Probability

>>339348
Assuming the deck is shuffled on each draw:
1) The odds of failing are (.97)^100, meaning 100 times in a row you fail to draw the right card. This multiplies to .0475 = 4.75%.
2) The odds of failing are ((.97)^10)^10).10 times in a row, you draw 10 and none of them are a winning card. This is exactly the same as .97 to the hundredth power, just multiplied a little differently.

>>339348
How many cards are in the deck? Assuming a finite deck, the second scenario (10 cards drawn with no shuffling in between) would give higher odds of winning by some amount.

>>339348
The exact same, it's just a different way to say the same thing
>draw 100 cards, see if any of them are winner cards
>draw 10 cards, see if any are winner cards, if not draw 10 more, etc

>>339428
You're assuming the cards aren't replaced and shuffled between draws. The op doesn't specify one way or the other, but given that it does specify drawing cards as singles, rather than 100 at once, I'd lean toward the opposite assumption.

>>339459
That is irrelevant.

>>339462
Say there's a hundred card deck with 3 winners. Drawing one card and replacing and reshuffling after each draw yields scenario 1 in >>339367

Drawing ten cards and replace+reshuffle after each draw yields odds of failing as:
(97/100 * 96/99 * 95/98 * 94/97 * 93/96 * 92/95 * 91/94 * 90/93 * 89/92 * 88/91)^10 = 0409769075

>>339479
You're using too much math.

>>339462
Say there's 100 cards, and 3 of them are winning cards, and you draw 100 of them at once.

>>339516
You're guaranteed to win in this degenerate case. But you're introducing more assumptions than what's required to answer the question.

I'm using a standard 52 card deck and 4 at a time because it's easier for me to visualize, but here's the math

Odds of getting the winning card by drawing one at a time, 4 times,
1/52+1/51+1/50+1/49 = about 7.92% chance

4 at a time
4/52=1/13 = about 7.69% chance

mathematically, you'd be better going one at a time, although, much like the monty hall problem, it's a matter of semantics

>>339520
>here's the math
>gets first scenario wrong
>gets second scenario right for faulty reasons
>gets final conclusion wrong
You shouldn't chime in on other people's math problems.

I fucked up the OP.
Let me rephrase.
Let's assume hitting a button gives you a 3% chance of winning.
If you press the button 10x10 versus 1x100, which has a better chance of you winning more?

>>339529
>>339428
100 independent trials = 100 independent trials.

>>339524
how is the first scenario wrong? Your starting draw has 1/52 chance, the next draw has a 1/51 chance, then 1/50, then 1/49

If you don't just add the probabilities of each of those, what's the formula?

>>339519
They're the same assumptions >>339479 is making.

Reducing to the pathological case is a classic way of illustrating tricky maths to the non-mathematically-minded. Instead of hearing "blah blah blah numbers blah so I'm right", they hear "you open one door. The host opens the other hundred billion trillion doors and none of them have the money behind them. Do you swap?"