * 0.2 * 0.2 * 0.2 * 0.2 * 0.2 * 0.2 * 0.2 * 0.2 * 0.2
= something too tiny to care about.

If you are 4 to 1 favourite you have a 0.8 chance, so
0.8 * 0.8 * 0.8 * 0.8 * 0.8 * 0.8 * 0.8 * 0.8 * 0.8 * 0.8 * 0.8 * 0.8 * 0.8 * 0.8 * 0.8 * 0.8 = about 0.03

So if you are 4 to 1 favourite you will win all of your 16 goes about 3 times every 100 sets of 16.

Dunno why you'd care though. I would think this would be more relevant to roulette.

[swiggidy]

my math expertise ends at simple multiplication.

so the amount of times you win sixteen 4 to 1 favs is 256/625?

your fraction is for winning 4 in a row

(4/5) ^ 4 = 4/5 * 4/5 * 4/5 * 4/5

(4/5)^16 = 0.8^16 = 0.028

To answer your question. Loosing 4 (4:1 fav) in a row is 0.1%. That would be winning 30 in a row.

If you're playing SnGs I feel ya, but...
if you make it 60:40 (1: 1.5)
loosing 4, 1%
same as winning 10

I think the problem is, you loose 4, you get knocked out of 4 tournaments. You win 10 and you steam-roll your way to 1st in one tournament.

[swiggidy]

you will win all of your 16 goes about 3 times every 100 sets of 16.

or 16 in a row once in a 48 hand trial

[Xioustic]

my math expertise ends at simple multiplication.

All the math involved is decimal multiplication and (some) fractions if you want to take shortcuts. Calculators make this easy.

I'll take my shot for those that don't quite understand it. This is taught in statistics class and pre-algebra in the states, so I'm sure any Euros in this discussions should know it as well.

What we Know
You have 4:1 odds. In other words, for every 4 times you win, you lose once.
This also means you win 4 times every 5 times. You also lose 1 out of 5 times.

Turn it Into Percentages
To calculate something happening in a row you have to turn it all into percentages FIRST.

To turn 4:1 to a percent, you pick a side. Are you winning? Take the 4/5. Or are you losing? 1/5.

Winning = 4/5 = 0.8 = 80%.
Losing = 1/5 = 0.2 = 20%.

Find Probability of a Sequence of Events
If you want a probability of something happening in a sequence you multiply the percentages however many times that it happens in a row.

Winning twice in a row = 80%*80%. What's 80% of 80%? 0.8*0.8 = .64 = 64% of the time.

Losing twice in a row = 20%*20%. What's 20% of 20%? 0.2*0.2 = .04 = 4% of the time.

Solve the Problem
So, you want your odds of losing two 4:1 in a row and your odds of winning eight 4:1 in a row.

According to our maths:
Losing twice in a row = 20%*20% = 0.2*0.2 = 0.4 = 4% of the time.
Winning eight times in a row = 80%*80%*80%*80%*80%*80%*80%*80% = .8*.8*.8*.8*.8*.8*.8*.8 = (.8)^8 = 0.16777 = 0.168 = 16.8%.

So the answer to your question is no, winning two 4 to 1 dog hands is no the same as losing eight 4 to 1 hands. An underdog in this situation wins 4% of the time twice in a row, and the shoe-in wins 8 times in a row 16.8% of the time.

This is all simple multiplication. If you're good with decimals it's really easy to do in your head and on the fly at a poker table.

OK, so I may be asking a REALLY stupid question, but what the $%^&* does this question have to do with poker? When would I need to calculate all of that when playing a hand?

This allows you to make fun calculations, like getting dealt pocket aces twice in a row. Or getting dealt two monster pairs (QQ+) in a row. Or getting dealt something in the top 10% of the deck twice in a row.

I used to do this kind of stuff to figure out my odds of getting dealt a better hand and thus finding a better spot to push in the next orbit of a SnG if the table was tight and I faced a push. Now I usually just guess.

This is where it should be getting a bit over some heads and becomes theory rather than practice.
Basic Poker Theory
For example, at a table of 10 players, every one orbit you're supposed to (in theory) get something in the top 10% of the deck. This would be 88+, A9s+, KTs+, QTs+, AJo+, and KQo. The math on this is .1(pct of deck you want)*10(number of players) = 1 = 100% of the time you should be getting something in the top 10% in a 10 man table.

Let's say you drop one player. 9 handed table. You still want something in that top 10%. .1(pct of deck you want)*9(number of players) = .9 = 90% of the time you should be getting something in the top 10%.

In a 9-man table, what's the best that we are pretty much guaranteed (in theory) to get dealt? x(pct of deck we get)=1/9(number of players) = 11.1%. This means every orbit at a 9-man table you should be getting one of the top 11% hands, which just opens you up to 77+, ATo in addition to what is above.

At a 2-man table (HU), you're going to get something dealt in the top 50% half of the time in theory. This is obvious. 1/2(number of players) = .5 = 50%. So half of the time your holding is better than what the opponent should have.

Thus, as you remove players you have to settle for lesser hands because they're more than likely better than the opponents holding.

Common Applications in Poker

• What hands to raise with, what qualifies a poker player as tight or loose based on table size. If a player raises 20% of the time in a series of 2 orbits we assume he's raising with (66+,A4s+,K8s+,Q9s+,J9s+,T9s,A9o+,KTo+,QTo+,JTo), but in actuality he could be raising QQ+ and just be getting lucky. We assume that's not the case because of this math. It's improbable.
• When and how often to make moves. The odds an opponent doesn't have something he'd be willing to call with 2 or 3 times in a row that we push while he's in the BB and we're in the SB, etc.
• Our Risk of Ruin. The chance that our bankroll will take a downswing given the consistency of our play. For example, if I'm ITM 40% of the time, losing 10 buy-ins in a row is 60% multiplied by itself 10 times = .6^10 = 0.6% of the time. So it's unlikely that will happen, but now we know the likeliness it will happen.

Danger of This Thinking
This is all theory. It's not guaranteed by anything but the random number generator on the site. You could still win your 80% every time for 100 times (suffer no bad beats in 100 cases in a row), but the chances of that are (.8)^100 = .00000002037%. So pretty fckn unlikely. But it COULD still happen.

Randomness is random. If this theory worked perfect every time it wouldn't be random, but if it didn't work every time it would be a broken random number generator. Funny, huh?

Wheeee, maths!
BTW, please let me know if I'm doing any of this wrong.

[GatorJH]

You guys scare me