Overall odds of Spades Royal Flush requiring that AK in hand
July 16th, 2012 at 11:04:28 AM
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My local casino offers a super progressive jackpot for getting a Royal with the AK of spades in your hand. People are always arguing over the true odds of this event, and the number that people come up with is always different.
I've calculated getting AK of spades in your hand is 1/1321. I know the next part is simply choosing 3 cards from the combination of choosing 5 cards from 50.
So the flop combinations total to 2,118,760.
This is where I'm stuck. You need 10, J, Q in any order from 5 cards.
This is really bothering me, and I know it's not that hard.
I've calculated getting AK of spades in your hand is 1/1321. I know the next part is simply choosing 3 cards from the combination of choosing 5 cards from 50.
So the flop combinations total to 2,118,760.
This is where I'm stuck. You need 10, J, Q in any order from 5 cards.
This is really bothering me, and I know it's not that hard.
July 16th, 2012 at 11:47:51 AM
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You need the other 3 of the royal, then any 47 C 2 = 1081 choices for the other two cards. Since we are ignoring the order for the board combos, I'm pretty sure we can ignore it here too. So, probability of making the spade royal given you have AKs is 1081/2118760 = 0.00051 (1 in 1960)
Total prob is 0.0000003847, or 1 in 2598960.
My card probability is a little rusty, I could have messed up the ordering argument...
Total prob is 0.0000003847, or 1 in 2598960.
My card probability is a little rusty, I could have messed up the ordering argument...
Wisdom is the quality that keeps you out of situations where you would otherwise need it
July 16th, 2012 at 12:07:53 PM
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First of all, I will assume that it does not pay off if you have AK of spades in your hand and the five cards on the table are AKQJT of, say, hearts.
There are C(52,2) = 1326 possible sets of two cards for your hand; only one (AK of spades) is good.
You need QJT of spades, plus two other cards, on the table; there are C(50,5) = 2,118,760 combinations, of which C(47,2) = 1081 of them contain QJT of spades..
The chance of you winning the jackpot = 1/1326 x 1081/2118760 = 1/2,598,960.
Not particularly coincidentally, that is also the probability of being dealt a royal flush in spades with five cards.
There are C(52,2) = 1326 possible sets of two cards for your hand; only one (AK of spades) is good.
You need QJT of spades, plus two other cards, on the table; there are C(50,5) = 2,118,760 combinations, of which C(47,2) = 1081 of them contain QJT of spades..
The chance of you winning the jackpot = 1/1326 x 1081/2118760 = 1/2,598,960.
Not particularly coincidentally, that is also the probability of being dealt a royal flush in spades with five cards.
July 16th, 2012 at 12:14:22 PM
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I don`t understand how the odds can be the same for being dealt a royal in spades as to the op`s chances when he gets to use 7 cards?
I realize he has to start with the AK or KA of spades but after that he has 5 chances to get 3 cards. what am i missing?
I realize he has to start with the AK or KA of spades but after that he has 5 chances to get 3 cards. what am i missing?
Happy days are here again
July 16th, 2012 at 12:40:58 PM
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While I don't have an elegant argument for why they are equal yet, I am pretty sure the math is right. Here's another way to get it:
The probability of getting a spade royal in 7 cards is 1 in 123760. Now draw 2 of those randomly to get your start hand. You have a 1 in 21 chance of picking the AKs. Total prob: 1 in 2598960.
A smiley face sticker for whoever comes up with a nice explanation for why the probabilities are equal.
The probability of getting a spade royal in 7 cards is 1 in 123760. Now draw 2 of those randomly to get your start hand. You have a 1 in 21 chance of picking the AKs. Total prob: 1 in 2598960.
A smiley face sticker for whoever comes up with a nice explanation for why the probabilities are equal.
Wisdom is the quality that keeps you out of situations where you would otherwise need it
July 16th, 2012 at 1:00:00 PM
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That is the original number I got and realized I must have done something wrong to reach the same value.
Anyway, that's good to know. I made the argument, however, the whatever the calculated odds are, the true odds are likely slightly higher, as all 5 community cards are not always dealt out (due to folding).
Anyway, that's good to know. I made the argument, however, the whatever the calculated odds are, the true odds are likely slightly higher, as all 5 community cards are not always dealt out (due to folding).
