# Why can't one add the (outs/possible) fractions of hitting flush on turn or river for the overall chance given 4 suited on flop?

"This means that the total odds for completing a flush - which should matter for example if you're going all in after the flop - are (19.14%) + (19.5%*(1-19.14%)) = 34.96% (the odds of completing on the turn, plus the odds of completing on the river times the odds of not completing on the turn)." -quoted from related question.

Vs. 9/47+9/46 = 39.13%

Thanks for the discrete structures review in advance!

You can't simply add them because there is overlap between the two cases. Both the turn and the river could be a card of that suit and in this case the flush gets counted twice. That's why you multiply the chance of making a flush on the river with the chance with the chance that you do not make a flush on the turn before adding it to the chance of making a flush on the turn.

• This makes sense because the river is dependent on not getting the flush on the turn; therefore, you multiply the chance of getting the flush on the river by the chance of not getting the flush on the turn. Thank you! – Charley Erd Feb 12 at 19:59

This means that the total odds for completing a flush - which should matter for example if you're going all in after the flop - are (19.14%) + (19.5%*(1-19.14%)) = 34.96% (the odds of completing on the turn, plus the odds of completing on the river times the odds of not completing on the turn).

• Your odds of hitting the flush on the turn are 9/47.
• If you miss it, your odds of hitting the flush on the river are 9/46.

So to combine the odds, we take the different parts of the equation: (odds of hitting the flush on the turn) + (odds of missing the flush on the turn * odds of hitting the flush on the river). Otherwise we'd be counting the same even twice - when hitting the flush, you don't care about the next card (for this calculation), and even if you did, the math would be wrong.

Let's take an easier example to explain it better: let's flip a fair coin three times. What are the odds at least one of them would be tails?

• The odds for the first flip are 1/2, or 50%.
• The odds for the second flip are 1/2.
• The odds for the third flip are 1/2.

How do you combine them?

We could just add them, and reach a mathematical impossibility - 150% of hitting at least one tails.

Or we could make a more educated calculation:

• If the first flip is tails, we've succeeded, so that's 50%.
• If the first flip is heads and the second flip is tails, we've also succeeded. That's 50% * 50% (heads1 * tails2), or 25%.
• If the first flip is heads and the second flip is heads and the tails flip is tails, we've also succeeded. That's 50% * 50% * 50% (heads1 * heads2 * tails3), or 12.5%.

Now we can total them to reach the correct answer: 50% + 25% + 12.5% = 87.5%.

Discrete structures?

34.96% is the odds of hitting exactly 1 flush card on turn or river.

If you hold the ace you don't care if the board has a 4 flush. But even then 39.13% is wrong as if the first is a flush there are only 8 flush cards left.

I think odds of the board getting a 3 or 4 flush after getting a 2 flush on the flop is 38.3%.

• It's the odds of hitting at least 1 flush card on turn or river. Not exactly. – configurator Dec 13 '18 at 13:07