Tough math probability question

This is the correct answer. I have a post grad degree in this shit :slight_smile:


Right. So let’s do the coun problem with a 20d dice. 19/20^3 is .0.8574…

Oh, wait that would be .1426.

Actually that does seem right.

So yes, going back with the .58 # you came up with.

The last troll that challenged me to come find him wound up missing. When his body was eventually found it was badly burned and mutilated, tortured. And he had been raped.

If we think of this as a deck of cards - we want to pull a heart. 25% chance we pull a heart. Now we want to pull two hearts - which means we can’t pull from the other 77% of the deck (less the heart we just pulled).

1/4 * 12/51 = 1/17 or 5% we pull a second heart on our next shot.

Inversely here, we have to pull from the larger group three times in a row, whereas we only have to pull from the smaller group once. This was where I erred and why @MrColdCock was right.

So yes, 23.8% to start us off - and as the other poster mentioned above, we now have to calculate from the group we do not want to pull from.

76.2% we WILL pull from the bigger group before we start. And we pull and we do pull from that group.
.762 * 15/20 = .5715 or 57.15% chances we pull from the bigger group twice in a row. We give er a go and we miss again, drawing from the bigger group.

We still want to pull from the 5 but now our shot is .5715 * 14/19 = 42.11% (you pick from the larger pool all times) as previously said. I AM A LOSER!

RIP Homeslice

1 Like

I think that is right. It’s the old chance of rain where you have a certain percentage chsncr in consecutive days

1 Like

100%. In golf probably around 1%

This is what I got- about a 57.89% chance of picking.

I did this in my head and can’t quite remember the train of thought so you’ll have to just trust me on this but the answer is 66% chance

Ask Scott Steiner to answer it.

1 Like

I’m on my desktop now, I’ll show how I got my answer:

Calculated odds of NOT getting one of the five:

Round 1 - 16/21
Round 2 - 15/20
Round 3 - 14/19

This leaves 7,980 possible outcomes (21 * 20 * 19).
You would NOT get one of the five in 3,360 instances (16 * 15 * 14)

3360 / 7980 = 0.421052…
Inverted is 0.578947 or roughly a 57.89% of getting one of the balls.

1 Like

Yep. Makes perfect sense.

My brother made a bet with this guy, $100, when drawing numbers for an event. There were 21 numbers left and he says I bet $100 number 1 or 2 or 3 or 4 or 5 come up over the next three draws. He won $100 as it did

1 Like

This is what I would go with.

I am 99% at a 95/95 confidence level that the answer is the 58% one

1 Like

“this is what i said, but i’m starting to question the product rule as part of it. i think it’s the sum of each probability (5/21) + (5/20) + (5/19) = .7512”

This doesn’t account for overlap where balls are picked more than once, and will give a higher probability than reality.


Balls are removed after each draw. And that 75% can’t be right by looking at it another way. Say you drew four balls. Then it would show over a 100% chance you got one of the five. Not possible. Four would be

1 - ( (16/21)x(15/20)x(14/19)x(13/18) )

Not (16/21) + (15/20) + (14/19) + (13/18)

Yes of not drawing it

I mean there are 5 balls that you are looking at. You have to account for getting a ball in the first round, then getting one of the 4 remaining balls in a later round. This is why it’s easier to calculate the probability of NOT getting a ball and inverting the percentage.


this is the correct answer