PLO Strat Poll (3 Viewers)

[MrCatPants]

Can you show me the math on the king having a second spade 60% of the time? Intuitively that feels high to me but I’m not the strongest on these calcs. Genuinely curious on this.

Rough version i would guess - 3 other cards for their omaha hand, and given three spades are already accounted for like a 20% chance per card that it will also be a spade (versus the normal 25% chance per card.)

 

[raynmanas]

Can you show me the math on the king having a second spade 60% of the time? Intuitively that feels high to me but I’m not the strongest on these calcs. Genuinely curious on this.

If we already have As10s, and somebody else already has Ks, their odds of having a second spade (knowing we have exactly 2 spades and they have 1, which leaves 10 spades unaccounted for and 47 total cards unaccounted for) are:

(10/47) + (10/47) + (10/47) = 0.638 = 63.8% chance of them having at least one more spade.

 

[CrazyEddie]

All three answers can be defended given how ambiguously worded the poll was. If you got the answer "wrong" you shouldn't let anyone give you any crap over it, assuming your reasoning was valid.

 

[Highli99]

If we already have As10s, and somebody else already has Ks, their odds of having a second spade (knowing we have exactly 2 spades and they have 1, which leaves 10 spades unaccounted for and 47 total cards unaccounted for) are:

(10/47) + (10/47) + (10/47) = 0.638 = 63.8% chance of them having at least one more spade.

Thanks for this. I’m embarrassed to say I forgot we were calculating PLO and hot holdem. Makes more sense now.

 

[Poker Zombie]

All three answers can be defended given how ambiguously worded the poll was. If you got the answer "wrong" you shouldn't let anyone give you any crap over it, assuming your reasoning was valid.

I'll do you one better - if you pulled the chips, it was the correct decision for that hand at that time.

 

[Sunshine]

If we already have As10s, and somebody else already has Ks, their odds of having a second spade (knowing we have exactly 2 spades and they have 1, which leaves 10 spades unaccounted for and 47 total cards unaccounted for) are:

(10/47) + (10/47) + (10/47) = 0.638 = 63.8% chance of them having at least one more spade.

This is not the way to do this kind of probability calculation - doesn't account for depletion and adding the probabilities neglects the fact that if the first comes through, the 2nd/3rd don't matter. Assuming your starting point, receiving three cards from a pool of 47 of which 10 are spades, the probability of NOT getting an additional spade would be:
37/47 * 36/46 * 35/45 = 0.48
So the chances that there WOULD be an additional spade is 52%.

Did I do this right?

 

[pltrgyst]

This is not the way to do this kind of probability calculation - doesn't account for depletion and adding the probabilities neglects the fact that if the first comes through, the 2nd/3rd don't matter. Assuming your starting point, receiving three cards from a pool of 47 of which 10 are spades, the probability of NOT getting an additional spade would be:
37/47 * 36/46 * 35/45 = 0.48
So the chances that there WOULD be an additional spade is 52%.

You are correct about depletion, but the fact that "if the first comes through, the second and third don't matter" -- doesn't matter.

 

[raynmanas]

This is not the way to do this kind of probability calculation - doesn't account for depletion and adding the probabilities neglects the fact that if the first comes through, the 2nd/3rd don't matter. Assuming your starting point, receiving three cards from a pool of 47 of which 10 are spades, the probability of NOT getting an additional spade would be:
37/47 * 36/46 * 35/45 = 0.48
So the chances that there WOULD be an additional spade is 52%.

Did I do this right?

i'm no stats expert.