2. My relatives like to play a raffle every Christmas. Each participant pays som
ID: 1248218 • Letter: 2
Question
2. My relatives like to play a raffle every Christmas. Each participant pays some set price to enter the raffle and puts an ID card of theirs into a bag. One by one, IDs are pulled out of the bag. The owner of the last ID card in the bag wins the pot.
Let the price=P, and the number of players = n.
i) write out the EV equation for purchasing one raffle entrance for yourself. Without designating a specific price of number of players, is this +,- or 0 EV?
ii) What happens to the EV if a one player buys a second raffle spot before the game begins? (so that he has 2 ID’s in the bag at the start of the game) Show the modified equation, and show how its value changes, or doesn’t change.
iii) What would happen if one person were allowed to re-enter the raffle after it has begun? Modify the equation in some way to illustrate your point. (The easiest example would be to show what happens if someone can re-enter when there are only 2 participants left). What is the EV? Would a risk neutral person re-enter?
iv) Suppose it is down to the final two contestants. Before the final ID is drawn, a proposal is made: Both players decide that if they win, they will split the pot with the other player. Explain how a risk averse, risk neutral, and risk seeking person would respond to this suggestion.
Hint: If the only way you know how to solve these questions is by using concrete prices and numbers of players, go ahead and do that.
Explanation / Answer
Let the price=P, and the number of players = n.
i) write out the EV equation for purchasing one raffle entrance for yourself. Without designating a specific price of number of players, is this +,- or 0 EV?
Number of players= N, Price of each ticket= P
Prize money= NP.
Probability to win= 1/N
Probability to loose= 1- 1/N
= (N-1)/N
Pay out for a win, X win= NP
Pay out for a loss, X loss= 0
EV = Pw . Xwin + Pl . X loose
= 1/N x NP + (N-1)/N x 0
= P +0
EV = P
EV (X-P)= P-P
EV(X-P)= 0
S0, expected value is zero.
ii) What happens to the EV if a one player buys a second raffle spot before the game begins? (so that he has 2 ID’s in the bag at the start of the game) Show the modified equation, and show how its value changes, or doesn’t change.
Number of players= N, Price of each ticket= P
Prize money= NP
Probability to win for first ticket = 2/N
Probability to win for second ticket = (N- 2)/N
Probability to loose= 1- 2/N + (N-2)/N
= (2N-4)/N
Pay out for a win, X win= NP
Pay out for a loss, X loss= 0
EV = Pw1 . Xwin + Pw2 . Xwin + Pl . X loose
= [2/N x NP] + [(N-2)/N x NP] + [(2N-4)/N x 0]
= 2P + NP -2P +0
EV= NP
EV(X-P) = NP- P
iii) What would happen if one person were allowed to re-enter the raffle after it has begun? Modify the equation in some way to illustrate your point. (The easiest example would be to show what happens if someone can re-enter when there are only 2 participants left). What is the EV? Would a risk neutral person re-enter?
With only two participants left the probability to win is ½
EV = ½ NP
EV(X-P) = NP/2 – P
If third participant enters
Probability to win = 1/3 for three people.
EV = (1/3 NP) + (1/3 x 0) + (1/3 x 0)
Net EV, EV(X-P) = NP/3 - P
iv) Suppose it is down to the final two contestants. Before the final ID is drawn, a proposal is made: Both players decide that if they win, they will split the pot with the other player. Explain how a risk averse, risk neutral, and risk seeking person would respond to this suggestion.
A risk averse person will take the offer of dividing the price money.
Risk neutral person is more concerned about the expected return. His expected value return in 0, as calculated in the first case, so this amount of parting is equal to NP/2.
So, he will also take the option of dividing the prize money.
Risk taker will take the risk, to wait for the final card.
As per hint You are allowed to do the problem by taking some value of P and N
P= $10 N= 10 persons.
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