What is the largest poker bet that cannot be made?

[Ed Aboud]

Homework Statement

The local poker club plays with blue chips worth €5 and red chips worth €8.
What is the largest bet that cannot be made? (Hint: Use induction)

Homework Equations

The Attempt at a Solution

I honestly have no idea how to start this question.

Thanks in advance for any help!

 

[HallsofIvy]

What bets CAN be made? Obviously you cannot make bets of less that €5. What about €6 or €7, etc.

 

[Pud]

The ability to bet increases as the bet goes up, as there are more ways to combine chips to achieve a certain bet.

Any bet ending in 0, 2, 4, 5, 6, 8 can be made once over a certain number. The aim is to use $8 chips until the number ends in 0 or 5 then use $5 chips.

Ending in 0 or 5: Any bet ending in 0 or 5 can be made, no minimum.
Ending in 2: Any bet over (and including) $32 can be made. (Since $32 is the lowest bet ending in 2 that can be made using $8 chips)
Ending in 4: Any bet over (and including) $24 can be made. (Since $24 is the lowest bet ending in 2 that can be made using $8 chips)
Ending in 6: Any bet over (and including) $16 can be made. (Since $16 is the lowest bet ending in 2 that can be made using $8 chips)
Ending in 8: Any bet ending in 8 can be made, no miniumum. (Since an $8 chip will reduce the number to a multiple of 5)

For bets ending in 1, 3, 7 and 9 the aim is to reduce it to an even number, then use the above principle. To achieve this you have to minus 5 (use a $5 chip) from the number first.

Ending in 1: Minus 5 results in bet ending in 4, refer to above.
Ending in 3: Minus 5 results in bet ending in 8, refer to above.
Ending in 7: Minus 5 results in bet ending in 2, refer to above.
Ending in 9: Minus 5 results in bet ending in 4, refer to above.

Going by this we can see (well with enough looking you can XD) that any and all bets above $32 can be made.

The highest bet that cannot be made is $27. As it is odd, you minus 5 to give $22. As it ends in 2 you try to minus $32 however you cannot.

I'm sorry this reply is very rambling and probably quite hard to follow, however I wrote it as I played with the problem in my head.

I also sketched out a flow chart of what to do for any bet, summarised below.

1. Does the bet end in 0, 5 or 8? >> Yes-Go to Step10 No-Go to Step 2
2. Is the bet Even? >> Yes-Go to Step 3 No-Go to Step 9
3. Does the bet end in 2? >> Yes-Go to Step 6 No-Go to Step 4
4. Does the bet end in 4? >> Yes-Go to Step 7 No-Go to Step 5
5. Does the bet end in 6? >> Yes-Go to Step 8 No-You've gone wrong!
6. Minus 32. >> Go to Step 1. If you cannot minus 32 the bet cannot be made.
7. Minus 24. >> Go to Step 1. If you cannot minus 24 the bet cannot be made.
8. Minus 16. >> Go to Step 1. If you cannot minus 16 the bet cannot be made.
9. Minus 5. >> Go to Step 1.
10. The bet can be made.

 

[HallsofIvy]

Any bet that can be made using 5€ and 8€ chips must be of the form c= 5n+ 8m for integers n and m. 3- 2= 1 but 2= 5-3 so 3- (5-3)= 2(3)- 5= 1. 3= 8- 5 so 2(8- 5)- 5= 2(8)- 3(5)= 1. Then for any c, (2c)8+ (-3c)(5)= c. One possible "solution" to "c= 5n+ 8m" is m= 2c, n= -3c. But it is also true then that m= 2c- 5k, n= -3c+ 8k is a solutions for any integer k because then 5(-3c+ 8k)+ 8(2c- 5k)= -15c+ 40k+ 15c- 40k= c. Since we are talking about "numbers of chips" here, m and n must both be positive. So the question is "for what values of c is there no k making both 2c-5k and -3c+ 8k positive?" Both positive means we 2c- 5k> 0 and -3c+ 8k> 0 so k< 2c/5 and k> 3c/8. That is, 3c/8< k< 2c/5.

It is certainly true that if 2c/5- 3c/8$\ge$ 1, there must be such a k. That is, if 2c/5- 3c/8= (16-15)c/40= c/40$\ge$ 1. If c$\ge$ 40, there is always such an integer and always a way of getting c by "8"s and "5"s.

I had originally posted this declaring that it would be sufficient to have that difference less than 1 to see that there could not be such a sum of "8"s and "5"s and so 39 is the largest bet that cannot be made- that's not true. I quickly found that 3(8)+ 3(5)= 24+ 15= 39! Yes, 2(39)/5- 3(39)/8 is less than one. But 2(39)/5 is slightly larger than 15 and 3(39)/8 is slightly less than 15. k= 15 gives m= n= 3.

But, we know that the maximum "impossible bet" must be less than 39. It was not too difficult, though tedious, to work back through the numbers less than 39 and find that the largest "impossible bet" is _____ (Well, I'm not going to just give you the number. For c= 38, 37, 36, ... you can calculate 3c/8 and 2c/5 and find the largest c for which there is NO integer between them.)