[rsk111]

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I thought about it in exactly this way and I don't think it is right. Instead of the classic "do you see why?" I'll say that upon request I think I can provide a specific counterexample to refute the lottery example. While my counterexample can refute the lottery example I am *not* sure that it generalizes to apply to the football pool example (I am still thinking about how to do that).

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Noobie:

After hanging around here for a while, you'll realize that I'm wrong most of the time. Therefore, do not wait for me to request a counterexample to prove me wrong, just do it like everyone else does.

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I just didn't want to type it all out, unless someone really cared or if you figured it out. Anyway, here it is.

Let's say you have a lottery with one number (1-10). You can buy a ticket with any number from 1-10 printed on it and each week one number from 1-10 will be pulled out of a hat. Each ticket costs $1 and you can buy as many tickets as you want (and for each ticket you can choose whatever number you want). The prize pool is the total receipts collected from ticket sales. Any player who has a ticket with the winning number is entitled to a share of the prize pool (i.e., if there is only one winning ticket, that person gets all of the prize money, if there are two winning tickets then each person gets 1/2 of the prize pool, etc).

Now let's say three players decide to play this lottery. Player A decides to buy 10 tickets with so that he has one ticket with each number. Player B and Player C decided that they will just buy one ticket and they randomly choose which number to play (it really doesn't matter what they choose, all that is important is that they just pick one number).

In this scenario, total prize pool = $12 ($10 from Player A, $1 from Player B, and $1 from Player C)

According to your reasoning, Player A should have no advantage, because even though he has 10x the chance of winning, he paid 10 times as much, so it should even out. Let's see if this is the case with some simple EV calculations.

EV for Player A

*81% of the time Player A will have the only winning ticket and win 100% of the prize pool for a net win of $2 (i.e., no tie)
*18% of the time Player A and only one of player B or player C will have the same winning ticket so Player A will win 50% of the prize pool for a net loss of $4 (i.e., two way tie)
*1% of the time Player A, B, an C will have a winning ticket and each get 1/3 (i.e., three way tie), so player A will have a net loss of $6

So, therefore the EV equals

0.81*2 + 0.18*(-4) + 0.01*(-6) = 0.84,

So Player A has a EV of $0.84. Similar calculations will yield an EV of -$0.42 for the other players.

So in this instance, by buying multiple entries, player A has managed to gain an advantage in the lottery.

However, for this same lottery game, it is easy to also come up with examples where it is -EV to buy multiple tickets. So, in the end, we can show that buying multiple tickets can be either +EV or -EV, depending on the strategy employed.

I may have made a mistake, somewhere. If so, let me know. Furthermore, as I said previously, I am not sure how to apply this to the football pool case.