Implied odds on OESD

Alright… so I just tried to calculate how big of a raise you should call when you have an OESD. This is meant for situations where you are pretty sure you will be able to win a good amount if you hit.
i.e.: you think your opponent has a set or 2pairs, which most people at low stakes won’t usually fold.


the odds of hitting the straight on the turn..

8/47
.170
1 in 5.88

Chance of hitting on the river if you didn’t hit on the turn

39/47 * 8/46
.144
1 in 6.94

Odds of hitting your straight

1 - ( 39/47 * 38/46 )
1 - ( 0.8298 * 0.8261)
1 – 0.6855
0.3145


So over 100 of those situations, lets assume you’ll hit the straight 17times on the turn and 14times on the river.

So you’ll call looking for the straight 100 times on the flop, and 83 times on the turn. (17 times you’ll already have the straight on the turn)

You’ll win 31 hands.

Now let’s calculate the maximum amount to call.

x = amount you can win… pot + your opponents stack (or yours if its smaller)
y = total of calls

31x = 183y
0.169x = y

Or

y = 5.91x


Would it be correct to make a rule such as “call a raise only if the amount you can win is 7x bigger. “ ? (7 for profit and chance of your opponent folding)

Originally Posted by JCMoney

Odds of hitting your straight

1 - ( 39/47 * 38/46 )
1 - ( 0.8298 * 0.8261)
1 – 0.6855
0.3145

I don't have this number memorized, but your math looks right

Originally Posted by JCMoney

So over 100 of those situations, lets assume you’ll hit the straight 17times on the turn and 14times on the river.

So you’ll call looking for the straight 100 times on the flop, and 83 times on the turn. (17 times you’ll already have the straight on the turn)

You’ll win 31 hands.

You are not allowing for having to call extra money on the turn.

Originally Posted by JCMoney

Now let’s calculate the maximum amount to call.

x = amount you can win… pot + your opponents stack (or yours if its smaller)
y = total of calls

31x = 183y
0.169x = y

Or

y = 5.91x

Would it be correct to make a rule such as “call a raise only if the amount you can win is 7x bigger. “ ? (7 for profit and chance of your opponent folding)

Where did 183 come from?

On the flop, you need ~6:1 to call for the turn. If you are looking for implied odds, it doesn't really matter how big opponents stack is, as long as it's big enough that pot + stack = 6 * bet. That is, unless he's a super shorty this shouldn't be an issue.

The odds of hitting on the turn or river is 1:2. I don't see a correlation between your 7x rule and the 10x rule for sets (which I'm assuming this idea originated from).

It's simple enough, call your open-ended straight draw with pot odds, if you are a little behind it's probably still OK to call because of implied odds.

id call a lot more if i know my opponent will pay it

Originally Posted by swiggidy

On the flop, you need ~6:1 to call for the turn.

If you're referring to odds as in "returns for one's bet", shouldn't that be
5-to-1 and not 6-to-1? If the probability of hitting something is 1 out of 6,
i.e. if one will hit it once in every six tries, then on that hit, one needs only
win 5x to make up for the other five times one does not hit.

Originally Posted by aqn

Originally Posted by swiggidy

On the flop, you need ~6:1 to call for the turn.

If you're referring to odds as in "returns for one's bet", shouldn't that be
5-to-1 and not 6-to-1? If the probability of hitting something is 1 out of 6,
i.e. if one will hit it once in every six tries, then on that hit, one needs only
win 5x to make up for the other five times one does not hit.

If the probability of hitting something is 1/6, then the odds of hitting are 1:5, so the payout can be 5:1.

If the odds of hitting are 1:6, then the payout has to be 6:1

Originally Posted by swiggidy

Originally Posted by aqn

Originally Posted by swiggidy

On the flop, you need ~6:1 to call for the turn.

If you're referring to odds as in "returns for one's bet", shouldn't that be
5-to-1 and not 6-to-1? If the probability of hitting something is 1 out of 6,
i.e. if one will hit it once in every six tries, then on that hit, one needs only
win 5x to make up for the other five times one does not hit.

If the probability of hitting something is 1/6, then the odds of hitting are 1:5, so the payout can be 5:1.

If the odds of hitting are 1:6, then the payout has to be 6:1

Precisely: the odds of hitting an OESD on the turn is 8 / 47 or 17.02%
or about 1 out of every 5.87 times, as JCMoney already pointed out.
That's hitting once out of every six times and the "payout" when
hit needs to be five to one, not 6:1. Yes? No? Maybe?
Just wanna make sure I'm not missing something...

Originally Posted by aqn

Precisely: the odds of hitting an OESD on the turn is 8 / 47 or 17.02%
or about 1 out of every 5.87 times, as JCMoney already pointed out.
That's hitting once out of every six times and the "payout" when
hit needs to be five to one, not 6:1. Yes? No? Maybe?
Just wanna make sure I'm not missing something...

This is correct.

This post is really confusing, I don't think there is anything to take from it (hence the lack of replies).

I ignored the 1 in 5.88 at the beginning. Clearly (now) that is 1/5.88 not 1:5.88.

I was using "y = 5.91 x". I don't even know what it means, but if x = 1, then y = 6, thus 6:1.

Originally Posted by JCMoney

the odds of hitting the straight on the turn..

8/47
.170
1 in 5.88

Chance of hitting on the river if you didn’t hit on the turn

39/47 * 8/46
.144
1 in 6.94

Odds of hitting your straight

1 - ( 39/47 * 38/46 )
1 - ( 0.8298 * 0.8261)
1 – 0.6855
0.3145

Hos is this possible? If you don't hit on the turn, your chances of hitting your desired card on the river should be HIGHER, right? Because there is one less card that can't help you still in the deck.

http://www.flopturnriver.com/chart_pot_odds.html

Originally Posted by ensign_lee

the odds of hitting the straight on the turn..

8/47
.170
1 in 5.88

Chance of hitting on the river if you didn.t hit on the turn

39/47 * 8/46
.144
1 in 6.94

Odds of hitting your straight

1 - ( 39/47 * 38/46 )
1 - ( 0.8298 * 0.8261)
1 . 0.6855
0.3145

then

Originally Posted by JCMoney

Hos is this possible? If you don't hit on the turn, your chances of hitting your desired card on the river should be HIGHER, right? Because there is one less card that can't help you still in the deck.

You're both correct.

1/6.94 is the probability of both events happening:
not hitting on the turn and hitting on the river.

If for some reasons you don't care about the turn, and regard hitting the
OESD on the river as an independent event, then yes, the probability of
hitting is 8/46, which is higher than that on the the turn (8/47).