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In a given lottery, let's say my odds of winning a grand prize are $p_0$, and the grand prize has value $V$. I can either buy more tickets, at a cost $C_1$, or for each ticket, I can buy a multiplier for each ticket, at a cost $C_2$ per ticket. The multipliers are structured as follows: 

5X      
1 in 10
4X  
1 in 10
3X  
1 in 3.33
2X  
1 in 2

That is, if I pay $C_2$ extra for a ticket for a multiplier, then I have a $1/2$ chance of doubling my prize, a $1/3$ chance of tripling, etc.

For simplicity let's assume there is only the grand prize of value $V$. Should I buy more tickets, or buy multipliers on all my tickets?

My Thoughts

My average payout per ticket is $Vp_0$, which costs $C_1N$, if I buy $N$ tickets. So if I buy multipliers on $M \leq N$ tickets, my profit equation takes the form: $$ \text{Profit} = NVp_0 - C_1N - C_2M $$ I suppose the question is, at what point does the incremental payout from an additional ticket become less than buying a multiplier on an existing ticket. If $m$ is the multiplier, and $P_m$ is the probability of that multiplier, then the question is at what point does $Vp_0 - C_1 \lessgtr mP_mVp_0 - C_2$

Moderat
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1 Answers1

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On the assumption that there are lots of tickets sold (so your buying one more does not change the value of the existing tickets), each ticket has some expected value $v$. This includes all the prizes. If you buy another ticket, you get $v$. If you buy a multiplier, you get $v(\frac 12 \cdot 1 + \frac 3{10}\cdot 2 + \frac 1{10}\cdot 3 + \frac 1{10}\cdot 4)=\frac {18}{10}v$ added to your value. Your loss on buying another ticket is $C_1-v.$ Your loss on buying a multiplier is $C_2-\frac {18}{10}v.$ Compare these and you will know which way you are less badly off.

Ross Millikan
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  • Actually I double with chance $1$ in $2$ so isn't it ($1/2 \cdot 2 + 1/3 \cdot 3 + \cdots$)? But thank you this is the way I was thinking about it – Moderat Feb 10 '15 at 23:45
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    No, because you were already getting $v$. When you double, your increase is $v$, so that is the value of the multiplier. – Ross Millikan Feb 10 '15 at 23:46