question about pot odds

[trumpman84]

If I have a flush draw on the flop and someone goes all-in, assuming I have no other outs, I have about a 33% chance to win the hand.

First off, is this a 3 to 1 shot or a 4 to 1 shot? So, would I need to be getting 3 to 1 or 4 to 1 on the pot size?

Say the all-in is for 4 dollars and the pot is 12 dollars (including his bet). Could I call or would the pot have to be 16 dollars?

I've always been confused converting %'s to "x to 1" chance of winning. I've always thought since you win once out of every 3 hands, you have 3 to 1 odds, but my friends who is a good poker player told me you add one to that so I have 4 to 1. Just looking for some clarification.

[mondo]

The odds are roughly 2:1 against making the flush by the river. The pot is offering you $12 for a $4 bet. Thus you are getting 3:1 pot odds on a 2:1 proposition. Call.

This does not include the possibility of pairing one or both of your cards which MAY be winners, as well as neglecting possible str8s, etc.

Converting % to odds is 100% - (% of event occurring) divided by % of event occuring.

In your example, using 32%, 100-32 = 68 divided by 32 is 2.125 : 1 odds against hitting the flush.

peace, mondo

[Isura]

It depends on what game you are playing, how many cards are left to come. As an example, say you are playing hold'em and have a flush draw with 1 card to come. There are 52-6 = 46 "unknown" cards left, and 9 of them make your flush. So, 37 cards don't make your flush. Since 37/46 is about 0.8, about 80% of the remaining cards miss your flush. In other words, you miss the flush 4 out of 5 times, and make it 1 out of 5 times. We say then that your odds are 4 to 1 to make the flush. If there was $4 in the pot, and you bet $1, you lose $1 4 times, and win $4 1 time. Hence, you break even in the long run. Now, if there was say $5 in the pot, you still lose $1 4 times, but now win $5 1 time. So you net $1 profit every 5 times (in the long run). So basically, you first determine your real odds of making your draw (4 to 1 in our example). Then you compare this to the money you will win that is already in the pot ($4 and $5). If there is more in the pot, your pot odds are higher than your real odds, so it is profitable to call a bet.

To answer your question then, 33% chance of winning is 2 to 1 odds, so the pot must have more than 2 bets in it for each bet you must make to call. (The pot doesn't actually need 3 bets to call 1 bet, 2.1 or 2.2 or anymore more than 2 gives you good odds to call). I hope that makes sense.

[Isura]

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It depends on what game you are playing, how many cards are left to come. As an example, say you are playing hold'em and have a flush draw with 1 card to come. There are 52-6 = 46 "unknown" cards left, and 9 of them make your flush. So, 37 cards don't make your flush. Since 37/46 is about 0.8, about 80% of the remaining cards miss your flush. In other words, you miss the flush 4 out of 5 times, and make it 1 out of 5 times. We say then that your odds are 4 to 1 to make the flush. If there was $4 in the pot, and you bet $1, you lose $1 4 times, and win $4 1 time. Hence, you break even in the long run. Now, if there was say $5 in the pot, you still lose $1 4 times, but now win $5 1 time. So you net $1 profit every 5 times (in the long run). So basically, you first determine your real odds of making your draw (4 to 1 in our example). Then you compare this to the money you will win that is already in the pot ($4 and $5). If there is more in the pot, your pot odds are higher than your real odds, so it is profitable to call a bet.

To answer your question then, 33% chance of winning is 2 to 1 odds, so the pot must have more than 2 bets in it for each bet you must make to call. (The pot doesn't actually need 3 bets to call 1 bet, 2.1 or 2.2 or anymore more than 2 gives you good odds to call). I hope that makes sense.

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Offcourse, this is assuming that you always win when you make the draw, and always lose if you dont. Otherwise, the calculation must be slightly modified based on any information you have offcourse.

[binions]

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If I have a flush draw on the flop and someone goes all-in, assuming I have no other outs, I have about a 33% chance to win the hand.

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Assuming you have 9 outs (ie the other guy doesn't have a flush draw), you have 34.97% chance to hit at least one of your flush outs.

The formula is to figure out the chance of NOT hitting a flush out on either turn or river. 38/47 * 37/46 = 65.03% of missing your flush.

Which means 34.97% if hitting at least one of your flush outs (including the times you hit runner runner flush outs)

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First off, is this a 3 to 1 shot or a 4 to 1 shot? So, would I need to be getting 3 to 1 or 4 to 1 on the pot size? I've always been confused converting %'s to "x to 1" chance of winning. I've always thought since you win once out of every 3 hands, you have 3 to 1 odds, but my friends who is a good poker player told me you add one to that so I have 4 to 1. Just looking for some clarification.

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This is a 65.03 to 34.97 shot, or 1.86:1. In other words, for every 1 time you make your flush, you will not make it 1.86 times.

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Say the all-in is for 4 dollars and the pot is 12 dollars (including his bet). Could I call or would the pot have to be 16 dollars?

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If you have to call $4 to win $12, you are getting 3:1 (12:4) on the call. If a flush will win, you should call, because you will hit the flush more often than 3:1.

To be precise, for every time you win $12, you will lose $4 1.86 times. 12-7.44 = $4.56 you win every 2.86 times you make that call. So the call has an expected value of $4.56/2.86 = $1.59 every time you make it.

Assuming, of course, you have 9 live outs (i.e. the other guy doesnt have a flush draw).



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