Caribbean Stud Poker : probability for the player winning with a royal flush

In the Caribbean Stud Poker table, for the player winning with a royal flush, I can see there are 16759740 combinations. As the total number of combinations is 19933230517200, the probability is 8.40794E-07.

I should have calculated this probability this way :
The odds of a royal flush are 4 in 2,598,960. The probability is 1.53908E-06
If the player gets a royal flush then the odds of the dealer getting a royal flush is reduced to 3 in 1,533,939. The probability is 1.95575E-06
The probability of this happening in the same time (for a push) is 3.01E-12 (The product of the two prior values).
In conclusion, I should have said that the probability for player winning with royal flush is : 1.53908E-06 - 3.01E-12 = 1.53907E-06 far from the initial 8.40794E-07 value.

Please, could you tell me where I am making the mistake ?
Thanks

The probability of a player getting and winning with a royal flush are (odds of player royal)-(odds of coincidental dealer royal). The odds of the player winning the hand once the player's royal is already known are 1-(odds of coincidental dealer royal). I'm not sure what your subtraction is all about at the end there...

Thanks for the reply.

Quote: rdw4potus

The probability of a player getting and winning with a royal flush are (odds of player royal)-(odds of coincidental dealer royal).

That's exactly what I meant:
The odds of player royal are 4 in 2,598,960, which is 1 in 649,740.
The odds of dealer royal is reduced to 3 in 1,533,939, or 1 in 511,313.
The odds of coincidental dealer royal is 1 in 332,220,508,620 (The product of the two prior values).
The probability of the push hand is 1/332,220,508,620 =3.01E-12 (very weak)

So the probability for player winning with royal flush should be near 4/ 2,598,960 because the coincidence is weak, isn't it ?