The question states:- "Which ticket offers the best value for money if you buy as much tickets until you win."
To recap: we have 3 kinds of tickets:-
option 1: 15% chance of success, cost $120
option 2: 9% chance of success, cost $60
option 3: 6% chance of success, cost $20
Assumptions.
1. there is no limit to the number of tickets available (none is stated in the question, and a previous post from OP in this thread states that is the case);
2. we have unlimited funds available;
3. the ticket try events are independent, like rolls of a single die (no dependency between events is stated in the question); such that the theory of cumulative probability can be used to model the scenario.
Let P be the probability of losing any one try.
The cumulative probability of losing after N tries for any specific ticket type is P^N, and hence that of winning is 1-(P^N).
Let us assume that (to a first approximation) we need a 99% (or 0.99) probability of winning to qualify as "success" (100% certainty can never be achieved).
So we need to find the number of trys required to yield a cumulative probability of losing of 1%, or 0.01 (since 1-0.99=0.01).
Then P has the following values:-
option 1($120 per ticket) : 1-0.15 = 0.85
option 2 ($60 per ticket) : 1-0.09 = 0.91
option 3 ($20 per ticket) : 1-0.06 = 0.94
In each case, what power of P will yield a cumulative probability of losing of 0.01?
Then we need to solve for x in the equation:
P^x = 0.01
Taking logs:-
log(P^x) = log(0.01)
x. log(P) = -2
x = -2 / log(P)
So the number of trys required to obtain a 99% chance of winning with each ticket option is:-
option 1:
x = -2 / log(0.85)
= 28 tries
option 2:
x = -2 / log(0.91)
= 49 tries
option 3:
x = -2 / log(0.94)
= 74 tries
Hence the total cost if you buy as many tickets as you need until you win (with a 99% chance of winning) is:-
option 1: cost = 28 * 120 = $3360
option 2: cost = 49 * 60 = $2940
option 3: cost = 74 * 20 = $1480
Similarly we can calculate the number of tries and spend required to give a 99.9% chance of winning:-
option 1: -3/log(0.85) = 43 => cost = 43*120 = $5160
option 2: -3/log(0.91) = 73 => cost = 73*60 = $4380
option 3: -3/log(0.94) = 112 => cost = 112*20 = $2240
Therefore option 3 is the best value for money; with the proviso that we have assumed the ticket try events are independent and that cumulative probability can therefore be applied.
Reference.
Probability is the measure of the possibility that a given event will occur. Cumulative probability is the measure of the chance that two or more events will happen. Usually, this consists of events in a sequence, such as flipping "heads" twice in a row on a coin toss, but the events may also be...
www.sciencing.com
www.quantamagazine.org
