Probability problem.

(35/5000)*(35/5000)*(35/5000)*(35/5000)

Yes and therefore the answer is

4*(35/5000)*(35/5000)*(35/5000)*(4965/5000)

= 4*.007 x .007 x .007 x .993

= .000001372 x .993

= .000001362396

or about 1.36 in a million.

(solution, sum of four probabilities, that you win first three and not fourth, 1,2,4 and not third, 1,3,4 and not second, or second, third and fourth).

The chance of winning two would be 6*.007*007*.993*.993 = .000289898406

The chance of winning one would be 4*.007*.993*.993*.993 = .027416106396

The chance of winning none would be .993*.993*.993*.993 = .972292630401

The chance of winning four would be .007*.007*.007*.007 = .000000002401

It can be seen that these probabilities add up to unity

. 0 0 0 0 0 0 0 0 2 4 0 1 (4)
. 0 0 0 0 0 1 3 6 2 3 9 6 (3)
. 0 0 0 2 8 9 8 9 8 4 0 6 (2)
. 0 2 7 4 1 6 1 0 6 3 9 6 (1)
. 9 7 2 2 9 2 6 3 0 4 0 1 (0)**

1.0 0 0 0 0 0 0 0 0 0 0 0

** It can be added that your chance of winning at least one prize is therefore .027707369599 or about 2.8 per cent. That number is 1 minus the (0) win.

Your chance of winning at least three is the sum of lines (3) and (4) above, not that much greater than 3 but not 4.

M.T. Cranium wrote: »

4*(35/5000)*(35/5000)*(35/5000)*(4965/5000)

Just for my own learning, I'm wondering why this isn't:

(35/5000)*(1/5000)*(1/5000)*(4965/5000)

I assumed because it was one particular ticket that you had a 1:5000 chance of that coming out in each of the other two draws after you've chosen the ticket in the first draw?

Roaming Doc:

The two terms you show as (1/5000) should be (35/5000) because in the other two draws where you win, you have that same 35/5000 chance of winning some prize. Note also your formula is missing the 4 at the beginning of my formula. That applies because there are four orders in which you can win the three prizes (first, second, third times ... first, second, fourth times drawn ... first, third, fourth and second, third, fourth).

Here's an easier to visualize analogy to this question using cards.

What are my chances of randomly drawing a spade from the deck three times out of four? Assume that the deck is full each time (the cards you draw go back in for rounds 2, 3 and 4).

Your chance of getting a spade is 13/52 or 1/4 each time.

So the answer is 4 x (1/4) x (1/4) x (1/4) x (3/4) since you could draw the non-spade at each of the four opportunities.

(That reduces to 12/256 or 3/64 or about 5%). Each of these comes with three ways to fill the non-spade (that was not a term for the raffle because the non-win was simply a non-win, not three kinds of non-wins).

To see that this is correct, now add up the odds of drawing 3 of 4 for the other suits ... that gets us to 48/256. Add to those the chances of drawing all four of one suit, each is 1/256, bringing the total to 52/256. How about 2 of 4 being one suit, 2 of 4 being another? That one is 36 times (1/4) to the fourth, as there are six combinations of suits (DH, DC, DS, HC, HS, CS) and six ways they can come out in each case.This adds 36 bringing us to 88/256. Another option is two of one suit, and two of each of the others. That can happen 3 ways per two-suit, and those can each come out six ways, for a total of 18, for each of four suits, or 72 total variants. That adds up to a grand total of 160/256.

The only other outcome is one of each suit. The chance that any given draw will be this result is 4 x 1 x 3/4 x 1/2 x 1/4 or 96/256. Why the use of 4 in that? The formula does not specify which suit begins the process, so there are four sets of 1 x 3/4 x 1/2 x 1/4. Once you specify the start, the order of the other three is handled by the final three terms.

======================

To get back to your suggested formula, what you have there is the probability that your ticket will win any prize on first draw, one specific prize on the second draw, and one specific prize on the third draw and will then miss the prize table on the fourth draw. (this assumes the 35 prizes are all distinct from each other).