I got quad sevens five times in 3 hours . The first three times were within the same round of ten hands. Without a technical explanation what are the odds of three times in ten hands and five times in three hours...ten players for all hands dealt. The dealer For the first three quads said in all his years of dealing he had never seen that..
2 Answers
I will have to get technical, sorry.
There are 35 possible ways to make quad sevens. They are all equally likely. You can get this number by writing out all possibilities like this:
XXXXOOO
XXXOXOO
XXXOOXO
etc.
Using the binomial coefficient with n = 7 and k = 4 will get you there as well and also a lot quicker. Odds of quad sevens in a single hand therefore is the odds of a single way to get this multiplied by 35. One way to get quad sevens is get all sevens in the first 4 cards. This is simply equal to: 4/52 * 3/51 * 2/50 * 1/49
. Multiplied by 35 is approximately equal to 0.000129282
.
Now to get quads 3 times in 10 hands we can again write out all the possiblities, or we can use the binomial coefficient again. Turns out there are 120 possible ways to get quads 3 times in 10 hands. This means we just have to calculate the odds of one possible way, for example in the first 3 hands. This is equal to our first number cubed, which is approximately 2.1608*10^-12
. Now we can multiply this by 120 and we get: 2.5930*10^-10
.
NOTE: this assumes the player always gets to the river and four sevens on the board also counts as quads.
So the odds of getting quad sevens 3 times in 10 hands is equal to 2.5930*10^-10
or 1 : 3,856,599,742
.
Note: To be clear, these are the odds of getting at least 3 times quads out of the ten hands. To calculate the odds of exactly 3 times you need to multiply 2.1608*10^-12
by the odds of not getting quad sevens to the power of 7 and then follow the same steps. Answer will, though a little rarer, be rougly the same.
Also note: these are the odds of getting quad sevens, not quads in general. To calculate the latter instead of 4/52 * 3/51 * 2/50 * 1/49
use 3/51 * 2/50 * 1/49
.
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This makes the (probably necessary) assumption that we go to the river every time we have a starting hand with at least one seven in it, right? Edit: in fact, it assumes we go to the river every single hand, as it includes the times when all 4 sevens are in the board, doesn’t it?– 3N1GM4Commented Dec 1, 2017 at 9:02
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No, I realise the calculations become unmanageable any other way, but it’s just worth making people aware of the assumptions so they know the true odds of what happened are probably a fair bit longer than the answer given.– 3N1GM4Commented Dec 1, 2017 at 12:58
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