Deuces Wild odds

Hello Wizrard of Odds,

A friend of mine not too long ago on a FIVE cent super times play three handed deuces wild, received a deuce on her hand, then got 10 times from the times play spinner and one of the three hands made FOUR DEUCES.

WHAT ARE THE ODDS OF THAT HAPPENING?

Thank You,
Rick

Any thoughts on this?

1) Probability of redrawing 3 deuces in the remaining 47 cards in 4 chances.
2) Probability of redrawing 3 deuces on one of the remaining 3 hands. (This would be easy though, it would just be #1 + #1 + #1)
2) Probability of getting 10x on the draw (which I believe any mult is 1/15 odds, but they're weighted such that the avg mult I think is somewhere like 3x or 4x). I'm not in the mood to lookup the 10x probability of the draw DSTP so without that can't give you an exact answer (1/15 I think to get "any" multiplier, and they're weighted such that something like 3x is the avg multiplier, which means the 10x is only a fraction of that fraction).

It's a fun math exercise I'd encourage you to step through yourself too. For example, what's the odds of getting a deuce on the first draw card? 3/47 which is ~.064, aka ~6.4%. Then play it out from there... =).

Basically, all of these factors together this was quite improbable... however to someone that understands the law of large numbers it is also inevitable.

Quote: Romes

I'm not in the mood to lookup the 10x probability of the draw DSTP so without that can't give you an exact answer (1/15 I think to get "any" multiplier, and they're weighted such that something like 3x is the avg multiplier, which means the 10x is only a fraction of that fraction).

link to original post

If someone else is in the mood, perhaps https://wizardofodds.com/games/video-poker/tables/super-times-pay/ is a useful place to start, and then check out the other page https://wizardofodds.com/games/video-poker/tables/double-super-times-pay/

Super Times Pay and Double Super Times pay are similar, but different. Double Super Times Pay has at least one extra feature.

Appreciate all replies,,,didn’t realize it was as involved as it is to find out the answer, kind of figured the wizard would have the answer in short order on the basis he has figured the odds on something similar.

Okay:

Odds of getting a single deuce on your initial hand is: 4*c(48,4)/c(52,5)= 0.299473636


Odds of getting a 10 multiplier = 0.04 from Wizard's table.

Odds of redraw and getting 4 deuces on a single hand: c(3,3)*c(44,1)/c(47,4) = 0.000246685
-----So odds of redraw and getting 4 deuces on at least one of three hands is 3x that: 0.000740056
*****************************
Total probability of this occurring = 0.299473636 * 0.04 * 0.000740056 = 0.000008865 or one in 112,802 (approx)

Edit: I left out the odds of getting any multiplier at all which is 1/15.

So Total prob = 0.000008865 * 1/15 = 0.000000591 or one in 1,692,031 (approx)

Thank You

Quote: gordonm888

-----So odds of redraw and getting 4 deuces on at least one of three hands is 3x that: 0.000740056
link to original post

not quite right....(the common mistake is odds * # of hands)

it should be 1-(1-0.000246685)^3 = 0.0007398724545439883

Consider another simplified problem - assume one event occurs at the probability of 1/3, what's the probability of it happens at least 1 time after doing it 3 times? It's not 3 * (1/3) = 1. It should be 1-(1-2/3)^3 = 19/27

Quote: HopHoofer

Quote: gordonm888

-----So odds of redraw and getting 4 deuces on at least one of three hands is 3x that: 0.000740056
link to original post

not quite right....(the common mistake is odds * # of hands)

it should be 1-(1-0.000246685)^3 = 0.0007398724545439883

Consider another simplified problem - assume one event occurs at the probability of 1/3, what's the probability of it happens at least 1 time after doing it 3 times? It's not 3 * (1/3) = 1. It should be 1-(1-2/3)^3 = 19/27
link to original post

Welcome to the forum, hophoofer.

Yes, of course, my method did not take into account that sometimes one will gets 4 deuces on the redraw 2 or 3 times on the same hand. For the small probability I was working with, the difference in calculating the probability of getting 4 deuces as 1-(1x)^3 where x is the probability of it not happening is only about 0.00000019. I was just doing a fast calculation to try to be helpful. But your objection is certainly correct.