Let’s say everyone in the world has a deck of 100 cards, of which, 99 are white and 1 is red.
Everyone has 5000 attempts at drawing a single card from their deck of 100 to try and get the red one (each time replacing the drawn card). If and when they draw the red card, they stop drawing and have won.
After everybody has either drawn the red card or used up all 5000 attempts, what percentage of people, statistically, will have won?
My instinct (and limited knowledge of probability) tells me around 50% (5000 * 1/100), but I know that instinct can be very, very wrong when talking about probability.
If I am correct, would it also follow that for 90% of the population to have won (again statistically speaking) then we would have to raise the attempts to 9000?
Odds of drawing a white card - P(W): 0.99
Odds of drawing a red card - P®: 0.01
If we run this ‘experiment’ twice, the odds of a white card on both the first and second attempt (with replacement) is P(W) X P(W) = 0.99 x 0.99 = .9801.
Put another way - the odds that we get a red card in at least one of those attemps is 1 - (odds both are not white) = 1 - 0.9801, or 0.0199, or just slightly under 2%.
The odds of running the experiment 10 times and never getting a red card is P(W)x P(W) x …10 times, or P(W)^10 = 90.44%. The odds you get a red card one time in those 10 attemps = 9.6%.
At 100 times, there is a 63% chance we get a red card at least once. And at 1000 times the odds of getting 1000 white cards is 0.00004, which means the odds of getting a red card at least once is 99.99%!
If everyone in the world gets 5,000 attempts, I think 99.99999% of the population will have ‘won’.
Well, first of all 5000 draws times 1/100 chance of getting a red card in a single draw is not 50%, it’s 5 red cards (expected, on average). So your math was off.
But what you calculated was “If everyone drew 5000 times, how many red cards would the average person end up drawing?” The answer to that is 5.
But that’s not what the people are doing; what they’re doing is seeing if they can draw 5000 cards without getting a single red card, which is the same as drawing 5000 cards, and getting white every time. Which is pretty easily 99/100 (chance of getting a white card first time) times 99/100 (white card second time) times… times 99/100 (white card 5,000th time), or (99/100)^5000. Which ends up being a very very small number. So small you’d need a billion people to even have a one in a million chance of someone drawing all 5,000 without getting a red card.
The correct answer, as has been mentioned is 1-0.99^5000.
That is 99.999999999999999999985% (and change) of people would draw a red card. That’s one in ten thousand billion billion people that would draw all white cards.
50 would be the break-even point if the white cards drawn are discarded rather than re-inserted. Clearly the given problem should break even at a higher point.
The way I thought it out was obviously wrong and I think can now see how.
I took each ‘set’ of 100 draws as a unique entity. Whereby probability would suggest that only 1 red card would be drawn in that set. :smack:
The way it went in my head was that for each person to have a least a 50/50 chance of drawing a red card OVERALL they would have to draw 50 ‘sets’ of 100 draws, therefore the 5000 figure.
Like I said in the OP, my knowledge of probability is rather limited; however I do know that if Monty offers me the chance to swap doors…I always will
Or, look at it another way: Suppose everyone drew 50 cards, rather than stopping when they hit the red one. On average, you would get 1 red card per two people, as you’d expect. But some people would get lucky and draw the red card twice, or even more often. To balance this out, you have to have more than half of the people who never draw the red at all (if you will, the multi-red folks have “stolen some of their luck”).
With 69ish draws, meanwhile, half the population has no red cards, less than half has one, and a few have two or more, so the average is now more than one red card per two people.
Good explanation. It’s related to Chevelier de Mere’s problem a French gentleman who liked to fleece his friends at dice games. How many rolls of a pair of six sided dice does it take before you can (on average) expect to see 6-6 (boxcars)? There’s a one in thirty-six probability of rolling 6-6 on any given roll so at first blush one might think it would take 36 rolls. Doing the math, the probability of not rolling double sixes is 35/36 so the probability of not rolling a series of double sixes is (35/36)^n. Playing around with that, one will quickly see that the fifty percent point lies between 24 and 25 rolls. 24.6050977177 to be precise.
