Number of Cards Ted Dropped

Here is a problem by J.A.H. Hunter.

Ted paused in his dealing. He sneezed violently, and some of the cards in his hand fell to the floor. 'There goes all my luck,' he explained. But Sam looked under the table. 'All face down, so pick them up and go right on,' he said. 'Anyway you dropped only one less than you've dealt to my partner and myself together.' Bob laughed. 'That's right,' he agreed, 'and you've still got just three times as many undealt cards in your hand as you've dealt me.' Len didn't object, so Ted picked up the fallen cards and went on with the deal. It was only a friendly game of bridge, but how many cards he dropped?

Please give the answer and the method to get the answer.

Peter Heichelheim

Spoiler . . . . . . . . . . . . . . . . . . . . . . . . . .

Okay, as Ted was dealing, he may have dealt me & my partner the same number of cards, or we may be one off. Let X be the number of cards in my hand.

Now, let Y be the number of cards Ted, dropped, which is one less than what me & my partner have. The three possibilities for Y are: 2X-2 (when I have one fewer card than my partner) 2X-1 (when I have the same number of cards) 2X (when my partner has one more card)

The number of cards left to deal is 3X. This gives three possible equations:

X + X-1 + 2X-2 + 3X = 52 or 7X - 3 = 52 X + X + 2X-1 + 3X = 52 or 7X - 1 = 52 X + X+1 + 2X + 3X = 52 or 7X + 1 = 52

The only one of these with an integral solution is 7X-3=52, so I have 8 cards, my partner has 7, Ted dropped 14 cards and still holds 24 in his hand.

Err.. when I have one MORE card than my partner

says...

SPOILER

v

v

v

v

v

It's a little tough to tell from the wording, but assume Sam and Bob are partners, and Len and Ted are partners. Assume the deal goes S - L - B - T.

Using a simple table, add up the cards dealt to each player, plus the number dropped (S+B-1) plus the number remaining (B*3). Check to see if this is 52.

So ...

S L B T Dropped Held Total 1 0 0 0 0 0 1 (First card dealt) 1 1 0 0 0 0 1 (second card dealt) . . . 6 6 6 5 11 18 52

Er, uh, bridge has FOUR hands.

Let's say Bob has B cards. Then everyone else has either B, B+1, or B-1. Sam and his partner have somewhere between 2B-2 (happens only when Sam's partner is dealer and Bob is at dealer's left) to 2B+1 (2B+2 requires Bob to be dealer). So there are somewhere between 2B-3 and 2B cards on the floor, undealt, and 3B other undealt cards in Ted's hands.

Cases: 2B-2. Total dealt is then 4B-3 with 5B-3 undealt so 9B-6=52, false.

2B-1. Here again dealing stopped with Bob, since one of Sam and his partner have B and the other has B-1. Ted certainly has B-1. The other player has B or B-1, so total dealt is 4B-{2,3}, undealt is 5B-2, so 9B-{4,5}=52, false.

2B. Sam and his partner have B. The nondealer between them has B, the last player (presumably dealer) may have B or B-1. Total dealt is 4B-{0,1}, undealt is 5B-1, so 9B-{1,2}=52. We can have B=6 here if we have a B-1 player: here the deal out is 6-6-6-5 with 18 undealt cards held and 11 on the floor.

2B+1. Here dealing stopped before Bob, so dealer certainly has B, Sam and partner have B+1 and B in some order. If Sam and Bob are not partners, this is everyone; if Sam and Bob are partners the person left is Len in chair 2, who has either B+1 or B. So total dealt is 4B+{1,2} and undealt is 5B, so 9B+{1,2}=52, false.

Hence, the only situation is: 6-6-6-5, with Sam not in chair 2 and 11 cards dropped.

Forgot to mention that you don't need to work the whole table. Each person has approximately the same number of cards (4x total). About 2x -1 were dropped; about 3x remain to be dealt. So, about 9x = 52. x is about 6. 9*6=54. Aloow one for the '-1' in the number dropped and another one for an 'uneven deal' leads directly to the right solution.

Congratulations. You got J.A.H. Hunter's answer which was 11 cards dropped.

I tried for ages to work this one out, but I suppose knowing not a lot about cards did not help.