What are the chances of this happening?

Question

1) What is the probabilty of the same number being drawn twice in succession from a bag containing balls numbered 1 to 49 where the drawn ball is immediately replaced.
2) What is the probabilty of the same number being drawn 4 times in 36 draws.
Some woman keeps winning the Church's draw and I m wondering if she has supernatural assistance.
Please give the calculation.

Answer 1

1) since each ball has equal chances of being drawn, the probability for a particular ball being drawn is 1/49. As the ball is replaced, there are again 49 balls. The chances of two consecutive draws of the same ball are (1/49) squared.

2) the probability of drawing a particular ball is 1/49 and that of drawing any other ball is 48/49. The probability distribution is (1/49 + 48/49) power n, where n is the number of draws. Now in this case we want exactly four draws of the chosen ball in total 36 draws, the required probability is 5th term (4+1) in the binomial distribution (1/49 + 48/49) power 36, i.e., C(36,4) x (1/49)power 4 x (48/49) power 32
This approximates to 0.005282176

Answer 2

There are two right answers to the first question depending upon the perspective taken by the observer. If no draws have occurred and we are saying "What are the chances that in the next two draws the same number will be drawn twice?" the answer is (1/49)^2 or one in 2401. However, if the perspective is after the first draw and before the second such as "She drew 14 in her first draw, what is the probability she will draw 14 the next time around?" the answer is 1 in 49. In fact, had she drawn 14 each of the past 5 times, the probability of her drawing 14 the sixth go around is still 1 in 49.

So, the answer that is the most correct of the two is based upon the exact question being asked.

Others have answered the second question well enough that I can't add much.

By the way, I was amused by The Truth's answer. It is witty. And wrong. It's wrong because the two answers aren't equally weighted. In other words, while there are only two choices (it either happens or doesn't) the choices aren't equally probable to occur. If I had 100 coins, 99 American pennies and one Canadian penny (which feels the same as an American) and I ask "What is the probability that I will blindly choose the Canadian penny?" the answer is clearly NOT 50% despite there being only two types of coins to choose from. That is because the American pennies far outnumber the Canadian one and, thus, are more probable to be chosen.

EDIT EDIT EDIT!!

Upon further reflection I am forced to say that regardless of perspective, the possibility of the same ball being drawn twice consecutively is 1/49. The reason is that we aren't saying "What is the probability that ball number 20 will be drawn twice" but rather "what is the probability that the same ball will be drawn twice. Thus, the first draw merely tells us what ball we are looking for in the second draw. In other words, the probability of picking the "right" ball in the first draw is 100% since any ball chosen in the first draw is correct. The question then becomes "What is the chance that we will draw the first ball again?" Because there are 49 balls and we are looking for one it is 1/49. So, since we haven't decided which ball we are looking for prior to the first draw, the first draw is merely deciding which ball we are looking for on the second draw which makes the probability of choosing the same ball twice 1/49. Does this make sense to people?

Answer 3

#1: 1/49. This is trivial
#2: There are 49^36 different possible sequences of draws. The number of sequences that has at least 4 of some particular number in it (say 1) is given by the number of ways to choose 4 positions for the number in the 36 possible draws, times the number of ways the remaining numbers can be permuted. This is given by 36!/(32!4!)*49^32. Since there are 49 different numbers that could be repeated, we multiply this by 49 to get the total number of successful outcomes, and then divide by 49^36 to find the probability. This gives us (36*35*34*33)/(24*49^3) = 1,413,720/2,823,576 = 8,415/16,807, which is just slightly greater than 1/2. Thus for the same number to be drawn at least four times in 36 draws is not unusual, and should be expected to occur about half the time.

Note: The error in anonymous_dave's calculation is that he implicitly assumes that the same number being drawn four times must happen in the first four draws. This is false.

Answer 4

[1] The probability of the second number drawn is the same as the first, given the first has already been drawn is 1 / 49 (or about 2.04%).

[2] The probability of a number with a theoretical probability of 1 / 49 being drawn 4 out of 36 tries is:
P(4 winners) × P(32 losers) × (number of ways 4 wins and 32 losses can be ordered)
(1 / 49)^4 × (48 / 49)^32 × C(36, 4)
= (0.00000017346652555743) × (0.5169458284691) × 58905
= approx. 0.528%, or about once in every 189 times 36 numbers are drawn.

Not terribly likely, but within the realm of believability.

Add to that Pascal's (correct) assertion that this is the probability of your given woman at church winning 4 of 36 draws... but the fact that any of the 49 numbers can be drawn the desired number of times means the 0.528% must be multiplied by 49... giving a final answer of 25.883%, or about once every 4 times.

Answer 5

pascal, louise, and aguy have all provided good accurate information, and some of the others have either provided comic relief or kept the curve low in stats class for the rest of us.

i do have a few suggestions for you douglas.

first, please clarify your question if you want exact answers, though you may already have the answer if one of the three good answerers guessed right about what exactly you are asking.

second, dont assume that there is or is not some "intervention", be it supernatural or maybe of a more dubious nature until you have more information. just because the odds are high or low, doesnt mean that something is random or not. the most unlikely events are sometimes just a matter of chance. and sometimes, even cheaters or gods do a bad job and get results that seem "ordinary"

lastly, try to bet on the 4 time winner, im guessing she's gonna win again

Answer 6

1)
If you have a specific number in mind then:
1/49 * 1/49 = 1 in 2401
If you just want to match the first number drawn a second time then:
1/49
2) (1/49) ^ 4 * (48/49) ^ 32 = 1 in 11,151,654

Answer 7

Not sure about second one, but the first one's 1/49.

The first draw doesn't matter, but the probability that the second number will match the first number is 1/49. Therefore, 1 * 1/49 = 1/49. Ta da!

Good luck on the whole draw thing.

Answer 8

the probability of bieng drawn successfully in two consecutive drawings is 4.16 x 10^(-4), or (1/49) squared. This can also be said as 1 in 2401.

The probability of being drawn 4 out of 36 draws is .01.

Answer 9

25 trillion to 1

Answer 10

1 in 49 for the first.

I feel a lottery answer coming on!