3500 Rotary Club tickets total sold @ $ 10 each
3 drawings one after another out of the the barrel of 3500 sold tickets
1st pick wins $10,000
2nd pick wins $1000
3rd pick wins $500
Tickets sell for $10 each and I purchased 30 tickets
is it not 30/3500 + 30/3499 + 30/2498? (about 2.5%)
Assuming the tickets are randomly mixed up.
(30/3500) * 3 = 2.6%
I’d actually run it like this:
The chances of NOT winning anything with the first draw is 3470/3500 (.9914)
If you don’t win that one, then the chances of not winning anything with the second become 3469/3499, then 3468/3498 for the third. (There really isn’t that much difference made from subtracting the winning tickets, out of a pool of 3500)
.9914 cubed works out to an overall chance of 97.45% of not winning anything.
Which means a 2.55% chance of at least one win.
The difference between this and One And Only Wanderers approach, (or ColdPhoenix’s) is probably minimal, because his would double-count for the possibility of more than one win, but that’s a vanishingly small factor in this case.
Yeah, because my (incorrect) method would give a 300% chance of winning if you bought all the tickets…
The number of times you win follows a hypergeometric distribution with N = 3500, n = 3 and m = 30.
n Prob(Winning n times)
0 0.9744983
1 0.0252897
2 0.0002114
3 0.0000005
Your expected winnings are much trickier to calculate because the order of the draws matters. I’d do it by simulation rather than trying to get exact figures.
I’m not sure what you mean by the “order”. The draw is three tickets pulled out of a set of 3500 with the first, second and third tickets drawn out yielding 10,000, 1000 and 500 dollars respectively.
There are three ways you can have one winning ticket: on the first draw, on the second draw, and on the third draw. I can ignore that when calculating the win probabilities, but I can’t when calculating the probabilities of various prizes because there are different prizes on each draw.
Doesn’t seem too bad…
T=30
N=3500
S1=$10000
S2=$1000
S3=$500
win1 + !win1 * { win2 + !win2 * win3 }
expected = T/N*S1 + [1-T/N] * { [T/(N-1)]*S2 + [1-T/(N-1)] * [T/(N-2)]*S3 }
… which is $98.43
So, your net expectation value for the game is -$201.57.
There are 8 possible scenarios:
(w = win; l = loss; displaying the the drawings in order)
www
wwl
wlw
wll
lww
lwl
llw
lll
There are 9 chances of winning, depending on where you are in the question, based on 3 numerators and 3 denominators: 30, 29, and 28; and 3500,3499, 3498, respectively.
The numerator will be 30 until you win once, at which time it will become 29 (I’m assuming that once a ticket ‘wins’ it cannot be drawn again). If you were to win both of the first drawings, you would then have 28 tickets to use for the last drawing.
The denominator is always 3500 in the first round, 3499 in the 2nd, and 3498 in the last.
In every scenario, the chance of losing is equal to 1 minus the chance of winning.
So, for each of the 8 possible outcomes, you multiply the probability of the specific outcomes in each rounds in order to find the probablity of that scenario manifesting.
For example, for lww, you would multiply (1-(30/3500))(30/3499)(29/3498), which equals 0.007%.
To come up with the chance of winnign at all, you add all of the scenarios except for lll, which equals 2.5502% by my calcs.
To find the expected value, you sum the products of the probability of each scenario occuring by the value fo the scenario. My calculated expected value equals $98.57.
In Excel, it looks liek this:
3500 3499 3498
30 29 28 30 29 28 30 29 28
0.008571429 0.008285714 0.008 0.008573878 0.008288082 0.008002286 0.008576329 0.008290452 0.008004574
0.991428571 0.991714286 0.992 0.991426122 0.991711918 0.991997714 0.991423671 0.991709548 0.991995426
w w w
0.0086 0.0083 0.0080 0.0001%
10000.0000 1000.0000 500.0000 $11,500.00
w w l $0.01
0.0086 0.0083 0.9920 0.0070%
10000.0000 1000.0000 $11,000.00
w l w $0.78
0.0086 0.9917 0.0083 0.0070%
10000.0000 500.0000 $10,500.00
w l l $0.74
0.0086 0.9917 0.9917 0.8430%
10000.0000 $10,000.00
l w w $84.30
0.9914 0.0086 0.0083 0.0070%
1000.0000 500.0000 $1,500.00
l w l $0.11
0.9914 0.0086 0.9917 0.8430%
1000.0000 $1,000.00
l l w $8.43
0.9914 0.9914 0.0086 0.8430%
500.0000 $500.00
l l l $4.21
0.991428571 0.991426122 0.991423671 97.4498% 2.5502%
100.0000%
$98.57
Question for statisticians: The OP asks about “statistical chances,” but is there really any statistics (as opposed to probability) involved in this question?
Not at all.
No, but I knew what he meant.
The assumption up-thread has been that this is a lottery drawing without replacement. If you win, you’re done.
Not if you have more than 1 ticket, based on my reading. Please indicate where you see otherwise - I’ve been drinking, may have missed it.
Yeah, you can’t win with a winning ticket, but you can with the rest. Unless the Rotary Club sez otherwise.
Righto - that’s how I read it - which is is why the chance for winning on the second draw after winning on the first draw is 29/3499 in my calcs.
Although everyone seems to be coming to the basic result of 2.55% chance of winning anything at all, and a PV of about 94 dollars and change - which I have to assume is sufficient for the OP.
ETA: about $84 of the $94 PV comes from the wll scenario - which indicates that 90%-ish of the value comes from the chances of winning the big prize alone - the odds of the little prizes are no better than the big prize and the smallish dollar amounts combined with the low odds make them contribute little to EV, while the odds of winnign multiple prizes are so remote as to be nearly irrelevant, all of the stuff only adds to about $10, or the cost of 1 of his 30 tickets.
Wow – the Rotary Club is making out like a bandit here!
Assuming they don’t hold the drawing until they have sold all 3500 tickets, they have an income of $35,000 but only pay out 32.8% of that ($11,500) as prizes. No casino in Las Vegas would dare to have a game with such a low payout.
agreed - thought to mention that I was calculating stuff for my first post
Ah, yes, of course. I was interpreting the problem in a silly way. But the conclusion of bandit-out-making seems solid either way!